Fundamental Concepts Of 3-dimensional Geometry

(i) direction ratios are:- (2, -6, 3)

So, the direction cosines are- (l, m, n), where, ,

So, l, m, and n are:-

The direction cosines are:-

(ii) direction ratios are:- (2, -1, -2)

So, the direction cosines are:- (l, m, n), where, ,

So, l, m, and n are:-

The direction cosines are:-

(iii) direction ratios are:- (-9, 6, -2)

So, the direction cosines are- (l, m, n), where, ,

So, l, m, and n are:-

The direction cosines are:-

Given two line segments , we have the direction ratios,

Of the line joining these 2 points as,

AB = - + + k, (direction ratio)

The unit vector in this direction will be the direction cosines, i.e.,

Unit vector in this direction is:- (- + + k)/

The direction cosines are ()

(ii) Given two line segments , we have the direction ratios,

Of the line joining these 2 points as,

AB = -4 + (-12) + 6k

The direction ratio in the simplest form will be, (2, 6, -3)

The unit vector in this direction will be the direction cosines, i.e.,

Unit vector in this direction is:- (2 + 6 -3k)/

The direction cosines are ()

(iii) Given two line segments , we have the direction ratios,

Of the line joining these 2 points as,

AB = 2 - 3 + 3k, (direction ratio)

The unit vector in this direction will be the direction cosines, i.e.,

Unit vector in this direction is:- (2 -3 + 3k)/

The direction cosines are ()

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

The line joining these two points is given by,

AB = 2i + 5j -4k

C(0, 3, 2) and D(3, 5, 6),

The line joining these two points,

CD = 3i + 2j +4k

To prove that the two lines are perpendicular we need to show that the angle between these two lines is

So, AB.CD = 0 (dot product)

Thus, (2i + 5j -4k). (3i + 2j +4k) = 6 + 10 – 16 = 0.

Thus, the two lines are perpendicular.

Given: O(0, 0, 0) and A(2, 1, 1)

The line joining these two points is given by,

OA = 2i + j +k

B(3, 5, -1) and D(4, 3, -1),

The line joining these two points,

BC = i - 2j +0k

To prove that the two lines are perpendicular we need to show that the angle between these two lines is

So, OA.BC = 0 (dot product)

Thus, (2i + j +k). (i - 2j +0k) = 2 - 2 + 0 = 0.

Thus, the two lines are perpendicular.

Given: A(3, 5, -1) and B(5, p, 0)

The line joining these two points is given by,

AB = 2i + (p-5)j + k

C(2, 1, 1) and D(3, 3, -1),

The line joining these two points,

CD = i + 2j - 2k

As the two lines are perpendicular, we know that the angle between these two lines is

So, AB.CD = 0 (dot product)

Thus, (2i + (p-5)j + k) .(i + 2j - 2k) = 0.

ð

ð p = 5

Thus, p = 5.

Given O(0, 0, 0), P(2, 3, 4) So, OP = 2i + 3j + 4k

Q(1, -2, 1), So, OQ = i – 2j + k

To prove that OPOQ we have,

OP.OQ = 0, i.e. the angle between the line segments is

So, the dot product i.e. |OP||OQ|cos = 0,cos = 0,

OP.OQ = 0

Thus, (2i + 3j + 4k).( i – 2j + k) = 2 – 6 + 4 = 0

Hence, proved.

Given A(1, 2, 3), B(4, 5, 7), the line joining these two points will be

AB = 3i + 3j + 4k

And the line segment joining, C(-4, 3, -6) and D(2, 9, 2) will be

CD = 6i + 6j + 8k

If CD = r(AB), where r is a scalar constant then,

The two lines are parallel.

Here, CD = 2(AB),

Thus, the two lines are parallel.

Given: A(7, p, 2) and B(q, -2, 5), line segment joining these two points will be, AB = (q-7)i + (-2-p)j + 3k

And the line segment joining C(2, -3, 5) and D(-6, -15, 11) will be, CD = -8i – 12j + 6k

Then, the angle between these two line segments will be 0 degree. So, the cross product will be 0.

ABCD = 0

ð ((q-7)i + (-2-p)j + 3k)( -8i – 12j + 6k) = 0

Thus, solving this we get,

p = 4 and q = 3

We have to show that the three points are colinear , i.e. they all lie on the same line,

If we define a line which is having a parallel line to AB and the points A and B lie on it, if point C also satisfies the line then, the three points are colinear,

Given A(2, 3, 4) and B(-1, -2, 1), AB = -3i – 5j -3k

The points on the line AB with A on the line can be written as,

R = (2, 3, 4) +a(-3, -5, -3)

Let C = (2-3a, 3-5a, 4-3a)

ð (5, 8, 7) = (2-3a, 3-5a, 4-3a)

ð If a = -1, then L.H.S = R.H.S, thus

The point C lies on the line joining AB,

Hence, the three points are colinear.

