Three Dimensional Geometry

We are given that,

is inclined at 60° to and at 45° to .

We need to find the position vector of point A in space.

That is,

Position vector of A =

So, we need to find .

We know that, space contains three axes, namely, X, Y and Z.

Let OA be inclined at angle α with OZ.

And we know that direction cosines are associated by the relation,

l² + m² + n² = 1 …(i)

And here, direction cosines are cosines of the angles inclined by on , and . So,

l = cos 60°

m = cos 45°

n = cos α

Substituting these values of l, m and n in equation (i), we get

(cos 60°)² + (cos 45°)² + (cos α)² = 1

We know the values of cos 60° and cos 45°.

That is,

We get,

So is given as,

…(ii)

We have,

Putting these values of l, m and n in equation (ii), we get

Also, put .

Thus, position vector of point A in space is.

We are given with,

Vector =

Point = (1, -2, 3)

This point can be written in the form of vector as .

Let,

We need to find the vector equation of the line which is parallel to the vector and passes through the point .

We know that,

Vector equation of a line passing through a point and parallel to a given vector is given as,

Where, λ ∈ ℝ

To re-phrase, we need to find .

Just substitute values of the vectors and in the above equation. We get,

This can be further rearranged or just be represented as it is.

On rearranging,

Thus, the required vector equation of the line isor can be written as.

We are given with lines,

Let these lines be L₁ and L₂, such that

Where λ, μ ∈ ℝ

We need to show that lines L₁ and L₂ intersect.

In order to show this, let us find out any point on line L₁ and line L₂.

For line L₁:

We need to find the values of x, y and z. So,

Take .

⇒ x – 1 = 2λ

⇒ x = 2λ + 1

Take .

⇒ y – 2 = 3λ

⇒ y = 3λ + 2

Take .

⇒ z – 3 = 4λ

⇒ z = 4λ + 3 …(i)

∴, any point on line L₁ is (2λ + 1, 3λ + 2, 4λ + 3).

For line L₂:

We need to find the values of x, y and z. So,

Take .

⇒ x – 4 = 5μ

⇒ x = 5μ + 4

Take .

⇒ y – 1 = 2μ

⇒ y = 2μ + 1

Take

⇒ z = μ …(ii)

∴, any point on line L₂ is (5μ + 4, 2μ + 1, μ).

Since, if line L₁ and L₂ intersects then there exists λ and μ such that,

(2λ + 1, 3λ + 2, 4λ + 3) ≡ (5μ + 4, 2μ + 1, μ)

⇒ 2λ + 1 = 5μ + 4 …(iii)

3λ + 2 = 2μ + 1 …(iv)

4λ + 3 = μ …(v)

Substituting value of μ from equation (v) in equation (iv), we get

3λ + 2 = 2(4λ + 3) + 1

⇒ 3λ + 2 = 8λ + 6 + 1

⇒ 3λ + 2 = 8λ + 7

⇒ 8λ – 3λ = 2 – 7

⇒ 5λ = -5

⇒ λ = -1

Putting λ = -1 in equation (v), we get

4(-1) + 3 = μ

⇒ μ = -4 + 3

⇒ μ = -1

To check, put λ = -1 and μ = -1 in equation (iii),

2(-1) + 1 = 5(-1) + 4

⇒ -2 + 1 = -5 + 4

⇒ -1 = -1

∴, λ and μ also satisfy equation (iii).

So, z-coordinate from equation (i),

z = 4λ + 3

⇒ z = 4(-1) + 3 [∵, λ = -1]

⇒ z = -4 + 3

⇒ z = -1

And z-coordinate from equation (ii),

z = μ

⇒ z = -1 [∵, μ = -1]

So, the lines intersect.

Their point of intersection is (5μ + 4, 2μ + 1, μ) = (5(-1) + 4, 2(-1) + 1, -1)

Or (5μ + 4, 2μ + 1, μ) = (-5 + 4, -2 + 1, -1)

Or (5μ + 4, 2μ + 1, μ) = (-1, -1, -1)

Thus, the given lines intersect, and the point of intersection is (-1, -1, -1).

We have lines,

We need to find the angle between the lines.

The line is parallel to vector .

Let . Then, we can say that

The line is parallel to vector .

Similarly, let . Then, we can say that

The line is parallel to vector .

If θ is the angle between the lines, then cosine θ is given by

Substituting values of and in the above equation, we get

Here,

…(i)

Also,

…(ii)

Substituting equation (i) and (ii) in cos θ, we get

Thus, the angle between the lines is .

Given: Points A(0, -1, -1), B(4, 5, 1), C(3, 9, 4) and D(-4, 4, 4)

To Prove: Line through A and B intersects the line through C and D.

Proof: We know that, the equation of a line passing through two points (x₁, y₁, z₁) and (x₂, y₂, z₂) is,

So,

The equation of line passing through points A(0, -1, -1) and B(4, 5, 1) is

Where, x₁ = 0, y₁ = -1 and z₁ = -1

And, x₂ = 4, y₂ = 5 and z₂ = 1

Let

We need to find the value of x, y and z. So,

Take .

⇒ x = 4λ

Take .

⇒ y + 1 = 6λ

⇒ y = 6λ – 1

Take

⇒ z + 1 = 2λ

⇒ z = 2λ – 1

This means, any point on the line L₁ is (4λ, 6λ – 1, 2λ – 1).

The equation of line passing through points C(3, 9, 4) and D(-4, 4, 4) is

Where, x₁ = 3, y₁ = 9 and z₁ = 4

And, x₂ = -4, y₂ = 4 and z₂ = 4

Let

We need to find the value of x, y and z. So,

Take .

⇒ x – 3 = -7μ

⇒ x = -7μ + 3

Take .

⇒ y – 9 = -5μ

⇒ y = -5μ + 9

Take .

⇒ z – 4 = 0

⇒ z = 4

This means, any point on the line L₂ is (-7μ + 3, -5μ + 9, 4).

