Introduction To 3-d Coordinate Geometry

Given: Points are given

To find: name of the octant

Formula used:

Notation of octants:

If x, y and z all three are positive, then octant will be XOYZ

If x is negative and y and z are positive, then the octant will be X’OYZ

If y is negative and x and z are positive, then the octant will be XOY’Z

If z is negative and x and y are positive, then the octant will be XOYZ’

If x and y are negative and z is positive, then the octant will be X’OY’Z

If z and y are negative and x is positive, then the octant will be XOY’Z’

If x and z are negative and x is positive, then the octant will be X’OYZ’

If x, y and z all three are negative, then octant will be X’OY’Z’

(i) (5, 2, 3)

In this case, since x, y and z all three are positive then octant will be XOYZ

(ii) (-5, 4, 3)

In this case, since x is negative and y and z are positive then the octant will be X’OYZ

(iii) (4, -3, 5)

In this case, since y is negative and x and z are positive then the octant will be XOY’Z

(iv) (7, 4, -3)

In this case, since z is negative and x and y are positive then the octant will be XOYZ’

(v) (-5, -4, 7)

In this case, since x and y are negative and z is positive then the octant will be X’OY’Z

(vi) (-5, -3, -2)

In this case, since x, y and z all three are negative then octant will be X’OY’Z’

(vii) (2, -5, -7)

In this case, since z and y are negative and x is positive then the octant will be XOY’Z’

(viii) (-7, 2, -5)

In this case, since x and z are negative and x is positive then the octant will be X’OYZ’

(i) Given: Point is (-2, 3, 4)

To find: the image of the point in yz-plane

Since we need to find its image in yz-plane, a sign of its x-coordinate will change

So, Image of point (-2, 3, 4) is (2, 3, 4)

(ii) Given: Point is (-5, 4, -3)

To find: image of the point in xz-plane

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 4, -3) is (-5, -4, -3)

(iii) Given: Point is (5, 2, -7)

To find: the image of the point in xy-plane

Since we need to find its image in xy-plane, a sign of its z-coordinate will change

So, Image of point (5, 2, -7) is (5, 2, 7)

(iv) Given: Point is (-5, 0, 3)

To find: image of the point in xz-plane

Since we need to find its image in xz-plane, sign of its y-coordinate will change

So, Image of point (-5, 0, 3) is (-5, 0, 3)

(v) Given: Point is (-4, 0, 0)

To find: image of the point in xy-plane

Since we need to find its image in xy-plane, sign of its z-coordinate will change

So, Image of point (-4, 0, 0) is (-4, 0, 0)

Given: A cube has side 4 having one vertex at (1, 0, 1)

To find: coordinates of the other vertices of the cube.

Let Point A(1, 0, 1) and AB, AD and AE is parallel to –ve x-axis, -ve y-axis and +ve z-axis respectively

Since side of cube = 5

Point B is (-4, 0, 1)

Point D is (1, -5, 1)

Point E is (1, 0, 6)

Now, EH is parallel to –ve y-axis

⇒ Point H is (1, -5, 6)

HG is parallel to –ve x-axis

⇒ Point G is (-4, -5, 6)

Now, again GC and GF is parallel to –ve z-axis and +ve y-axis respectively

Point C is (-4, -5, 1)

Point F is (-4, 0, 6)

Given: Points are (3, 0, -1) and (-2, 5, 4)

To find: lengths of the edges of the parallelepiped formed

For point (3, 0, -1)

x₁ = 3, y₁ = 0 and z₁ = -1

For point (-2, 5, 4)

x₂ = -2, y₂ = 5 and z₂ = 4

Plane parallel to coordinate planes of x₁ and x₂ is yz-plane

Plane parallel to coordinate planes of y₁ and y₂ is xz-plane

Plane parallel to coordinate planes of z₁ and z₂ is xy-plane

Distance between planes x₁ = 3 and x₂ = -2 is 3 – (-2) = 3 + 2 = 5

Distance between planes x₁ = 0 and y₂ = 5 is 5 – 0 = 5

Distance between planes z₁ = -1 and z₂ = 4 is 4 – (-1) = 4 + 1 = 5

Hence, edges of parallelepiped is 5, 5, 5

Given: Points are (5, 0, 2) and (3, -2, 5)

To find: lengths of the edges of the parallelepiped formed

For point (5, 0, 2)

x₁ = 5, y₁ = 0 and z₁ = 2

For point (3, -2, 5)

x₂ = 3, y₂ = -2 and z₂ = 5

Plane parallel to coordinate planes of x₁ and x₂ is yz-plane

Plane parallel to coordinate planes of y₁ and y₂ is xz-plane

Plane parallel to coordinate planes of z₁ and z₂ is xy-plane

Distance between planes x₁ = 5 and x₂ = 3 is 5 – 3 = 2

Distance between planes x₁ = 0 and y₂ = -2 is 0 – (-2) = 0 + 2 = 2

Distance between planes z₁ = 2 and z₂ = 5 is 5 – 2 = 3

Hence, edges of parallelepiped is 2, 2, 3

Given: Point P(-4, 3, 5)

To find: distances of the point P from coordinate axes

The distance of the point from x-axis will be given by,

The distance of the point from y-axis will be given by,

The distance of the point from z-axis will be given by,

= 5

Given: Point (3, -2, 5)

To find: the coordinates of 7 more points such that the absolute values of all 8 coordinates are the same

Formula used:

Absolute value of any point(x, y, z) is given by,

We need to make sure that absolute value to be the same for all points

In the formula of absolute value, there is square of the coordinates. So when we change the sign of any of the coordinates, it will not affect the absolute value.

Let point A(3, -2, 5)

Remaining 7 points are:

Point B(3, 2, 5) (By changing the sign of y coordinate)

Point C(-3, -2, 5) (By changing the sign of x coordinate)

Point D(3, -2, -5) (By changing the sign of z coordinate)

Point E(-3, 2, 5) (By changing the sign of x and y coordinate)

Point F(3, 2, -5) (By changing the sign of y and z coordinate)

Point G(-3, -2, -5) (By changing the sign of x and z coordinate)

Point H(-3, 2, -5) (By changing the sign of x, y and z coordinate)

Given: P(1, -1, 0) and Q(2, 1, 2)

To find: Distance between given two points

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between (1, -1, 0) and (2, 1, 2) is

= 3

Hence, Distance between P and Q is 3 units

Given: A(3, 2, -1) and Q(-1, -1, -1)

To find: Distance between given two points

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between (3, 2, -1) and (-1, -1, -1) is

= 5

Hence, Distance between A and B is 5 units

Given: Points are (-2, 3, 1) and (2, 1, 2)

To find: Distance between given two points

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between (-2, 3, 1) and (2, 1, 2) is

Hence, Distance between two given points is √21 units

Given: A(4, -3, -1), B(5, -7, 6) and C(3, 1, -8)

To prove: Points A, B and C are collinear

Formula used:

Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(4, -3, -1) and B(5, -7, 6) is AB,

Distance between B(5, -7, 6) and C(3, 1, -8) is BC,

Distance between A(4, -3, -1) and C(3, 1, -8) is AC,

Clearly,

AB + AC

= BC

Hence, A, B and C are collinear

Given: P(0, 7, -7), Q(1, 4, -5) and R(-1, 10, -9)

To prove: Points P, Q and R are collinear

Formula used:

Points P, Q and R are collinear if PQ + QR = PR or PQ + PR = QR or PR + QR = PQ

Distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between P(0, 7, -7) and Q(1, 4, -5) is PQ,

Distance between Q(1, 4, -5) and R(-1, 10, -9) is QR,

Distance between P(0, 7, -7) and R(-1, 10, -9) is PR,

Clearly,

PQ + PR

= QR

Hence, P, Q and R are collinear

Given: A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6)

To prove: Points A, B and C are collinear

Formula used:

Points A, B and C are collinear if AB + BC = AC or AB + AC = BC or AC + BC = AB

Distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(3, -5, 1) and B(-1, 0, 8) is AB,

Distance between B(-1, 0, 8) and C(7, -10, -6) is BC,

Distance between A(3, -5, 1) and C(7, -10, -6) is AC,

Clearly,

AB + AC

= BC

Hence, A, B and C are collinear

(i) xy-plane

Given: Points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)

To find: the point on xy-plane which is equidistant from the points

As we know z = 0 in xy-plane.