We have to show that the three points are colinear , i.e. they all lie on the same line,

If we define a line which is having a parallel line to AB and the points A and B lie on it, if point C also satisfies the line then, the three points are colinear,

Given A(-2, 4, 7) and B(3, -6, -8), AB = 5i – 10j -15k

The points on the line AB with A on the line can be written as,

R = (-2, 4, 7) +a(5, -10, -15)

Let C = (-2+5a, 4-10a, 7-15a)

ð (1, -2, -2) = (-2+5a, 4-10a, 7-15a)

ð If a = 3/5, then L.H.S = R.H.S, thus

The point C lies on the line joining AB,

Hence, the three points are colinear.

We have to show that the three points are colinear , i.e. they all lie on the same line,

If we define a line which is having a parallel line to AB and the points A and B lie on it, as the points are colinear so C must satisfy the line,

Given A(-1, 3, 2) and B(-4, 2, -2), AB = -3i – j -4k

The points on the line AB with A on the line can be written as,

R = (-1, 3, 2) +a(-3, -1, -4)

Let C = (-1-3a, 3-1a, 2-4a)

ð (5, 5, p) = (-1-3a, 3-1a, 2-4a)

ð As L.H.S = R.H.S, thus

ð 5 = -1 – 3a, a = -2

Substituting a = -2 we get, p = 2-4(-2) = 10

Hence, p = 10.

Let

R₁ =

And R₂ =

R₁.R₂ = |R₁||R₂|cos

Here, as R1 and R2 are the unit vectors with a direction given by the direction cosines hence, |R1| and |R2| are 1.

So, cos = R₁.R₂ / 1

ð cos

ð =

The angle between the lines is

The angle between the two lines is given by

cos =

where R₁ an R₂ denote the vectors with the direction ratios,

So, here we have,

R₁ = ai + bj + ck and R₂ = (b-c)i +(c-a)j + (a-b)k

cos =

cos

Hence,

The angle between the two lines is given by

cos =

where R₁ and R₂ denote the vectors with the direction ratios,

So, here we have,

R₁ = 2i - 3j + 4k and R₂ = i +2j + k

cos =

cos

Hence,

The angle between the two lines is given by

cos =

where R₁ and R₂ denote the vectors with the direction ratios,

So, here we have,

R1 = i + j + 2k and R2 = ()i -()j + (4)k

cos =

cos

Hence,

The angle between the two lines is given by

cos =

where R₁ and R₂ denote the vectors with the direction ratios,

So, here we have,

R1 = 3i - 2j + k and R2 = 4i + 5j + 7k

cos =

cos

Hence,

(i) The angle between the two lines is given by

cos =

where R₁ and R₂ denote the vectors with the direction ratios,

So, here we have,

R1 = i - j + k and R2 = i for x- axis

cos =

cos

Hence,

With y- axis, i. e. R2 = j

cos =

cos

Hence,

With z- axis, i. e. R2 = k

cos =

cos

Hence,

(ii) The angle between the two lines is given by

cos =

where R₁ and R₂ denote the vectors with the direction ratios,

So, here we have,

R1 = j - k and R2 = i for x- axis

cos =

cos

Hence,

With y- axis, i. e. R2 = j

cos =

cos

Hence,

With z- axis, i. e. R2 = k

cos =

cos

Hence,

(iii) The angle between the two lines is given by

cos =

where R₁ and R₂ denote the vectors with the direction ratios,

So, here we have,

R1 = i - 4j + 8k and R2 = i for x- axis

cos =

cos

Hence,

With y- axis, i. e. R2 = j

cos =

cos

Hence,

With z- axis, i. e. R2 = k

cos =

cos

Hence,

Given: A(1, 8, 4)

Line segment joining B(0, -1, 3) and C(2, -3, -1) is

BC = 2i – 2j – 4k

Let the foot of the perpendicular be R then,

As R lies on the line having point B and parallel to BC,

So, R = (0, -1, 3) + a(2, -2, -4)

R(2a, -1-2a, 3-4a)

The line segment AR is

AR = (2a-1)i + (-1-2a-8)j + (3-4x-4)k

As the lines AR and BC are perpendicular thus, (as R is the foot of the perpendicular on BC)

AR.BC = 0

ð 2(2a-1) + (-2)(-9-2a) + (-4)(-1-4a) = 0

ð 24a + 20 = 0

ð a =

Substituting a in R we get,

R()