If lines intersect then there exist a value of λ, μ for which

(4λ, 6λ – 1, 2λ – 1) ≡ (-7μ + 3, -5μ + 9, 4)

⇒ 4λ = -7μ + 3 …(i)

6λ – 1 = -5μ + 9 …(ii)

2λ – 1 = 4 …(iii)

From equation (iii), we get

2λ – 1 = 4

⇒ 2λ = 4 + 1

⇒ 2λ = 5

Putting the value of λ in equation (i), we get

⇒ 2 × 5 = -7μ + 3

⇒ 10 = -7μ + 3

⇒ 7μ = 3 – 10

⇒ 7μ = -7

⇒ μ = -1

Putting the values of λ and μ in equation (ii), we get

⇒ 3 × 5 – 1 = 5 + 9

⇒ 15 – 1 = 14

⇒ 14 = 14

Since, these values of λ and μ satisfy equation (ii), this implies that the lines intersect.

Hence, the lines through A and B intersects line through C and D.

Given: Lines x = py + q, z = ry + s and x = p’y + q’, z = r’y + s’ are perpendicular.

To Prove: pp’ + rr’ + 1 = 0

Proof: Take x = py + q and z = ry + s

From x = py + q,

⇒ py = x – q

From z = ry + s,

⇒ ry = z – s

So,

Or,

…(i)

Now, take x = p’y + q’ and z = r’y + s’

From x = p’y + q’,

⇒ p’y = x – q’

From z = r’y + s’,

⇒ r’y = z – s’

So,

Or,

…(ii)

From (i),

Line L₁ is parallel to . [From the denominators of the equation (i)]

From (ii),

Line L₂ is parallel to . [From the denominators of the equation (ii)]

According to the question, the lines are perpendicular.

Then, the dot product of the vectors must be equal to 0.

That is,

⇒ pp’ + 1 + rr’ = 0

[∵, by vector dot multiplication, ]

Or,

Thus, the given lines are perpendicular if pp’ + rr’ + 1 = 0.

Given that,

There is a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.

We need to find the equation of such plane.

For this, let us find the mid-point of line AB.

Since, mid-point is the halfway between two end-points. So,

⇒ Midpoint of AB = (3, 4, 6)

This can be represented into position vector,

Also, we need to find normal of the plane, .

And we know that, equation of the plane which bisects perpendicularly the line joining two points, here, A and B is

Where,

Substitute the values of , and in the above equation, we get

⇒ 2(x – 3) + 2(y – 4) + 4(z – 6) = 0

Further simplifying it,

⇒ 2x – 6 + 2y – 8 + 4z – 24 = 0

⇒ 2x + 2y + 4z – 6 – 8 – 24 = 0

⇒ 2x + 2y + 4z – 38 = 0

⇒ 2(x + y + 2z – 19) = 0

⇒ x + y + 2z – 19 = 0

⇒ x + y + 2z = 19

Thus, required equation of the plane is x + y + 2z = 19.

We are given that,

A plane is at a distance of 3√3 units from the origin.

Also, the normal is equally inclined to coordinate axis.

We need to find the equation of the plane.

We know that, vector equation of a plane at a distance d from the origin is given by

lx + my + nz = d …(i)

where,

l, m and n are direction cosines of the normal of the plane.

And since, normal is equally inclined to coordinate axis, then

l = m = n

⇒ cos α = cos β = cos γ …(ii)

Also, we know that,

cos² α + cos² β + cos² γ = 1

⇒ cos² α + cos² α + cos² α = 1 [from (ii)]

⇒ 3cos² α = 1

This means,

Substituting values of l, m and n in equation (i), we get

Here, d = 3√3

So,

⇒ x + y + z = 3√3 × √3

⇒ x + y + z = 3 × 3

⇒ x + y + z = 9

Thus, the required equation of the plane is x + y + z = 9.

Given that,

The line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3).

We need to find the equation of the plane.

We must understand that,

Any line which is perpendicular to the plane is the normal.

Let the points be P(-2, -1, -3) and Q(1, -3, 3), so the line PQ is perpendicular to the plane.

This also means, PQ is normal to the plane.

Therefore, PQ = (1 – (-2), -3 – (-1), 3 – (-3))

⇒ PQ = (1 + 2, -3 + 1, 3 + 3)

⇒ PQ = (3, -2, 6)

⇒ Normal to the plane =

The vector equation of a plane is given by

Put these values in the above equation,

We get

⇒ 3(x – 1) + (-2)(y + 3) + 6(z – 3) = 0

⇒ 3(x – 1) – 2(y + 3) + 6(z – 3) = 0

⇒ 3x – 3 – 2y – 6 + 6z – 18 = 0

⇒ 3x – 2y + 6z – 3 – 6 – 18 = 0

⇒ 3x – 2y + 6z – 9 – 18 = 0

⇒ 3x – 2y + 6z – 27 = 0

⇒ 3x – 2y + 6z = 27

Thus, required equation of the plane is 3x – 2y + 6z = 27.

We are given with points, (2, 1, 0), (3, -2, -2) and (3, 1, 7).

We know that, the equation of a line passing through three non-collinear points (x₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given as

Here,

(x₁, y₁, z₁) ≡ (2, 1, 0)

(x₂, y₂, z₂) ≡ (3, -2, -2)

(x₃, y₃, z₃) ≡ (3, 1, 7)

So,

x₁ = 2, y₁ = 1 and z₁ = 0

x₂ = 3, y₂ = -2 and z₂ = -2

x₃ = 3, y₃ = 1 and z₃ = 7

Putting these values in the equation of the line, we get

Take 1^st row and 1^st column, multiply the first element of the row (a₁₁) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 1^st column.

Here,

Now take 1^st row and 2^nd column, multiply the second element of the row (a₁₂) with the difference of multiplication of opposite elements (a₂₁ × a₃₃ – a₂₃ × a₃₁), excluding 1^st row and 2^nd column.

Here,

Similarly, take 1^st row and 3^rd column, multiply the third element of the row (a₁₃) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 3^rd column.

Here,

Solving it further,

Now, since

⇒ -21x – 9y + 3z + 51 = 0

⇒ -21x – 9y + 3z = -51

⇒ -3(7x + 3y – z) = -3 × 17

⇒ 7x + 3y – z = 17

Thus, required equation of the plane is 7x + 3y – z = 17.

We are given the equation of a line,

We need to find equations of the two lines through the origin which intersect the given line.

We know the theorem that says, equation of a line with direction ratios d = (b₁, b₂, b₃) that passes through the point (x₁, y₁, z₁) is given by the formula,

Also, we know that

Angles between two lines whose direction ratio are d₁ and d₂ is θ given by,

Using these theorems, we will find equation of the two lines.