Let P(x, y, 0) any point in xy-plane

According to the question:

PA = PB = PC

⇒ PA² = PB² = PC²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between P(x, y, 0) and A(1, -1, 0) is PA,

The distance between P(x, y, 0) and B(2, 1, 2) is PB,

Distance between P(x, y, 0) and C(3, 2, -1) is PC,

As PA² = PB²

(x – 1)²+ (y + 1)² = (x – 2)² + (y – 1)² + 4

⇒ x²+ 1 – 2x + y² + 1 + 2y = x²+ 4 – 4x + y² + 1 – 2y + 4

⇒ – 2x + 2 + 2y = 9 – 4x – 2y

⇒ – 2x + 2 + 2y – 9 + 4x + 2y = 0

⇒ 2x + 4y – 7 = 0

⇒ 2x = - 4y + 7……………………(1)

As PA² = PC²

(x – 1)²+ (y + 1)² = (x – 3)² + (y – 2)² + 1

⇒ x²+ 1 – 2x + y² + 1 + 2y = x²+ 9 – 6x + y² + 4 – 4y + 1

⇒ – 2x + 2 + 2y = 14 – 6x – 4y

⇒ – 2x + 2 + 2y – 14 + 6x + 4y = 0

⇒ 4x + 6y – 12 = 0

⇒ 2(2x + 3y – 6) = 0

Put the value of 2x from (1):

⇒ 7 – 4y + 3y – 6 = 0

⇒ – y + 1 = 0

⇒ y = 1

Put this value of y in (1):

2x = 7 – 4y

⇒ 2x = 7 – 4(1)

⇒ 2x = 3

Hence point in xy-plane is equidistant from A, B and C

(ii) yz-plane

Given: Points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)

To find: the point on yz-plane which is equidistant from the points

As we know x = 0 in yz-plane.

Let Q(0, y, z) any point in yz-plane

According to the question:

QA = QB = QC

⇒ QA² = QB² = QC²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between Q(0, y, z) and A(1, -1, 0) is QA,

The distance between Q(0, y, Z) and B(2, 1, 2) is QB,

Distance between Q(0, y, z) and C(3, 2, -1) is QC,

As QA² = QB²

1 + z²+ (y + 1)² = (z – 2)² + (y – 1)² + 4

⇒ z²+ 1 + y² + 1 + 2y = z²+ 4 – 4z + y² + 1 – 2y + 4

⇒ 2 + 2y = 9 – 4z – 2y

⇒ 2 + 2y – 9 + 4z + 2y = 0

⇒ 4y + 4z – 7 = 0

⇒ 4z = –4y + 7

As QA² = QC²

1 + z²+ (y + 1)² = (z + 1)² + (y – 2)² + 9

⇒ z²+ 1 + y² + 1 + 2y = z²+ 1 + 2z + y² + 4 – 4y + 9

⇒ 2 + 2y = 14 + 2z – 4y

⇒ 2 + 2y – 14 – 2z + 4y = 0

⇒ –2z + 6y – 12 = 0

⇒ 2(–z + 3y – 6) = 0

Put the value of z from (1):

⇒ 12y + 4y – 7 – 24 = 0

⇒ 16y – 31 = 0

Put this value of y in (1):

Hence point in yz-plane is equidistant from A, B and C

(iii) xz-plane

Given: Points A(1, -1, 0), B(2, 1, 2) and C(3, 2, -1)

To find: the point on xz-plane which is equidistant from the points

As we know y = 0 in xz-plane.

Let R(x, 0, z) any point in xz-plane

According to the question:

RA = RB = RC

⇒ RA² = RB² = RC²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between R(x, 0, z) and A(1, -1, 0) is RA,

Distance between R(x, 0, z) and B(2, 1, 2) is RB,

Distance between R(x, 0, z) and C(3, 2, -1) is RC,

As RA² = RB²

1 + z²+ (x – 1)² = (z – 2)² + (x – 2)² + 1

⇒ z²+ 1 + x² + 1 – 2x = z²+ 4 – 4z + x² + 4 – 4x + 1

⇒ 2 – 2x = 9 – 4z – 4x

⇒ 2 + 4z – 9 + 4x – 2x = 0

⇒ 2x + 4z – 7 = 0

⇒ 2x = –4z + 7……………………………(1)

As RA² = RC²

1 + z²+ (x – 1)² = (z + 1)² + (x – 3)² + 4

⇒ z²+ 1 + x² + 1 – 2x = z²+ 1 + 2z + x² + 9 – 6x + 4

⇒ 2 – 2x = 14 + 2z – 6x

⇒ 2 – 2x – 14 – 2z + 6x = 0

⇒ –2z + 4x – 12 = 0

⇒ 2(2x) = 12 + 2z

Put the value of 2x from (1):

⇒ 2(–4z + 7) = 12 + 2z

⇒ –8z + 14 = 12 + 2z

⇒ 14 – 12 = 8z + 2z

⇒ 10z = 2

Put this value of z in (1):

Hence point in xz-plane is equidistant from A, B and C

Given: Points are A(1, 5, 7), B(5, 1, -4)

To find: the point on z-axis which is equidistant from the points

As we know x = 0 and y = 0 on z-axis

Let R(0, 0, z) any point on z-axis

According to the question:

RA = RB

⇒ RA² = RB²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between R(0, 0, z) and A(1, 5, 7) is RA,

Distance between R(0, 0, z) and B(5, 1, -4) is RB,

As RA² = RB²

26+ (z – 7)² = (z + 4)² + 26

⇒ z²+ 49 – 14z + 26 = z²+ 16 + 8z + 26

⇒ 49 – 14z = 16 + 8z

⇒ 49 – 16 = 14z + 8z

⇒ 22z = 33

Hence point on z-axis is equidistant from (1, 5, 7) and (5, 1, -4)

Given: Points are A(3, 1, 2) and B(5, 5, 2)

To find: the point on y-axis which is equidistant from the points

As we know x = 0 and z = 0 on y-axis

Let R(0, y, 0) any point on the y-axis

According to the question:

RA = RB

⇒ RA² = RB²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between R(0, y, 0) and A(3, 1, 2) is RA,

Distance between R(0, y, 0) and B(5, 5, 2) is RB,

As RA² = RB²

13+ (y – 1)² = (y – 5)² + 29

⇒ y²+ 1 – 2y + 13 = y²+ 25 – 10y + 29

⇒ 10y – 2y = 54 – 14

⇒ 8y = 40

⇒ y = 5

Hence point R(0, 5, 0) on y-axis is equidistant from (3, 1, 2) and (5, 5, 2)

Given: Points A(1, 2, 3)

To find: the point on z-axis which is at distance of from the given point

As we know x = 0 and y = 0 on z-axis

Let R(0, 0, z) any point on z-axis

According to question:

⇒ RA² = 21

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between R(0, 0, z) and A(1, 2, 3) is RA,

As RA² = 21

5 + (z – 3)² = 21

⇒ z²+ 9 – 6z + 5 = 21

⇒ z² – 6z = 21 – 14

⇒ z²– 6z – 7 = 0

⇒ z²– 7z + z – 7 = 0

⇒ z(z– 7) + 1(z – 7) = 0

⇒ (z– 7) (z + 1) = 0

⇒ (z– 7) = 0 or (z + 1) = 0

⇒ z= 7 or z = -1

Hence points (0, 0, 7) and (0, 0, -1) on z-axis is equidistant from (1, 2, 3)

Given: Points are A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2)

To prove: the triangle formed by given points is an equilateral triangle

An equilateral triangle is a triangle whose all sides are equal

So we need to prove AB = BC = AC

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

l

Therefore,

The distance between A(1, 2, 3) and B(2, 3, 1) is AB,

Distance between B(2, 3, 1) and C(3, 1, 2) is BC,

The distance between A(1, 2, 3) and C(3, 1, 2) is AC,

Clearly,

AB = BC = AC

Thus, Δ ABC is a equilateral triangle

HenceProved

Given: Points are A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6)

To prove: the triangle formed by given points is an isosceles right-angled triangle

Isosceles right-angled triangle is a triangle whose two sides are equal and also satisfies Pythagoras Theorem

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(0, 7, 10) and B(-1, 6, 6) is AB,