Let the equation of a line be,

And it given that the lines pass through the origin.

⇒ (a₁, a₂, a₃) ≡ (0, 0, 0)

So, equation of both lines that passes through the origin will be of the form:

…(i)

Let

…(ii)

⇒ Direction ratio of the line = (2, 1, 1)

⇒ d₁ = (2, 1, 1) …(iii)

Where, d₁ = Direction ratio of the line (ii).

Let us represent the direction ratio in position vector.

So,

…(iv)

Any point of on the given line is given by (x, y, z). So, from (ii)

Take .

⇒ x – 3 = 2μ

⇒ x = 2μ + 3

Take .

⇒ y – 3 = μ

⇒ y = μ + 3

Take .

⇒ z = μ

So, any point on the line (ii) is

P(2μ + 3, μ + 3, μ)

Since, line (i) passes through origin, we can say that

(b₁, b₂, b₃) ≡ (2μ + 3, μ + 3, μ)

⇒ Direction ratio of line (i) = (2μ + 3, μ + 3, μ)

⇒ d₂ = (2μ + 3, μ + 3, μ) …(v)

Representing the direction ratio in position vector.

So,

…(vi)

We know, by theorem

Substituting the values of d₁ and d₂ from (iv) and (vi) in the above equation. Also, put θ = π/3 as mentioned in the question.

Solving numerator,

Solving denominator,

And,

Putting the values, we get

By cross-multiplying,

⇒ 6√(μ² + 3μ + 3) = 2(6μ + 9)

⇒ 6√(μ² + 3μ + 3) = 2 × 3(2μ + 3)

⇒ 6√(μ² + 3μ + 3) = 6(2μ + 3)

⇒ √(μ² + 3μ + 3) = 2μ + 3

Taking square on both sides,

⇒ (√(μ² + 3μ + 3))² = (2μ + 3)²

⇒ μ² + 3μ + 3 = (2μ)² + (3)² + 2(2μ)(3) [∵, (a + b)² = a² + b² + 2ab]

⇒ μ² + 3μ + 3 = 4μ² + 9 + 12μ

⇒ 4μ² – μ² + 12μ – 3μ + 9 – 3 = 0

⇒ 3μ²+ 9μ + 6 = 0

⇒ 3(μ² + 3μ + 2) = 0

⇒ μ² + 3μ + 2 = 0

⇒ μ² + 2μ + μ + 2 = 0

⇒ μ(μ + 2) + (μ + 2) = 0

⇒ (μ + 1)(μ + 2) = 0

⇒ (μ + 1) = 0 or (μ + 2) = 0

⇒ μ = -1 or μ = -2

So,

From (v):

Direction Ratio = (2μ + 3, μ + 3, μ)

Put μ = -1,

Direction Ratio = (2(-1) + 3, (-1) + 3, -1)

⇒ Direction Ratio = (-2 + 3, -1 + 3, -1)

⇒ Direction Ratio = (1, 2, -1) …(vi)

Now, put μ = -2,

Direction Ratio = (2(-2) + 3, (-2) + 3, -2)

⇒ Direction Ratio = (-4 + 3, -2 + 3, -2)

⇒ Direction Ratio = (-1, 1, -2) …(vii)

Using direction ratios in (vi) and (vii), in equation (i):

And

Thus, the required two lines are and .

We are given two equations that represents the direction cosines of two lines,

l + m + n = 0 …(i)

l² + m² – n² = 0 …(ii)

We need to find the angle between the lines whose direction cosines are given by the given equations.

Let us find the value of l, m and n.

From equation (i),

l + m + n = 0

⇒ n = -l – m

⇒ n = -(l + m) …(iii)

Substituting the value of n from (i) in (ii),

l² + m² – n² = 0

⇒ l² + m² – (-(l + m))² = 0

⇒ l² + m² – (l + m)² = 0

⇒ l² + m² – (l² + m² + 2lm) = 0

⇒ l² + m² – l² – m² – 2lm = 0

⇒ l² – l² + m² – m² – 2lm = 0

⇒ -2lm = 0

⇒ lm = 0

⇒ l = 0 or m = 0

First, put l = 0 in equation (i),

Equation (i) ⇒ 0 + m + n = 0

⇒ m + n = 0

⇒ m = -n

If m = λ, then

n = -m

⇒ n = -λ

∴, direction ratios (l, m, n) = (0, λ, -λ)

Now, put m = 0 in equation (i),

Equation (i) ⇒ l + 0 + n = 0

⇒ l + n = 0

⇒ l = -n

If n = λ, then

l = -n

⇒ l = -λ

∴, direction ratios (l, m, n) = (-λ, 0, λ)

By theorem, that says

Angles between two lines whose direction ratio are d₁ and d₂ is θ given by,

Substituting values of d₁ and d₂ in θ, we get

Solving numerator,

Solving denominator,

Substituting the values in θ, we get

[∵, ]

Thus, the required angle between the given lines is π/3.

We are given that,

l, m, n and l + δl, m + δm, n + δn are direction cosines of a variable line in two adjacent positions.

We need to show that, the small angle δθ between the two positions is given by

δθ² = δl² + δm² + δn²

We know the relationship between direction cosines, that is,

l² + m² + n² = 1 …(i)

Also,

(l + δl)² + (m + δm)² + (n + δn)² = 1

⇒ l² + (δl)² + 2(l)(δl) + m² + (δm)² + 2(m)(δm) + n² + (δn)² + 2(n)(δn) = 1

⇒ l² + m² + n² + (δl)² + (δm)² + (δn)² + 2lδl + 2mδm + 2nδn = 1

⇒ 1 + δl² + δm² + δn² + 2lδl + 2mδm + 2nδn = 1 [from (i)]

⇒ 2lδl + 2mδm + 2nδn + δl² + δm² + δn² = 1 – 1

⇒ 2(lδl + mδm + nδn) = -( δl² + δm² + δn²)

…(ii)

Let and are unit vectors along the line with direction cosines l, m, n and (l + δl), (m + δm), (n + δn) respectively.

That is,

We know that,

Angle between two lines is given by

Where, .

Here, the angle is very small as the line is variable in different but adjacent positions. According to the question, the small angle is δθ.

So,

Angle between two lines is given by δθ,

Substituting the values of and , we get

The dot multiplication of two vectors is calculated by summing up the multiplication of coefficients of , and .

⇒ cos δθ = l(l + δl) + m(m + δm) + n(n + δn)

⇒ cos δθ = l² + lδl + m² + mδm + n² + nδn

⇒ cos δθ = l² + m² + n² + lδl + mδm + nδn

⇒ cos δθ = 1 + lδl + mδm + nδn [∵, from (i)]

[∵, from (ii)]

Or,

Since, we know that

1 – cos 2θ = 2sin² θ

In L.H.S., the angle is 2θ. Then, in R.H.S, the angle becomes half, that is,

Similarly, first replace 2θ by δθ in L.H.S.

Then, make the angle half in R.H.S., that is, .

We get,

Note that, δθ is very small angle, so will be very smaller.

This also means, has a very small value.

⇒ δθ² = δl² + δm² + δn²

Thus, we have showed the required.

We are given with points,

Origin O(0, 0, 0) and point A(a, b, c).

Here, a, b, c are direction ratios.

We need to find:

(a). Direction cosines of line OA.

(b). Equation of plane through A at right angle to OA.

Let us start from (a).

(a). The given points are A(a, b, c) and O(0, 0, 0). Then,

We know that,

If (a, b, c) are direction ratios of a given vector, then its direction cosines are

As in the question,

Direction ratios are (a, b, c), then direction cosines of are

Now, let us solve (b).

(b). It is given in the question that the plane is perpendicular to OA.

And we know that,

A normal is an object such as a line or vector that is perpendicular to a given object.

So, we can say that,

[∵, ]

Also,

Vector equation of a plane where is the normal to the plane and passing through is,

Where,

Here, A(a, b, c) is the given point in the plane.

Substituting the respective vectors, we get

⇒ a(x – a) + b(y – b) + c(z – c) = 0

We can either further simplify or leave it as be.

From simplifying, we get

⇒ ax – a² + by – b² + cz – c² = 0

⇒ ax + by + cz – a² – b² – c² = 0

⇒ a² + b² + c² = ax + by + cz

Thus, the required equation of the plane is a² + b² + c² = ax + by + cz.

Given:

There are two systems of rectangular axis.

Both the systems have same origin.

There is a plane that cuts both of these systems.

One system is cut at a distance of a, b, c.

The other system is cut at a distance of a’, b’, c’.

To Prove:

Proof: Since, a plane cuts both systems at distances a, b, c and a’, b’, c’ respectively. Then, this plane would have different equations in these two different systems, according to their distances with the system.

Let the equation of the plane in the system having distances a, b, c be

And, let the equation of the plane in the system having distances a’, b’, c’ be

According to the question,

The plane cuts the systems from the origin.

We know that,

Perpendicular distance of a plane ax + by + cz + d = 0 (where, not all a, b, c are zero) from the origin is given by

So,

Perpendicular distance of the plane from the origin is,

And,

Perpendicular distance of the plane from the origin is,

We also know that,

If two systems of lines have the same origin then their perpendicular distance from the origin to the plane in both the systems are equal.

So,

By cross multiplication,

Now, take square on both sides,

Or,

Hence, proved.

It is given that,

We have a perpendicular from the point say C (2, 3, -8) to the line, of which equation is,

Or,

So, we can say that the vector equation of line is,

We need to find the foot of the perpendicular from the point C(2, 3, -8) to the given line.

Also, the perpendicular distance from the given point C to the line.

Let us find the point of intersection between the point C(2,3, -8) and the given line.

Take,

From ,

⇒ x – 4 = -2λ

⇒ x = 4 – 2λ

From

⇒ y = 6λ

From

⇒ z – 1 = -3λ

⇒ z = 1 – 3λ

We have,

x = 4 – 2λ, y = 6λ and z = 1 – 3λ.

So, coordinates of any point on the given line is (4 – 2λ, 6λ, 1 – 3λ).

Let the foot of perpendicular from the point C(2, 3, -8) on the line be L(4 – 2λ, 6λ, 1 – 3λ).

We know the direction ratio of any line segment CL, where C(x₁, y₁, z₁) and L(x₂, y₂, z₂), is given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

∴, Direction ratio of is given by

Direction ratio of = (4 – 2λ – 2, 6λ – 3, 1 – 3λ – (-8))

⇒ Direction ratio of = (4 – 2 – 2λ, 6λ – 3, 1 + 8 – 3λ)

⇒ Direction ratio of = (2 – 2λ, 6λ – 3, 9 – 3λ)

Also, direction ratio of the line is,

(-2, 6, -3)

Since, L is the foot of the perpendicular on the line, then

Sum of the product of these direction ratios (2 – 2λ, 6λ – 3, 9 – 3λ) and (-2, 6, -3) is 0.

-2(2 – 2λ) + 6(6λ – 3) + (-3)(9 – 3λ) = 0

⇒ -4 + 4λ + 36λ – 18 – 27 + 9λ = 0

⇒ (4λ + 36λ + 9λ) + (-4 – 18 – 27) = 0

⇒ 49λ – 49 = 0

⇒ 49λ = 49

⇒ λ = 1

Substituting λ = 1 in L(4 – 2λ, 6λ, 1 – 3λ). We get

L(4 – 2λ, 6λ, 1 – 3λ) = L(4 – 2(1), 6(1), 1 – 3(1))

⇒ L(4 – 2λ, 6λ, 1 – 3λ) = L(4 – 2, 6, 1 – 3)

⇒ L(4 – 2λ, 6λ, 1 – 3λ) = L(2, 6, -2)

Now, let us find the perpendicular distance from point C to the line, that is, point L.

⇒ We need to find .

We know that,

Put λ = 1,

To find ,

Thus, the foot of the perpendicular from the point to the given line is (2, 6, -2) and the perpendicular distance is 3√5 units.

It is given that,

The point is P(2, 4, -1).

The equation of the line is .

We need to find the distance of the point P from the given line.

Always note that,

To find distance between a point and a line, get foot of perpendicular from the point on the line.

We have let,

P(2, 4, -1) be the given point and,

be the given line.

Direction Ratio of the line L is (1, 4, -9). …(i)

Let us find any point on this line.