Distance between B(-1, 6, 6) and C(-4, 9, 6) is BC,

Distance between A(0, 7, 10) and C(-4, 9, 6) is AC,

= 6

Since, AB = BC

AB² + BC²

= 18 + 18

= 36

= AC²

As, AB = BC and AB² + BC² = AC²

Thus, Δ ABC is an isosceles-right angled triangle

HenceProved

Given: Points are A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7)

To prove: the quadrilateral formed by these 4 points is a square

All sides of a square are equal

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between A(3, 3, 3) and B(0, 6, 3) is AB,

Distance between B(0, 6, 3) and C(1, 7, 7) is BC,

Distance between C(1, 7, 7) and D(4, 4, 7) is CD,

The distance between A(3, 3, 3) and D(4, 4, 7) is AD,

Clearly,

AB = BC = CD = AD

Thus, Quadrilateral formed by ABCD is a square

HenceProved

Given: Points are A(1, 3, 0), B(-5, 5, 2), C(-9, -1, 2) and D(-3, -3, 0)

To prove: the quadrilateral formed by these 4 points is a parallelogram but not a rectangle

Opposite sides of both parallelogram and rectangle are equal

But diagonals of a parallelogram are not equal whereas they are equal for rectangle

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(1, 3, 0) and B(-5, 5, 2) is AB,

Distance between B(-5, 5, 2) and C(-9, -1, 2) is BC,

Distance between C(-9, -1, 2) and D(-3, -3, 0) is CD,

Distance between A(1, 3, 0) and D(-3, -3, 0) is AD,

Clearly,

AB = CD

BC = AD

Opposite sides are equal

Now, we will find length of diagonals

Distance between A(1, 3, 0) and C(-9, -1, 2) is AC,

Distance between B(-5, 5, 2) and D(-3, -3, 0) is BD,

Clearly,

AC BD

The diagonals are not equal, but opposite sides are equal

Thus, Quadrilateral formed by ABCD is a parallelogram but not a rectangle

HenceProved

Given: Points are A(1, 3, 4), B(-1, 6, 10), C(-7, 4, 7) and D(-5, 1, 1)

To prove: the quadrilateral formed by these 4 points is a rhombus

All sides of both square and rhombus are equal

But diagonals of a rhombus are not equal whereas they are equal for square

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(1, 3, 4) and B(-1, 6, 10) is AB,

= 7

Distance between B(-1, 6, 10) and C(-7, 4, 7) is BC,

= 7

Distance between C(-7, 4, 7) and D(-5, 1, 1) is CD,

= 7

Distance between A(1, 3, 4) and D(-5, 1, 1) is AD,

= 7

Clearly,

AB = BC = CD = AD

All sides are equal

Now, we will find length of diagonals

Distance between A(1, 3, 4) and C(-7, 4, 7) is AC,

Distance between B(-1, 6, 10) and D(-5, 1, 1) is BD,

Clearly,

AC BD

The diagonals are not equal but all sides are equal

Thus, Quadrilateral formed by ABCD is a rhombus but not square

HenceProved

Given: Points are O(0, 0, 0), A(0, 1, 1), B(1, 0, 1) and C(1, 1, 0)

To prove: given points are forming a regular tetrahedron

All edges of a regular tetrahedron are equal

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between O(0, 0, 0) and A(0, 1, 1) is OA,

Distance between O(0, 0, 0) and B(1, 0, 1) is OB,

Distance between O(0, 0, 0) and C(1, 1, 0) is OC,

Distance between A(0, 1, 1) and B(1, 0, 1) is AB,

Distance between B(1, 0, 1) and C(1, 1, 0) is BC,

Distance between A(0, 1, 1) and C(1, 1, 0) is AC,

Clearly,

AB = BC = AC = OA = OB = OC

All edges are equal

Thus, A, B, C and O forms a regular tetrahedron

HenceProved

Given: Points are A(3, 2, 2), B(-1, 4, 2), C(0, 5, 6), D(2, 1, 2)

To prove: given points lie on sphere whose centre is (1, 3, 4)

To find: radius of sphere

Let Center is O(1, 3, 4)

Since O is centre of sphere and A, B, C, D lie on a sphere

⇒ OA = OB = OC = OD = radius

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between O(1, 3, 4) and A(3, 2, 2) is OA,

= 3

Distance between O(1, 3, 4) and B(-1, 4, 2) is OB,

= 3

Distance between O(1, 3, 4) and C(0, 5, 6) is OC,

= 3

Distance between O(1, 3, 4) and D(2, 1, 2) is OD,

= 3

Clearly,

OA = OB = OC = OD = 3 units

Therefore, radius of sphere = 3 units and A, B, C, D lie on sphere having centre O

Given: Points are O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8)

To find: the coordinates of point which is equidistant from the points

Let required point P(x, y, z)

According to question:

PA = PB = PC = PO

⇒ PA² = PB² = PC² = PO²

Formula used:

Distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and O(0, 0, 0) is PO,

Distance between P(x, y, z) and A(2, 0, 0) is PA,

Distance between P(x, y, z) and B(0, 3, 0) is PB,

Distance between P(x, y, z) and C(0, 0, 8) is PC,

As PO² = PA²

x²+ y² + z² = (x – 2)² + y² + z²

⇒ x²= x²+ 4 – 4x

⇒ 4x = 4

⇒ x = 1

As PO² = PB²

x²+ y² + z² = x²+ (y – 3)² + z²

⇒ y²= y²+ 9 – 6y

⇒ 6y = 9

As PO² = PC²

x²+ y² + z² = x² + y² + (z – 8)²

⇒ z²= z²+ 64 – 16x

⇒ 16z = 64

⇒ z = 4

Hence point is equidistant from given points

Given: Points are A(-2, 2, 3) and B(13, -3, 13)

To find: the locus of point P which moves in such a way that 3PA = 2PB

Let the required point P(x, y, z)

According to the question:

3PA = 2PB

⇒ 9PA² = 4PB²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and A(-2, 2, 3) is PA,

The distance between P(x, y, z) and B(13, -3, 13) is PB,

As 9PA² = 4PB²

9{(x + 2)²+ (y – 2)² + (z – 3)²} = 4{(x – 13)² + (y + 3)² + (z – 13)²}

⇒ 9{x²+ 4 + 4x + y² + 4 – 4y + z² + 9 – 6z} = 4{x²+ 169 – 26x + y² + 9 + 6y + z² + 169 – 26z}

⇒ 9{x² + 4x + y² – 4y + z² – 6z + 17} = 4{x² – 26x + y² + 6y + z² – 26z + 347}

⇒ 9x² + 36x + 9y² – 36y + 9z² – 54z + 153 = 4x² – 104x + 4y² + 24y + 4z² – 104z + 1388

⇒ 9x² + 36x + 9y² – 36y + 9z² – 54z + 153 – 4x² + 104x – 4y² – 24y – 4z² + 104z – 1388 = 0

⇒ 5x² + 5y² + 5z² + 140x – 60y + 50z – 1235 = 0

Hence locus of point P is 5x² + 5y² + 5z² + 140x – 60y + 50z – 1235 = 0

Given: Points are A(3, 4, 5) and B(-1, 3, -7)

To find: the locus of point P which moves in such a way that PA² + PB² = 2k²

Let the required point P(x, y, z)

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and A(3, 4, 5) is PA,

Distance between P(x, y, z) and B(-1, 3, -7) is PB,

According to question:

PA² + PB² = 2k²

⇒ (x – 3)²+ (y – 4)² + (z – 5)² + (x + 1)² + (y – 3)² + (z + 7)² = 2k²

⇒ x²+ 9 – 6x + y² + 16 – 8y + z² + 25 – 10z + x²+ 1 + 2x + y² + 9 – 6y + z² + 49 + 14z = 2k²

⇒ 2x²+ 2y² + 2z² – 4x – 14y + 4z + 109 = 2k²

⇒ 2x²+ 2y² + 2z² – 4x – 14y + 4z + 109 – 2k² = 0

Hence locus of point P is 2x²+ 2y² + 2z² – 4x – 14y + 4z + 109 – 2k² = 0

Given: Points are A(a, b, c), B(b, c, a) and C(c, a, b)

To prove: the triangle formed by given points is an equilateral triangle

An equilateral triangle is a triangle whose all sides are equal

So we need to prove AB = BC = AC

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between A(a, b, c) and B(b, c, a) is AB,