Take the line L:

Take

⇒ x + 5 = λ

⇒ x = λ – 5

Take

⇒ y + 3 = 4λ

⇒ y = 4λ – 3

Take

⇒ z – 6 = -9λ

⇒ z = 6 – 9λ

∴, any point on the line L is (λ – 5, 4λ – 3, 6 – 9λ).

Let this point be Q(λ – 5, 4λ – 3, 6 – 9λ), the foot of the perpendicular from the point P(2, 4, -1) on the line L.

We know the direction ratio of any line segment PQ, where P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), is given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

So, the direction ratio of PQ is given by

Direction ratio of PQ = (λ – 5 – 2, 4λ – 3 – 4, 6 – 9λ – (-1))

⇒ Direction ratio of PQ = (λ – 7, 4λ – 7, 6 + 1 – 9λ)

⇒ Direction ratio of PQ = (λ – 7, 4λ – 7, 7 – 9λ) …(ii)

Also, we know that

If two lines are perpendicular, then the dot products of their direction ratio is 0.

Here, PQ is perpendicular to the line L. From (i) and (ii),

Direction ratio of L = (1, 4, -9)

Direction ratio of PQ = (λ – 7, 4λ – 7, 7 – 9λ)

So,

(1, 4, -9).(λ – 7, 4λ – 7, 7 – 9λ) = 0

⇒ 1(λ – 7) + 4(4λ – 7) + (-9)(7 – 9λ) = 0

⇒ λ – 7 + 16λ – 28 – 63 + 81λ = 0

⇒ λ + 16λ + 81λ – 7 – 28 – 63 = 0

⇒ 98λ – 98 = 0

⇒ 98λ = 98

⇒ λ = 1

So, the coordinate of Q, which is the foot of the perpendicular from the point on the given line, is

Q(λ – 5, 4λ – 3, 6 – 9λ) = Q(1 – 5, 4(1) – 3, 6 – 9(1))

⇒ Q(λ – 5, 4λ – 3, 6 – 9λ) = Q(1 – 5, 4 – 3, 6 – 9)

⇒ Q(λ – 5, 4λ – 3, 6 – 9λ) = Q(-4, 1, -3)

Now, let us find the perpendicular distance from point P to the line, that is, point Q.

⇒ We need to find .

We know that,

Put λ = 1,

To find ,

Thus, distance from the given point to the given line is 7 units.

It is given that,

The point is .

The plane is 2x – 2y + 4z + 5 = 0

We need to find the foot of the perpendicular from the Point P to the equation of the given plane.

Also, we need to find the distance from the point P to the plane.

Let Q be the foot of the perpendicular from the point .

Let the point Q be Q(x₁, y₁, z₁).

We know the direction ratio of any line segment PQ, where P(x₁, y₁, z₁) and Q(x₂, y₂, z₂), is given by (x₂ – x₁, y₂ – y₁, z₂ – z₁).

So, the direction ratio of PQ is given by

Direction ratio of PQ = (x₁ – 1, y₁ – 3/2, z₁ – 2)

Now, let be the normal to the plane 2x – 2y + 4z + 5 = 0.

is obviously parallel to the since, a normal is an object such as a line or vector that is perpendicular to a given object.

Direction ratio simply states the number of units to move along each axis.

For any plane, ax + by + cz = d

a, b and c are normal vectors to the plane.

And therefore, the direction ratios are (a, b, c).

So, direction ratio of = (2, -2, 4) for plane 2x – 2y + 4z + 5 = 0.

Cartesian equation of line PQ, where P(1, 3/2, 2) and Q(x₁, y₁, z₁) is

Let us find any point on this line.

From

⇒ x₁ – 1 = 2λ

⇒ x₁ = 2λ + 1

From

From

⇒ z₁ – 2 = 4λ

⇒ z₁ = 4λ + 2

Any point on the line is (2λ + 1, 3/2 – 2λ, 4λ + 2).

This point is Q.

Q(x₁, y₁, z₁) = Q(2λ + 1, 3/2 – 2λ, 4λ + 2) …(i)

And it was assumed to be lying on the given plane. So, substitute x₁, y₁ and z₁ in the plane equation, we get

2x₁ – 2y₁ + 4z₁ + 5 = 0

Simplifying it to find the value of λ.

⇒ 4λ + 2 – 3 + 4λ + 16λ + 8 + 5 = 0

⇒ 4λ + 4λ + 16λ + 2 – 3 + 8 + 5 = 0

⇒ 24λ + 12 = 0

⇒ 24λ = -12

Since, Q is the foot of the perpendicular from the point P.

The, substitute the value of λ in equation (i),

Also, we need to find .

Where,

P = (1, 3/2, 2)

Q = (0, 5/2, 0)

is found as,

Thus, foot of the perpendicular from the given point to the plane is (0, 5/2,0) and the distance is √6 units.

It is given that,

A line passes through the point P(3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.

We need to find the equation of the line.

Let the position vector of the point P(3, 0, 1) be

Or,

…(i)

Let be the normal vector to the given plane.

Then, is perpendicular to the normal of the plane x + 2y = 0 and 3y – z = 0.

Normal of the plane x + 2y = 0 is

Normal of the plane 3y – z = 0 is

So, is perpendicular to the vectors and .

So,

Take 1^st row and 1^st column, multiply the first element of the row (a₁₁) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 1^st column.

Here,

Now take 1^st row and 2^nd column, multiply the second element of the row (a₁₂) with the difference of multiplication of opposite elements (a₂₁ × a₃₃ – a₂₃ × a₃₁), excluding 1^st row and 2^nd column.

Here,

Similarly, take 1^st row and 3^rd column, multiply the third element of the row (a₁₃) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 3^rd column.

Here,

Further simplifying it,

So, the direction ratio of is (-2, 1, 3). …(ii)

We know that,

Vector equation of a line passing through a point and parallel to a vector is , where λ ∈ ℝ.

Here, from (i) and (ii),

Putting these vectors in , we get

But we know that,

Substituting it,

Thus, the required equation of the line is.

It is given that,

A plane passes through the points (2, 1, -1) and (-1, 3, 4) and perpendicular to the plane x – 2y + 4z = 10.

We need to find the equation of such plane.

We know that,

The cartesian equation of a plane passing through (x₁, y₁, z₁) having direction ratios proportional to a, b, c for its normal is

a(x – x₁) + b(y – y₁) + c(z – z₁) = 0

So,

Let the equation of the plane passing through (2, 1, -1) be

a(x – 2) + b(y – 1) + c(z – (-1)) = 0

⇒ a(x – 2) + b(y – 1) + c(z + 1) = 0 …(i)

Since, it also passes through point (-1, 3, 4), just replace x, y, z by -1, 3, 4 respectively.

⇒ a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0

⇒ -3a + 2b + 5c = 0 …(ii)

Since, a, b, c are direction ratios and this plane is perpendicular to the plane x – 2y + 4z = 10, just replace x, y, z by a, b, c (neglecting 10) respectively and equate to 0. So, we get

a – 2b + 4c = 0 …(iii)

If we need to solve two equations x₁a + y₁b + z₁c = 0 and x₂a + y₂b + z₂c = 0, the formula is:

Similarly, solve for equations (ii) and (iii).

That is,

⇒ a = 18λ

⇒ b = 17λ

⇒ c = 4λ

Substitute these values of a, b, c in equation (i),

a(x – 2) + b(y – 1) + c(z + 1) = 0

⇒ 18λ(x – 2) + 17λ(y – 1) + 4λ(z + 1) = 0

⇒ λ[18(x – 2) + 17(y – 1) + 4(z + 1)] = 0

⇒ 18(x – 2) + 17(y – 1) + 4(z + 1) = 0

⇒ 18x – 36 + 17y – 17 + 4z + 4 = 0

⇒ 18x + 17y + 4z – 36 – 17 + 4 = 0

⇒ 18x + 17y + 4z – 49 = 0

⇒ 18x + 17y + 4z = 49

Thus, equation of the required plane is 18x + 17y + 4z = 49.

We are given with two lines.

…(i)

…(ii)

Take equation (i),

…(iii)

We know that,

Vector equation of a line passing through a point and parallel to a vector is , where λ ∈ ℝ.

Comparing it with equation (iii), we get

Now, take equation (ii),

…(iv)

Similarly from (iv),

So,

Shortest distance between two lines is given by

Solve .

Take 1^st row and 1^st column, multiply the first element of the row (a₁₁) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 1^st column.

Here,

Now take 1^st row and 2^nd column, multiply the second element of the row (a₁₂) with the difference of multiplication of opposite elements (a₂₁ × a₃₃ – a₂₃ × a₃₁), excluding 1^st row and 2^nd column.

Here,

Similarly, take 1^st row and 3^rd column, multiply the third element of the row (a₁₃) with the difference of multiplication of opposite elements (a₂₂ × a₃₃ – a₂₃ × a₃₂), excluding 1^st row and 3^rd column.

Here,

Further, simplifying it.

…(v)

And,

…(vi)

Now, solving .

…(vii)

Substituting values from (v), (vi) and (vii) in d, we get

⇒ d = 14

Thus, the shortest distance between the given lines is 14 units.

We are given that,

A plane is perpendicular to another plane 5x + 3y + 6z + 8 = 0, and also contains line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.

We need to find the equation of such plane.

We know that,

The equation of a plane through the line of intersection of the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + λ(a₂x + b₂y + c₂z + d₂) = 0

Similarly, equation of a plane through the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 is,

(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0

⇒ x + 2y + 3z – 4 + 2λx + λy – λz + 5λ = 0

⇒ x + 2λx + 2y + λy + 3z – λz – 4 + 5λ = 0

⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 …(i)

So, direction ratio of plane in (i) is,

Direction ratio = (1 + 2λ, 2 + λ, 3 – λ)

Since, the plane in (i) is perpendicular to the plane 5x + 3y + 6z + 8 = 0.

Then, replace x, y, z by (1 + 2λ), (2 + λ), (3 – λ) respectively in plane 5x + 3y + 6z + 8 = 0 (neglecting 8) and equate to 0. We get

5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0

⇒ 5 + 10λ + 6 + 3λ + 18 – 6λ = 0

⇒ 10λ + 3λ – 6λ + 5 + 6 + 18 = 0

⇒ 7λ + 29 = 0

⇒ 7λ = -29

Putting this value of λ in equation (i), we get

⇒ -51x – 15y + 50z – 173 = 0

⇒ 51x + 15y – 50z + 173 = 0

Thus, equation of the required plane is 51x + 15y – 50z + 173 = 0.

Given: The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α.

To Prove:

Equation of the plane in its new position is given by

Proof: We have two planes given to us,

ax + by = 0 …(i)

z = 0 …(ii)

We know that,

Equation of plane passing through the line of intersection of planes (i) and (ii) is given by

ax + by + λz = 0 …(iii)

where, λ ∈ ℝ

The angle between the new plane and plane (i) is given to be α.

Since, angle between two planes is equal to the angle between their normal.

So,

Direction ratio of normal to ax + by = 0 or ax + by + 0z = 0 is (a, b, 0).

And,

Direction ratio of normal to ax + by + λz = 0 is (a, b, λ).

Also, we know that

Angle between normal vectors of two planes and is given as,

Substituting values of these vectors, we get

Multiplying √(a² + b²) by numerator and denominator on R.H.S.,

Squaring on both sides,

⇒ (a² + b² + λ²) cos² α = a² + b²

⇒ a² cos² α + b² cos² α + λ² cos² α = a² + b²

⇒ λ² cos² α = a² + b² – a² cos² α – b² cos² α

⇒ λ² cos² α = a² – a² cos² α + b² – b² cos² α

⇒ λ² cos² α = a²(1 – cos² α) + b²(1 – cos² α)

⇒ λ² cos² α = (a² + b²)(1 – cos² α)

⇒ λ² cos² α = (a² + b²) sin² α [∵, sin² α + cos² α = 1]

Since, .

⇒ λ² = (a² + b²) tan² α

Substitute the value of λ in equation (iii) to find the equation of the plane,

ax + by + λz = 0

Hence, proved.

We are given with two planes,

Also, Perpendicular distance of the plane from origin = 1

We need to find the equation of such plane.