The distance between B(b, c, a) and C(c, a, b) is AB,

The distance between A(a, b, c) and C(c, a, b) is AB,

Clearly,

AB = BC = AC

Thus, Δ ABC is a equilateral triangle

HenceProved

Given: Points are A(3, 6, 9), B(10, 20, 30) and C(25, -41, 5)

To check: the triangle formed by given points is a right-angled triangle or not

A right-angled triangle is a triangle which satisfies Pythagoras Theorem

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between A(3, 6, 9) and B(10, 20, 30) is AB,

Distance between B(10, 20, 30) and C(25, -41, 5) is BC,

Distance between A(3, 6, 9) and C(25, -41, 5) is AC,

AB² + BC²

= 686 + 4571

= 5257

AC²

AB² + AC²

= 686 + 2709

= 3395

BC²

AC² + BC²

= 2709 + 4571

= 7280

AB²

As, AB² + BC² AC²

AC² + BC² AB²

AB² + AC² BC²

Thus, Δ ABC is not a right angled triangle

Given: Points are A(0, 7, -10), B(1, 6, -6) and C(4, 9, -6)

To prove: the triangle formed by given points is an isosceles triangle

Isosceles right-angled triangle is a triangle whose two sides are equal

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(0, 7, -10) and B(1, 6, -6) is AB,

Distance between B(1, 6, -6) and C(4, 9, -6)is BC,

Distance between A(0, 7, -10) and C(4, 9, -6) is AC,

= 6

Clearly,

AB = BC

Thus, Δ ABC is an isosceles triangle

HenceProved

Given: Points are A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6)

To prove: the triangle formed by given points is a right-angled triangle

Right-angled triangle satisfies Pythagoras Theorem

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(0, 7, 10) and B(-1, 6, 6) is AB,

Distance between B(-1, 6, 6) and C(-4, 9, 6) is BC,

Distance between A(0, 7, 10) and C(-4, 9, 6) is AC,

= 6

AB² + BC²

= 18 + 18

= 36

= AC²

As, AB² + BC² = AC²

Thus, Δ ABC is a right angled triangle

HenceProved

Given: Points are A(-1, 2, 1), B(1,-2, 5), C(4, -7, 8) and D(2, -3, 4)

To prove: the quadrilateral formed by these 4 points is a parallelogram

Opposite sides of a parallelogram are equal, but diagonals are not equal

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(-1, 2, 1) and B(1,-2, 5) is AB,

= 6

Distance between B(1,-2, 5) and C(4, -7, 8) is BC,

Distance between C(4, -7, 8) and D(2, -3, 4) is CD,

= 6

Distance between A(-1, 2, 1) and D(2, -3, 4) is AD,

Clearly,

AB = CD

BC = AD

Opposite sides are equal

Now, we will find the length of diagonals

Distance between A(-1, 2, 1) and C(4, -7, 8) is AC,

Distance between B(1, -2, 5) and D(2, -3, 4) is BD,

Clearly,

AC BD

The diagonals are not equal, but opposite sides are equal

Thus, Quadrilateral formed by ABCD is a parallelogram

HenceProved

Given: Points are A(5, -1, 1), B(7, -4, 7), C(1, -6, 10) and D(-1, -3, 4)

To prove: the quadrilateral formed by these 4 points is a rhombus

All sides of both square and rhombus are equal

But diagonals of a rhombus are not equal whereas they are equal for square

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(5, -1, 1) and B(7, -4, 7) is AB,

= 7

Distance between B(7, -4, 7) and C(1, -6, 10) is BC,

= 7

Distance between C(1, -6, 10) and D(-1, -3, 4) is CD,

= 7

Distance between A(5, -1, 1) and D(-1, -3, 4) is AD,

= 7

Clearly,

AB = BC = CD = AD

All sides are equal

Now, we will find length of diagonals

Distance between A(5, -1, 1) and C(1, -6, 10) is AC,

Distance between B(7, -4, 7) and D(-1, -3, 4) is BD,

Clearly,

AC BD

The diagonals are not equal, but all sides are equal

Thus, Quadrilateral formed by ABCD is a rhombus

HenceProved

Given: Points are A(1, 2, 3) and B(3, 2, -1)

To find: the locus of points which are equidistant from the given points

Let the required point P(x, y, z)

According to the question:

PA = PB

⇒ PA² = PB²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and A(1, 2, 3) is PA,

The distance between P(x, y, z) and B(3, 2, -1) is PB,

As PA² = PB²

(x – 1)²+ (y – 2)² + (z – 3)² = (x – 3)² + (y – 2)² + (z + 1)²

⇒ x²+ 1 – 2x + y² + 4 – 4y + z² + 9 – 6z = x²+ 9 – 6x + y² + 4 – 4y + z² + 1 + 2z

⇒ x²+ 1 – 2x + y² + 4 – 4y + z² + 9 – 6z – x²– 9 + 6x – y² – 4 + 4y – z² – 1 – 2z = 0

⇒ 4x – 8z = 0

⇒ 4(x – 2z) = 0

⇒ x – 2z = 0

Hence locus of point P is x – 2z = 0

Given: Points are A(4, 0, 0) and B(-4, 0, 0)

To find: the locus of point P, the sum of whose distances from the given points is equal to 10, i.e. PA + PB = 10

Let the required point P(x, y, z)

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and A(4, 0, 0) is PA,

Distance between P(x, y, z) and B(-4, 0, 0) is PB,

According to question:

PA + PB = 10

Squaring both sides:

Squaring both sides:

⇒ 16x²+ 625 – 100x = 25x²+ 400 + 200x + 25y² + 25z²

⇒ 16x²+ 625 – 100x – 25x²– 400 – 200x – 25y² – 25z² = 0

⇒ -9x² – 25y² – 25z² – 300x + 225 = 0

⇒ 9x² + 25y² + 25z² + 300x – 225 = 0

Hence locus of point P is 9x² + 25y² + 25z² + 300x – 225 = 0

Given: Points are A(1,2, 3), B(-1, -2, -1), C(2, 3, 2) and D(4, 7, 6)

To prove: the quadrilateral formed by these 4 points is a parallelogram but not a rectangle

Opposite sides of both parallelogram and rectangle are equal

But diagonals of a parallelogram are not equal whereas they are equal for rectangle

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(1, 2, 3) and B(-1, -2, -1) is AB,

= 6

Distance between B(-1, -2, -1) and C(2, 3, 2) is BC,

Distance between C(2, 3, 2) and D(4, 7, 6) is CD,

= 6

The distance between A(1, 2, 3) and D(4, 7, 6) is AD,

Clearly,

AB = CD

BC = AD

Opposite sides are equal

Now, we will find the length of diagonals

The distance between A(1, 2, 3) and C(2, 3, 2) is AC,

Distance between B(-1, -2, -1) and D(4, 7, 6) is BD,

Clearly,

AC BD

The diagonals are not equal, but opposite sides are equal

Thus, Quadrilateral formed by ABCD is a parallelogram but not a rectangle

HenceProved

Given: Points are A(3, 4, -5) and B(-2, 1, 4)

To find: the equation of the set of the points, i.e. locus of points which are equidistant from the given points

Let the required point P(x, y, z)

According to the question:

PA = PB

⇒ PA² = PB²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and A(3, 4, -5) is PA,

The distance between P(x, y, z) and B(-2, 1, 4) is PB,

As PA² = PB²

(x – 3)²+ (y – 4)² + (z + 5)² = (x + 2)² + (y – 1)² + (z – 4)²

⇒ x²+ 9 – 6x + y² + 16 – 8y + z² + 25 + 10z = x²+ 4 + 4x + y² + 1 – 2y + z² + 16 – 8z

⇒ x²+ 9 – 6x + y² + 16 – 8y + z² + 25 + 10z – x²– 4 – 4x – y² – 1 + 2y – z² – 16 + 8z = 0

⇒ – 6x – 6y + 18z + 29 = 0

⇒ 6x + 6y – 18z – 29 = 0

Hence locus of point P is 6x + 6y – 18z – 29 = 0

Given: The vertices of the triangle are A(5, 4, 6), B(1, -1, 3) and C(4, 3, 2)

To find: the coordinates of D and the length AD

Formula used:

Distance Formula:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

We know angle bisector divides opposite side in the ratio of the other two sides.