We know,

Simplify the planes,

⇒ x + 3y – 6 = 0 …(i)

Also, for

⇒ 3x – y – 4z = 0 …(ii)

We know that,

The equation of a plane through the line of intersection of the planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c₂z + d₂ = 0 is given by

(a₁x + b₁y + c₁z + d₁) + λ(a₂x + b₂y + c₂z + d₂) = 0

Similarly, equation of a plane through the line of intersection of the planes x + 3y – 6 = 0 and 3x – y – 4z = 0 is

(x + 3y – 6) + λ(3x – y – 4z) = 0

⇒ x + 3y – 6 + 3λx – λy – 4λz = 0

⇒ x + 3λx + 3y – λy – 6 – 4λz = 0

⇒ (1 + 3λ)x + (3 – λ)y – 4λz – 6 = 0 …(iii)

Also, we know that

Perpendicular distance of a plane, ax + by + cz + d = 0 from the origin is P, such that

Similarly, perpendicular distance of a plane (iii), which is equal to 1 (according to the question) is

Squaring on both sides, we get

⇒ (1 + 3λ)² + (3 – λ)² + (-4λ)² = 36

⇒ (1)² + (3λ)² + 2(1)(3λ) + (3)² + (λ)² – 2(3)(λ) + 16λ² = 36

[∵, (a + b)² = a² + b² + 2ab and (a – b)² = a² + b² – 2ab]

⇒ 1 + 9λ² + 6λ + 9 + λ² – 6λ + 16λ² = 36

⇒ 9λ² + 16λ² + λ² + 6λ – 6λ = 36 – 1 – 9

⇒ 26λ² + 0 = 26

⇒ 26λ² = 26

⇒ λ² = 1

⇒ λ = ± 1

First, substitute λ = 1 in equation (iii) to find the equation of the plane.

(1 + 3λ)x + (3 – λ)y – 4λz – 6 = 0

⇒ (1 + 3(1))x – (3 – 1)y – 4(1)z – 6 = 0

⇒ 4x – 2y – 4z – 6 = 0

Now, substitute λ = -1 in equation (iii) to find the equation of the plane.

(1 + 3λ)x + (3 – λ)y – 4λz – 6 = 0

⇒ (1 + 3(-1))x + (3 – (-1))y – 4(-1)z – 6 = 0

⇒ (1 – 3)x + (3 + 1)y + 4z – 6 = 0

⇒ -2x + 4y + 4z – 6 = 0

Thus, equation of the required plane is 4x – 2y – 4z – 6 = 0 and -2x + 4y + 4z – 6 = 0.

We are given with points,

Also, .

Where,

So,

⇒ 5x + 2y – 7z + 9 = 0

We need to show that the points A and B are equidistant from the plane .

Also that, the points lie on the opposite side of the plane.

Normal of the plane 5x + 2y – 7z + 9 = 0 is,

We know that,

Perpendicular distance of the position vector of a point, to the the plane, P: ax + by + cz + d = 0 is given as

Where,

So, perpendicular distance of the point to the plane 5x + 2y – 7z + 9 = 0 having normal is,

So, perpendicular distance of the point to the plane 5x + 2y – 7z + 9 = 0 having normal is,

∴, |D₁| = |D₂|

But, since D₁ and D₂ have different signs.

⇒ The points A and B lie on the opposite sides of the plane.

Thus, we have shown that the given points are equidistant from the plane and lies on the opposite side of the plane.

Given,

and Position Vectors (Vector from origin to a point is called a position vector)

∴ line passing through A and along AB will have equation

and line passing through C and along CD will have equation

Now, PQ is a vector perpendicular to both AB and CD, such that P lies on AB and Q lies on CD

Therefore, coordinates of P and Q will have the form

P (6 + 3λ, 7 – λ, 4 + λ) (i)

Q (-3μ, 2μ – 9, 2 + 4μ) (ii)

Hence, vector PQ will be

Now, As PQ is perpendicular to both, therefore dot products

AB. PQ = 0 and CD. PQ = 0

AB. PQ = 3(-3μ – 6 – 3λ) – (2μ – 16 + λ) + (4μ – 2 – λ)

⇒ 0 = -9μ – 18 – 9λ – 2μ + 16 - λ + 4μ – 2 - λ

⇒ -7μ – 11λ - 4 = 0 …. (iii)

CD.PQ = 3(-3μ – 6 – 3λ) + 2(2μ – 16 + λ) + 4(4μ – 2 – λ)

⇒ 0 = -9μ – 18 – 9λ + 4μ - 32 + 2λ +16 μ – 8 - 4λ

⇒ 29μ + 7λ - 22 = 0 …. (iv)

Solving (iii) and (iv), we get

λ = -1 and μ = 1

Put value of λ in (i) to get,

P (6 + 3(-1), 7 – (-1), 4 + (-1))

P (6-3,7+1,4-1)

P (3,8,3)

Put value of μ in (ii) to get,

Q (-3(1), 2(1) – 9, 2 + 4(1))

Q (-3, 2 – 9, 2 +4)

Q (3, -7, 6)

Hence, position vector of P and Q will be

Given,

2l + 2m – n = 0 [1]

⇒ n = 2(l + m) [2]

and

mn + nl + lm = 0

⇒ 2m(l + m) + 2(l + m)l + lm = 0

⇒ 2lm + 2m² + 2l² + 2lm + lm = 0

⇒ 2m² + 5lm + 2l² = 0

⇒ 2m² + 4lm + lm + 2l² = 0

⇒ (2m + l)(m + 2l) = 0

So, we have two cases,

l = -2m

⇒ -4m + 2m – n = 0 [From 1]

⇒ n = 2m

Hence, direction ratios of one line is proportional to -2m, m, -2m or direction ratios are (2, 1, -2)

Another case is,

m = -2l

⇒ 2l + 2(-2l) – n = 0

⇒ 2l – 4l = n

⇒ n = -2l

Hence, direction ratios of another lines is proportional to l, -2l, -2l or direction ratios are (1, -2, -2)

Therefore, direction vectors of two lines are and

Also,

Angle between two lines, and is given by

Now,

⇒ cos θ = 0

⇒ θ = 90°

Hence, angle between given two lines is 90°

Let direction vector of three mutually perpendicular lines be

Let the direction vector associated with directions cosines l₁ + l₂ + l₃, m₁ + m₂ + m₃, n₁ + n₂ + n₃ be

As, lines associated with direction vectors a, b and c are mutually perpendicular, we have

[dot product of two perpendicular vector is 0]

⇒ l₁l₂ + m₁m₂ + n₁n₂ = 0 [1]