As AD is angle bisector of A and meets BC at D

⇒ BD : DC = AB : BC

Distance between A(5, 4, 6) and B(1, -1, 3) is AB,

The distance between A(5, 4, 6) and C(4, 3, 2) is AC,

AB : AC = 5:3

⇒ BD: DC = 5:3

Therefore, m = 5 and n = 3

B(1, -1, 3) and C(4, 3, 2)

Coordinates of D using section formula:

The distance between A(5, 4, 6) and is AD,

units

Hence, Coordinates of D areand the length of AD isunits

Given: A point C with z-coordinate 8 lies on the line segment joining the points A(2, -3, 4) and B(8, 0, 10)

To find: the coordinates of C

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let Point C(x, y, 8), and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(2, -3, 4) and B(8, 0, 10)

Coordinates of C using section formula:

On comparing:

⇒ 10k + 4 = 8(k + 1)

⇒ 10k + 4 = 8k + 8

⇒ 10k – 8k = 8 – 4

⇒ 2k = 4

⇒ k = 2

Here C divides AB in ratio 2:1

⇒ x = 6

⇒ y = -1

Hence, Coordinates of C are (6, -1, 8)

Given: A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

To prove: A, B and C are collinear

To find: the ratio in which C divides AB

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates

Therefore, m = k and n = 1

A(2, 3, 4), B(-1, 2, -3) and C(-4, 1, -10)

Coordinates of C using section formula:

On comparing:

⇒ -k + 2 = -4(k + 1)

⇒ -k + 2 = -4k – 4

⇒ 4k – k = - 2 – 4

⇒ 3k = -6

⇒ k = -2

⇒ 2k + 3 = k + 1

⇒ 2k – k = 1 – 3

⇒ k = – 2

⇒ -3k + 4 = -10(k + 1)

⇒ -3k + 4 = -10k – 10

⇒ -3k + 10k = -10 – 4

⇒ 7k = -14

⇒ k = -2

The value of k is the same in all three times

Hence, A, B and C are collinear

As k = -2

C divides AB externally in ratio 2:1

Given: points A(2, 4, 5) and B(3, 5, 4)

To find: the ratio in which the line joining given points is divided by the yz-plane

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

X coordinate is always 0 on yz-plane

Let Point C(0, y, z), and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(2, 4, 5) and B(3, 5, 4)

Coordinates of C using section formula:

On comparing:

⇒ 3k + 2 = 0(k + 1)

⇒ 3k + 2 = 0

⇒ 3k = – 2

Hence, C divides AB externally in ratio 2 : 3

Given: A(2, -1, 3) and B(-1, 2, 1)

To find: the ratio in which the line segment AB is divided by the plane x + y + z = 5

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let C(x, y, z) be any point on the given plane and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(2, -1, 3) and B(-1, 2, 1)

Coordinates of C using section formula:

On comparing:

Since, x + y + z = 5

⇒ 5(k + 1) = 4

⇒ 5k + 5 = 4

⇒ 5k = 4 – 5

⇒ 5k = – 1

Hence, the plane divides AB externally in ratio 1:5

Given: A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

To prove: A, B and C are collinear

To find: the ratio in which C divides AB

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates

Therefore, m = k and n = 1

A(3, 2, -4), B(9, 8, -10) and C(5, 4, -6)

Coordinates of C using section formula:

On comparing:

⇒ 9k + 3 = 5(k + 1)

⇒ 9k + 3 = 5k + 5

⇒ 9k – 5k = 5 – 3

⇒ 4k = 2

⇒ 8k + 2 = 4(k + 1)

⇒ 8k + 2 = 4k + 4

⇒ 8k – 4k = 4 – 2

⇒ 4k = 2

⇒ -10k – 4 = -6(k + 1)

⇒ -10k – 4 = -6k – 6

⇒ -10k + 6k = 4 – 6

⇒ -4k = -2

The value of k is the same in all three times

Hence, A, B and C are collinear

C divides AB externally in ratio 1:2

Given: The mid-points of the sides of the triangle are P(-2, 3, 5), Q(4, -1, 7) and R(6, 5, 3).

To find: the coordinates of vertices A, B and C

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

We know the mid-point divides side in the ratio of 1:1.

Therefore,

The coordinates of C is given by,

P(-2, 3, 5) is mid-point of A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

Therefore,

Q(4, -1, 7) is mid-point of B(x₂, y₂, z₂) and C(x₃, y₃, z₃)

Therefore,

R(6, 5, 3) is mid-point of A(x₁, y₁, z₁) and C(x₃, y₃, z₃)

Therefore,

x₁ + x₂ = -4……………………(4)

x₂ + x₃ = 8………………………(5)

x₁ + x₃ = 12……………………(6)

Adding (4), (5) and (6):

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 8 + 12 – 4

⇒ 2x₁ + 2x₂ + 2x₃ = 16

⇒ 2(x₁ + x₂ + x₃) = 16

⇒ x₁ + x₂ + x₃ = 8………………………(7)

Subtract (4), (5) and (6) from (7) separately:

x₁ + x₂ + x₃ – x₁ – x₂ = 8 – (-4)

⇒ x₃ = 12

x₁ + x₂ + x₃ – x₂ – x₃ = 8 – 8

⇒ x₁ = 0

x₁ + x₂ + x₃ – x₁ – x₃ = 8 – 12

⇒ x₂ = -4

y₁ + y₂ = 6……………………(8)

y₂ + y₃ = -2……………………(9)

y₁ + y₃ = 10……………………(10)

Adding (8), (9) and (10):

⇒ y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 10 + 6 – 2

⇒ 2y₁ + 2y₂ + 2y₃ = 14

⇒ 2(y₁ + y₂ + y₃) = 14

⇒ y₁ + y₂ + y₃ = 7………………………(11)

Subtract (8), (9) and (10) from (11) separately:

y₁ + y₂ + y₃ – y₁ – y₂ = 7 – 6

⇒ y₃ = 1

y₁ + y₂ + y₃ – y₂ – y₃ = 7 – (-2)

⇒ y₁ = 9

y₁ + y₂ + y₃ – y₁ – y₃ = 7 – 10

⇒ y₂ = -3

z₁ + z₂ = 10……………………(12)

z₂ + z₃ = 14……………………(13)

z₁ + z₃ = 6……………………(14)

Adding (12), (13) and (14):

⇒ z₁ + z₂ + z₂ + z₃ + z₁ + z₃ = 6 + 14 + 10

⇒ 2z₁ + 2z₂ + 2z₃ = 30

⇒ 2(z₁ + z₂ + z₃) = 30

⇒ z₁ + z₂ + z₃ = 15………………………(15)

Subtract (8), (9) and (10) from (11) separately:

z₁ + z₂ + z₃ – z₁ – z₂ = 15 – 10

⇒ z₃ = 5

z₁ + z₂ + z₃ – z₂ – z₃ = 15 – 14

⇒ z₁ = 1

z₁ + z₂ + z₃ – z₁ – z₃ = 15 – 6

⇒ z₂ = 9

Hence, vertices of sides areA(0, 9, 1) B(-4,-3, 9) and C(12, 1, 5)

Given: The vertices of the triangle are A(1, 2, 3), B(0, 4, 1) and C(-1, -1, -3)

To find: the coordinates of D

Formula used:

Distance Formula:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

We know angle bisector divides opposite side in the ratio of the other two sides.

As AD is angle bisector of A and meets BC at D

⇒ BD : DC = AB : BC

Distance between A(1, 2, 3) and B(0, 4, 1) is AB,

= 3

Distance between A(1, 2, 3) and C(-1, -1, -3) is AC,

= 7

AB : AC = 3:7

⇒ BD: DC = 3:7

Therefore, m = 3 and n = 7

B(0, 4, 1) and C(-1, -1, -3)

Coordinates of D using section formula:

Hence, Coordinates of D are

Given: A(12, -4, 8) and B(27, -9, 18)

To find: the ratio in which the line segment AB is divided by the sphere x² + y² + z² = 504

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let C(x, y, z) be any point on given plane and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(12, -4, 8) and B(27, -9, 18)

Coordinates of C using section formula:

On comparing:

Since, x² + y² + z² = 504

Hence, the sphere divides AB in ratio 2 : 3

Given: A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

To prove: the ratio in which the line segment AB is divided by the plane ax + by + cz + d = 0 is

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let C(x, y, z) be any point on given plane and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

Coordinates of C using section formula:

On comparing:

Since, ax + by + cz + d = 0

The plane divides AB in the ratio

Hence Provedco

Given: The mid-points of the sides of the triangle are P(1, 2, -3), Q(3, 0, 1) and R(-1, 1, -4).