Similarly,

⇒ l₁l₃ + m₁m₃ + n₁n₃ = 0 [2]

And

⇒l₂l₃ + m₂m₃ + n₂n₃ = 0 [3]

Now, let x, y, z be the angles made by direction vectors a, b and c respectively with p

Therefore,

⇒ cos x = l₁(l₁ + l₂ + l₃) + m₁(m₁ + m₂ + m₃) + n₁(n₁+ n₂ + n₃)

⇒ cos x = l₁² + l₁l₂ + l₁l₃ + m₁² + m₁m₂ + m₁m₃ + n₁² + n₁n₂ + n₁n₃

⇒ cos x = l₁² + m₁² + n₁² + (l₁l₂ + m₁m₂ + n₁n₂) + (l₁l₃ + m₁m₃ + n₁n₃)

As we know l₁² + m₁² + n₁² = 1 because sum of squares of direction cosines of a line is equal to 1

⇒ cos x = 1 + 0 = 1 [From, 1 and 2]

Similarly, cos y = 1 and cos z = 1

⇒ x = y = z = 0

Hence, angle made by vector p, with vectors a, b and c are equal!

_{If we draw a perpendicular from (α, β, γ) to y-axis, the foot of the perpendicular will have the coordinates (0, β, 0)}

_{Also, by distance formula, we have distance between two points}

_A(x1, y₁, z₁) and B(x₂, y₂, z₂) is

⇒ Required distance

_{We know, sum of squares of direction cosines of a line is equal to 1}

⇒ k² + k² + k² = 1

⇒ 3k² = 1

Given, Plane

Let,

⇒ n is a unit vector

Therefore, Given equation of plane as form

, where n is unit vector and d is distance from origin

On comparison, d = 1, hence distance of given plane from origin is 1.

Equation of line is

The line is having a direction vector

And Given equation of plane is 2x – 2y + z = 5

Normal to this plane is,

Also, we know the angle ϕ between the line with direction vector b and the plane with normal vector n is

The equation of XY plane is given by Z = 0

Given point (α,β,γ).

If we draw perpendicular from the point in the XY plane,

The coordinates of the point will be (α,β,0)

Let its reflection be (x,y,z)

So,

⇒ x = α

⇒ y = β

⇒ z = -γ

The reflection is:

(α , β ,-γ )

Given,

A(0,4,1), B (2, 3, –1), C(4, 5, 0) and D (2, 6, 2), therefore

Since, opposite vectors of this parallelogram are equal and opposite, therefore ABCD is a parallelogram and we know area of a parallelogram ABCD is

|AB × CD|

= 9 sq units

Given, xy + yz = 0

⇒ x(y + z) = 0

⇒ x = 0 and y + z = 0

Clearly, above equations are equation of planes [As they have form Ax + By + Cz + D = 0 form]

Also, x = 0 has normal vector

And y + z = 0 has normal vector

And dot product of these two is

= 0 + 0

= 0

Hence, planes are perpendicular [As the dot product of normal of two perpendicular planes is 0]

Given equation of plane is 2x – 3y + 6z = 11

Normal to this plane is,

Also,

x-axis has direction vector

Also, we know the angle ϕ between the line with direction vector b and the plane with normal vector n is

On comparison,

We know, Equation of a plane that cuts the coordinates axes at (a, 0, 0), (0, b, 0) and (0, 0, c) is

Here,

a = 2, b = 3 and c = 4

∴ required equation of plane is

We know,

If l, m, n are the direction cosines and a, b, c are the direction ratios of a line then,

Given,

a = 2, b = 2, c = -1

Hence, directions cosines are

Given, equation of line is

The following line clearly passes through a(5, -4, 6) and have direction ratios 3, 7 and 2

Also, the position vector of a is

And direction vector will of given line will be,

Position vector of (3, 4, -7) is

And position vector of (1, -1, 6) is

Also,

The vector equation of a line which passes through two points whose position vectors are and is

Hence, required equation of line is

By expanding the dot product, we will get the Cartesian form of given plane

Now, Given

Put

⇒ x + y – z = 2

(Required Cartesian equation)

True

Given equation of the plane is

x + 2y + 3z – 6 = 0

Clearly, Normal to the plane is

Unit vector of n is

True

First let’s convert the given equation to intercept form i.e.

Where, a, b and c are x, y and z intercepts respectively!

Given,

2x – 3y + 5z + 4 = 0

⇒ -2x + 3y – 5z = 4

Dividing by 4 both side

On comparing, we have intercepts as

False

we know the angle ϕ between the line with direction vector b and the plane with normal vector n is

Given,

Equation of line,

Here direction vector of line is

And

Equation of plane,

Normal to the plane is,

Therefore, we have

False

In the vector form, if θ is the angle between the two planes, and

Then,

Now, given planes are

and

Here,

and

Hence,

False

Given,

Equation of line,

Any point on this line should satisfy the plane equation to be on that plane,

Also any point of this line will have a position vector

Now, Given equation of plane

Putting a, in the above equation

= (2 + λ)(3) + (-3 – λ)(1) + (-1 + 2λ)(-1) + 2

= 6 – 3λ – 3 - λ + 1 – 2λ + 2

= 5 – 6λ ≠ RHS

Hence, Line doesn’t lies in the given plane

True

Given, equation of line is

The following line clearly passes through a(5, -4, 6) and have direction ratios 3, 7 and 2

Also, the position vector of a is

And direction vector will of given line will be,

Also, Vector equation of a line that passes through the given point whose position vector is and is

Hence, required equation of line is

False

We know that, equation of a line in Cartesian form is

Where, a, b and c are direction ratios and (x₁, y₁, z₁) is a particular point on the line.

Given, line is parallel to , therefore it has direction ratio as 2, 1 and 3

i.e. a = 2, b = 1, c = 3

And lines pass through (5, -2, 4)

Putting values, we get equation of line as

Which doesn’t match the given equation of line!

False

Let O be the origin, and P be the foot of the perpendicular drawn from the origin to plane

Then, position vector OP will be [which is normal of plane]

Unit vector of n is

Also,

By distance formula, we have distance between two points

A(x₁, y₁, z₁) and B(x₂, y₂, z₂) is

= √38

Now, equation of plane with unit normal vector and having perpendicular distance from origin d is

Equation of given plane is,

Which is given, hence statement is true!