To find: the coordinates of the centroid

Formula used:

Centroid of triangle ABC whose vertices are A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) is given by,

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

We know the mid-point divides side in the ratio of 1:1.

Therefore,

The coordinates of C is given by,

P(1, 2, -3) is mid-point of A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

Therefore,

Q(3, 0, 1) is mid-point of B(x₂, y₂, z₂) and C(x₃, y₃, z₃)

Therefore,

R(-1, 1, -4) is mid-point of A(x₁, y₁, z₁) and C(x₃, y₃, z₃)

Therefore,

x₁ + x₂ = 2……………………(1)

x₂ + x₃ = 6………………………(2)

x₁ + x₃ = -2……………………(3)

Adding (1), (2) and (3):

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 2 + 6 – 2

⇒ 2x₁ + 2x₂ + 2x₃ = 6

⇒ 2(x₁ + x₂ + x₃) = 6

⇒ x₁ + x₂ + x₃ = 3

y₁ + y₂ = 4……………………(4)

y₂ + y₃ = 0……………………(5)

y₁ + y₃ = 2……………………(6)

Adding (4), (5) and (6):

⇒ y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 4 + 0 + 2

⇒ 2y₁ + 2y₂ + 2y₃ = 6

⇒ 2(y₁ + y₂ + y₃) = 6

⇒ y₁ + y₂ + y₃ = 3

z₁ + z₂ = -6……………………(7)

z₂ + z₃ = 2……………………(8)

z₁ + z₃ = -8……………………(9)

Adding (7), (8) and (9):

⇒ z₁ + z₂ + z₂ + z₃ + z₁ + z₃ = -6 + 2 – 8

⇒ 2z₁ + 2z₂ + 2z₃ = -12

⇒ 2(z₁ + z₂ + z₃) = -12

⇒ z₁ + z₂ + z₃ = -6

Centroid of the triangle

= (1, 1, -2)

Hence, the centroid of the triangle is (1, 1, -2)

Given: The coordinates of the A and B of the triangle ABC are (3, -5, 7) and (-1, 7, -6) respectively. The centroid of the triangle is (1, 1, 1)

To find: the coordinates of vertex C

Formula used:

Centroid of triangle ABC whose vertices are A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) is given by,

Here A(3, -5, 7) and B(-1, 7, -6)

Centroid of the triangle

On comparing:

⇒ 2 + x₃ = 3

⇒ x₃ = 3 – 2

⇒ x₃ = 1

⇒ 2 + y₃ = 3

⇒ y₃ = 3 – 2

⇒ y₃ = 1

⇒ 1 + z₃ = 3

⇒ z₃ = 3 – 1

⇒ z₃ = 2

Hence, coordinates of vertex C(1, 1, 2)

Given: Points P(4, 2, -6) and Q(10, -16, 6)

To find: the coordinates of points which trisect the line PQ

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let Point R(x, y, z) and Point S(a, b, c) trisects line PQ

So, PR : RS : SQ = 1 : 1 : 1

Now, we will firstly apply section formula on PQ and find coordinates of R

Therefore, m = 1 and n = 2

P(4, 2, -6) and Q(10, -16, 6)

Coordinates of R using section formula:

Now, we will apply section formula on PQ and find coordinates of S

Therefore, m = 2 and n = 1

P(4, 2, -6) and Q(10, -16, 6)

Coordinates of R using section formula:

Hence, Coordinates of R and S are (6, -4, -2) and (8, -10, 2) respectively

Given: A(2, -3, 4), B(-1, 2, 1) and

To prove: A, B and C are collinear

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let C divides AB in ratio k: 1

Three points are collinear if the value of k is the same for x, y and z coordinates

Therefore, m = k and n = 1

A(2, -3, 4), B(-1, 2, 1) and

Coordinates of C using section formula:

On comparing:

⇒ 2 = -1(k + 1)

⇒ 2 = -k – 1

⇒ k = -1 – 2

⇒ k = -3

⇒ k – 9 = 6(k + 1)

⇒ k – 9 = 6k + 6

⇒ k – 6k = 6 + 9

⇒ -5k = 15

⇒ k = -3

⇒ 2k + 4 = 1(k + 1)

⇒ 2k + 4 = k + 1

⇒ 2k – k = 1 – 4

⇒ k = -3

The value of k is the same in all three times

Hence, A, B and C are collinear

Given: P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10) and P, Q and R are collinear

To find: the ratio in which Q divides PR

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let Q divides PR in ratio k : 1

Therefore, m = k and n = 1

P(3, 2, -4), Q(5, 4, -6) and R(9, 8, -10)

Coordinates of Q using section formula:

On comparing:

⇒ 9k + 3 = 5(k + 1)

⇒ 9k + 3 = 5k + 5

⇒ 9k – 5k = 5 – 3

⇒ 4k = 2

Q divides PR externally in ratio 1:2

Given: points A(4, 8, 10) and B(6, 10, -8)

To find: the ratio in which the line joining given points is divided by the yz-plane

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

the x coordinate is always 0 on yz-plane

Let Point C(0, y, z), and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(4, 8, 10) and B(6, 10, -8)

Coordinates of C using section formula:

On comparing:

⇒ 6k + 4 = 0(k + 1)

⇒ 6k + 4 = 0

⇒ 6k = – 4

Hence, C divides AB externally in ratio 2 : 3

Given: Points P(2, 3, 5)

To find: the distance of the point P from xy-plane

As we know z = 0 in xy-plane.

The shortest distance of the plane will be the z-coordinate of the point

Hence, the distance of point P from xy-plane is 5 units

Given: point P(3, 4, 5)

To find: distance of the point P from the z-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As, x and y coordinate on z-axis are zero

Let point D on z-axis is (0, 0, z)

Direction cosines of z-axis are (0, 0, 1)

Direction cosines of PD are (3 – 0, 4 – 0, 5 – z) = (3, 4, 5 – z)

Let are two vectors as shown in the figure:

The dot product of perpendicular vectors is always zero

Therefore,

⇒ 3 × 0 + 4 × 0 + (5 – z) × 1 = 0

⇒ 0 + 0 + 5 – z = 0

⇒ z = 5

Hence point D(0, 0, 5)

Distance between point P(3, 4, 5) and D(0, 0, 5) is d

= 5

Hence, the distanceof the point P from z-axis is 5 units

Given: distance between the points P(a, 2, 1) and Q(1, -1, 1) is 5 units

To find: the value of a

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

PQ = 5 units

The distance between points P(a, 2, 1) and Q(1, -1, 1) is PQ

Squaring both sides:

⇒ a² – 2a + 10 = 25

⇒ a² – 2a + 10 – 25 = 0

⇒ a² – 2a – 15 = 0

⇒ a² – 5a + 3a – 15 = 0

⇒ a(a – 5) + 3(a – 5) = 0

⇒ (a – 5) (a + 3) = 0

⇒ a = 5 or -3

Hence, the value of a is 5 or -3

Given: The mid-points of the sides of the triangle are P(1, 2, -3), Q(3, 0, 1) and R(-1, 1, -4).

To find: the coordinates of the centroid

Formula used:

Centroid of triangle ABC whose vertices are A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) is given by,

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

We know the mid-point divides side in the ratio of 1:1.

Therefore,

The coordinates of C is given by,

P(1, 2, -3) is mid-point of A(x₁, y₁, z₁) and B(x₂, y₂, z₂)

Therefore,

Q(3, 0, 1) is mid-point of B(x₂, y₂, z₂) and C(x₃, y₃, z₃)

Therefore,

R(-1, 1, -4) is mid-point of A(x₁, y₁, z₁) and C(x₃, y₃, z₃)

Therefore,

x₁ + x₂ = 2……………………(1)

x₂ + x₃ = 6………………………(2)

x₁ + x₃ = -2……………………(3)

Adding (1), (2) and (3):

⇒ x₁ + x₂ + x₂ + x₃ + x₁ + x₃ = 2 + 6 – 2

⇒ 2x₁ + 2x₂ + 2x₃ = 6

⇒ 2(x₁ + x₂ + x₃) = 6

⇒ x₁ + x₂ + x₃ = 3

y₁ + y₂ = 4……………………(4)

y₂ + y₃ = 0……………………(5)

y₁ + y₃ = 2……………………(6)

Adding (4), (5) and (6):

⇒ y₁ + y₂ + y₂ + y₃ + y₁ + y₃ = 4 + 0 + 2

⇒ 2y₁ + 2y₂ + 2y₃ = 6

⇒ 2(y₁ + y₂ + y₃) = 6

⇒ y₁ + y₂ + y₃ = 3

z₁ + z₂ = -6……………………(7)

z₂ + z₃ = 2……………………(8)

z₁ + z₃ = -8……………………(9)

Adding (7), (8) and (9):

⇒ z₁ + z₂ + z₂ + z₃ + z₁ + z₃ = -6 + 2 – 8

⇒ 2z₁ + 2z₂ + 2z₃ = -12

⇒ 2(z₁ + z₂ + z₃) = -12

⇒ z₁ + z₂ + z₃ = -6

Centroid of the triangle

= (1, 1, -2)

Hence, the centroid of the triangle is (1, 1, -2)

Given: point P(1, 2, 3)

To find: coordinates of the foot of the perpendicular from the point on the y-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As x and z coordinate on the y-axis is zero

Let point D is the point of the foot of perpendicular on the y-axis from point P be (0, y, 0)

Direction cosines of y-axis are (0, 1, 0)

Direction cosines of PD are (1 – 0, 2 – y, 3 – 0) = (1, 2 – y, 3)

Let are two vectors as shown in the figure:

The dot product of perpendicular vectors is always zero

Therefore,

⇒ 1 × 0 + (2 – y) × 1 + 3 × 0 = 0

⇒ 0 + 0 + 2 – y= 0

⇒ y = 2

Hence, coordinates of point D are (0, 2, 0)

Given: The coordinates of the A, B and C of the triangle ABC are (a, 1, 3), (-2, b, -5) and (4, 7, c) respectively. The centroid of the triangle is (0, 0, 0)

To find: the values of a, b, c

Formula used:

Centroid of triangle ABC whose vertices are A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) is given by,

Here A(a, 1, 3), B(-2, b, -5) and C(4, 7, c)

Centroid of the triangle

On comparing:

⇒ a + 2 = 0

⇒ a = –2

⇒ b + 5 = 0

⇒ b = –5

⇒ c – 2 = 0

⇒ c = 2

Hence, values of a, b and c are -2, -5, 2

Given: point P(3, 5, 12)

To find: length of the perpendicular drawn from the point P from the x-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As, y and z coordinate on x-axis are zero

Let point D on x-axis is (x, 0, 0)

Direction cosines of z-axis are (1, 0, 0)

Direction cosines of PD are (3 – x, 5 – 0, 12 – 0) = (3 – x, 5, 12)

Let are two vectors as shown in the figure:

The dot product of perpendicular vectors is always zero

Therefore,

⇒ (3 – x) × 1 + 5 × 0 + 12 × 0 = 0

⇒ 3 – x + 0 + 0 = 0

⇒ x = 3

Hence point D(3, 0, 0)

Distance between point P(3, 5, 12) and D(3, 0, 0) is d

= 13

Hence, the distanceof the point P from x-axis is 13 units

Given: The coordinates of the A and B of the triangle ABC are (3, -5, 7) and (3, 0, 1) respectively. The centroid of the triangle is (0, 0, 0)

To find: the coordinates of vertex C

Formula used:

Centroid of triangle ABC whose vertices are A(x₁, y₁, z₁), B(x₂, y₂, z₂) and C(x₃, y₃, z₃) is given by,

Here A(3, -5, 7) and B(3, 0, 1)

Centroid of the triangle

On comparing:

⇒ 6 + x₃ = 0

⇒ x₃ = –6

⇒ -5 + y₃ = 0

⇒ y₃ = 5

⇒ 8 + z₃ = 0

⇒ z₃ = –8

Hence, coordinates of vertex C(-6, 5, -8)

Locus is a moving point which satisfies given conditions

Here, conditions are y = 0 and z = 0

Hence, locus for this is x-axis whose equation is y = z = 0

Given: points A(2, 4, 5) and B(3, -5, 4)

To find: the ratio in which the line joining given points is divided by the yz-plane

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

the x coordinate is always 0 on yz-plane

Let Point C(0, y, z), and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(2, 4, 5) and B(3, -5, 4)

Coordinates of C using section formula:

On comparing:

⇒ 3k + 2 = 0(k + 1)

⇒ 3k + 2 = 0

⇒ 3k = –2

Hence, C divides AB externally in ratio 2 : 3

Given: point P(1, 2, 3)

To find: coordinates of the foot of the perpendicular from the point on the y-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As x and z coordinate on the y-axis is zero

Let point D any point on y-axis be (0, y, 0)

⇒ PD² = 10

Distance between P(1, 2, 3) and D(0, y, 0) is PD,

Now,

10 + (2 – y)² = 10

⇒ (2 – y)² = 10 – 10

⇒ (2 – y)² = 0

⇒ 2 – y = 0

⇒ y = 2

Hence, coordinates of point D are (0, 2, 0)

Given: points A(3, 2, 2) and B(5, 5, 4)

To find coordinates of a point on x-axis which is equidistant from given points.

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As, y and z coordinate on x-axis are zero

Let point D any point on x-axis be (x, 0, 0)

AD = BD

Distance between B(5, 5, 4) and D(x, 0, 0) is BD,

Distance between A(3, 2, 2) and D(x, 0, 0) is BD,

As, AD = BD

⇒ AD² = BD²

8 + (3 – x)² = 41 + (5 – x)²

⇒ 8 + 9 + x² – 6x= 41 + 25 + x² – 10x

⇒ 17 – 6x = 66 – 10x

⇒ 10x – 6x = 66 - 17

⇒ 4x = 49

Hence, coordinates of point D are

Given: Points are O(0, 0, 0), A(a, 0, 0), B(0, b, 0) and C(0, 0, c)

To find: the coordinates of point which is equidistant from the points

Let required point P(x, y, z)

According to question:

PA = PB = PC = PO

⇒ PA² = PB² = PC² = PO²

Formula used:

Distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between P(x, y, z) and O(0, 0, 0) is PO,

Distance between P(x, y, z) and A(a, 0, 0) is PA,

Distance between P(x, y, z) and B(0, b, 0) is PB,

Distance between P(x, y, z) and C(0, 0, c) is PC,

As PO² = PA²

x²+ y² + z² = (x – a)² + y² + z²

⇒ x²= x²+ a² – 2ax

⇒ 2ax = a²

As PO² = PB²

x²+ y² + z² = x²+ (y – b)² + z²

⇒ y²= y²+ b² – 2by

⇒ 2by = b²

As PO² = PC²

x²+ y² + z² = x² + y² + (z – c)²

⇒ z²= z²+ c² – 2cz

⇒ 2cz = c²

Hence point is equidistant from given points

Given: Points A(-2, 0, 6) and B(10, -6, -12)

To find: the coordinates of points P which is five-sixth of AB

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

Let AB = 6 units and Point C(x, y, z) is fifth-sixth of AB

⇒ CB = 6 – 5 = 1

Hence, AC : CB = 5 : 1

Now, we will firstly apply section formula on AB and find coordinates of C

Therefore, m = 5 and n = 1

A(-2, 0, 6) and B(10, -6, -12)

Coordinates of R using section formula:

Hence, Coordinates of C are (8, -5, -9)

Given: a parallelepiped is formed by the planes drawn through the points (2, 3, 5) and (5, 9, 7) parallel to the coordinates planes.

To find: length of edges of parallelepiped and length of diagonal

Planes parallel to (2, 3, 5) are:

x = 2, y = 3 and z = 5

Similarly, planes parallel to (5, 9, 7) are:

x = 5, y = 9 and z = 7

Now, let the length of the parallelepiped are L₁, L₂ and L₃

L₁ is the length of edge between planes x = 2 and x = 5

Clearly, L₁ = 5 – 3 = 2

L₂ is the length of an edge between planes y = 3 and y = 9

Clearly, L₂ = 9 – 3 = 6

L₃ is the length of an edge between planes z = 5 and z = 7

Clearly, L₃ = 7 – 5 = 2

Given: Points A(2, 0, 3), B(0, 3, 2) and C(0, 0, 1)

To find: the point on yz-plane which is equidistant from the points

As we know x = 0 in yz-plane.

Let Q(0, y, z) any point in yz-plane

According to the question:

QA = QB = QC

⇒ QA² = QB² = QC²

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

The distance between Q(0, y, z) and A(2, 0, 3) is QA,

The distance between Q(0, y, Z) and B(0, 3, 2) is QB,

Distance between Q(0, y, z) and C(0, 0, 1) is QC,

As QA² = QB²

4 + (z – 3)²+ y² = (z – 2)² + (y – 3)²

⇒ z²+ 9 – 6z + y² + 4 = z²+ 4 – 4z + y² + 9 – 6y

⇒ – 6z = – 4z – 6y

⇒ 6y – 6z + 4z = 0

⇒ 6y – 2z = 0

⇒ 6y = 2z

⇒ z = 3y…………………(1)

As QA² = QC²

4 + (z – 3)²+ y² = (z – 1)² + y²

⇒ z²+ 9 – 6z + y² + 4 = z²+ 1 – 2z + y²

⇒ 13 – 6z = 1 – 2z

⇒ 13 – 1 = 6z – 2z

⇒ 4z = 12

⇒ z = 3

Put the value of z from (1):

⇒ y = 1

Hence point Q(0, 1, 3) in yz-plane is equidistant from A, B and C

Given: points A(2, 4, 5) and B(3, 5, -9)

To find: the ratio in which the line joining given points is divided by the yz-plane

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

x coordinate is always 0 on yz-plane

Let Point C(0, y, z) and C divides AB in ratio k : 1

Therefore, m = k and n = 1

A(2, 4, 5) and B(3, 5, -9)

Coordinates of C using section formula:

On comparing:

⇒ 3k + 2 = 0(k + 1)

⇒ 3k + 2 = 0

⇒ 3k = –2

Hence, C divides AB externally in ratio 2 : 3

Given: points A(a, b, c) and B(-1, -c, -b)

To find: the ratio in which the line joining given points is divided by the xy-plane

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

z coordinate is always 0 on xy-plane

Let Point C(x, y, 0), and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(a, b, c) and B(-1, -c, -b)

Coordinates of C using section formula:

On comparing:

⇒ -bk + c = 0(k + 1)

⇒ -bk = -c

Hence, C divides AB internally in ratio c: b

Given: Points are P(0, 1, 2), Q(4, -2, 1) and O(0, 0, 0)

To check: the value of ∠POQ

Formula used:

Distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between P(0, 1, 2) and Q(4, -2, 1) is PQ,

Distance between Q(4, -2, 1) and O(0, 0, 0) is QO,

Distance between P(0, 1, 2) and O(0, 0, 0) is PO,

Now,

PO² = 5

QO² = 21

PQ² = 26

Clearly,

PO² + QO² = PQ²

A right-angled triangle is a triangle which satisfies Pythagoras Theorem

These points satisfy Pythagoras Theorem

Given: Extremities of the diagonal of a square are P(1, -2, 3) and Q(2, -3, 5)

To find: Length of the diagonal

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between P(1, -2, 3) and Q(2, -3, 5) is

Hence, Length of the diagonal isunits

Given: Points are A(5, -4, 2), B(4, -3, 1), C(7, 6, 4) and D(8, -7, 5)

To find: name of the quadrilateral formed by these 4 points

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between A(5, -4, 2) and B(4, -3, 1) is AB,

Distance between B(4, -3, 1) and C(7, 6, 4) is BC,

The distance between C(7, 6, 4) and D(8, -7, 5) is CD,

Distance between A(5, -4, 2) and D(8, -7, 5) is AD,

Clearly,

No two sides are equal

So, it cannot be square, rectangle or parallelogram.

Given: x² – 5x + 6 = 0

x² – 5x + 6 = 0

⇒ x² – 3x – 2x + 6 = 0

⇒ x(x – 3) – 2(x – 3) = 0

⇒ (x – 3)(x – 2) = 0

⇒ (x – 3) = 0 or (x – 2) = 0

⇒ x = 3 or x = 2

Both the results represents planes which are parallel to yz-plane

Hence, x² – 5x + 6 = 0 represents planes

Given: P(3, 4, -1) and Q(-1, 2, 3) represents diameter of sphere

To find: Radius of the sphere

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

Therefore,

Distance between P(3, 4, -1) and Q(-1, 2, 3), PQ is

= 6

So, Diameter = 6

We know that Diameter = 2 Radius

Hence, Radius of the sphere is 3 units

Given: points A(2, 3, 1) and B(6, 7, 1)

To find: the ratio in which the line joining given points is divided by the XOZ-plane

Formula used:

Section Formula:

A line AB is divided by C in m:n where A(x, y, z) and B(a, b, c).

The coordinates of C is given by,

the y coordinate is always 0 on XOZ-plane

Let Point C(x, 0, z), and C divides AB in ratio k: 1

Therefore, m = k and n = 1

A(2, 3, 1) and B(6, 7, 1)

Coordinates of C using section formula:

On comparing:

⇒ 7k + 3 = 0(k + 1)

⇒ 7k + 3 = 0

⇒ 7k = –3

Hence, C divides AB externally in ratio 3: 7

Locus is a moving point which satisfies given conditions

Here, conditions are y = 0 and z = 0

Hence, locus for this is x-axis whose equation is y = z = 0

Given: Points P(3, 4, 5)

To find: the coordinates of the foot of the perpendicular drawn from the point P on yz-plane

As we know x = 0 in yz-plane.

But y and z coordinates will remain the same for the foot of perpendicular.

Hence, the coordinates of the foot of the perpendicular from point P on yz-plane are (0, 4, 5)

Given: point P(6, 7, 8)

To find: coordinates of the foot of the perpendicular from a point P from the x-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As, y and z coordinate on x-axis are zero

Let point D on x-axis is (x, 0, 0)

Direction cosines of z-axis are (1, 0, 0)

Direction cosines of PD are (6 – x, 7 – 0, 8 – 0) = (6 – x, 7, 8)

Let are two vectors as shown in the figure:

The dot product of perpendicular vectors is always zero

Therefore,

⇒ (6 – x) × 1 + 7 × 0 + 8 × 0 = 0

⇒ 6 – x + 0 + 0 = 0

⇒ x = 6

Hence, coordinates of the foot of the perpendicular i.e. point D(6, 0, 0)

Given: Points P(6, 7, 8)

To find: the perpendicular distance of the point P from xy-plane

As we know z = 0 in xy-plane.

The shortest distance of the plane will be the z-coordinate of the point

Hence, the distance of point P from xy-plane is 8 units

Given: point P(3, 4, 5)

To find: length of the perpendicular from the point on the y-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As x and z coordinate on the y-axis is zero

Let point D is the point of the foot of perpendicular on the y-axis from point P be (0, y, 0)

Direction cosines of y-axis are (0, 1, 0)

Direction cosines of PD are (3 – 0, 4 – y, 5 – 0) = (3, 4 – y, 5)

Let are two vectors as shown in the figure:

The dot product of perpendicular vectors is always zero

Therefore,

⇒ 3 × 0 + (4 – y) × 1 + 5 × 0 = 0

⇒ 0 + 0 + 4 – y= 0

⇒ y = 4

Hence point D(0, 4, 0)

Distance between point P(3, 4, 5) and D(0, 4, 0) is d

Hence, the distanceof the point P from y-axis isunits

Given: point P(3, 3, 4)

To find: length of the perpendicular drawn from the point P from the x-axis

Formula used:

The distance between any two points (a, b, c) and (m, n, o) is given by,

As, y and z coordinate on x-axis are zero

Let point D on x-axis is (x, 0, 0)

Direction cosines of z-axis are (1, 0, 0)

Direction cosines of PD are (3 – x, 3 – 0, 4 – 0) = (3 – x, 3, 4)

Let are two vectors as shown in the figure:

The dot product of perpendicular vectors is always zero

Therefore,

⇒ (3 – x) × 1 + 3 × 0 + 4 × 0 = 0

⇒ 3 – x + 0 + 0 = 0

⇒ x = 3

Hence point D(3, 0, 0)

Distance between point P(3, 3, 4) and D(3, 0, 0) is d

= 5