Coordinates in Space Exercise: 9.2 Class 12 Basic Mathematics Solution [NEB UPDATED]

Syllabus: Content to study

1. Direction Cosine of a line
2. Direction Cosine of Coordinate axis
3. Direction cosines of the line through two given points
4. Projection of a point on a line or plane
5. Projection of a line segment on a line
6. Direction Ratios
7. Angle between two lines
   

Exercise 10.2

1. If a line makes an angle of $\frac{\pi }{3}$ and $\frac{\pi }{4}$ with positive x-axis and z-axis respectively. Find the acute angle made by the line with positive y-axis.

Solution:

α = $\frac{\pi }{3}$

, γ = $\frac{\pi }{4}$, β= ?

We have, cos²α + cos²β + cos²γ = 1

Or, $\frac{1}{4}$ + $\frac{1}{2}$ + cos²γ = 1

Or, cos²γ = 1 – $\frac{3}{4}$ = $\frac{1}{4}$.

So, γ = $\frac{\pi }{3}$.

 

2. Show that the direction cosines of a line equally inclined to the axes are ±$\frac{1}{{\sqrt 3 }}$, ± $\frac{1}{{\sqrt 3 }}$, ± $\frac{1}{{\sqrt 3 }}$

Solution:

Since, the line is equally inclined to the axes so let α be the equal angle made by the line with the axes. So, it’s dc’s are, l = cosα, m = cosα,n = cosα.

We have, l² + m² + n² = 1

Or, cos²α + cos²α + cos²α = 1

Or, 3cos²α = 1

So, cosα = ± $\frac{1}{{\sqrt 3 }}$.

So, the required dc’s are cosα, cosα, cosα, i.e. ±$\frac{1}{{\sqrt 3 }}$, ± $\frac{1}{{\sqrt 3 }}$, ± $\frac{1}{{\sqrt 3 }}$.

 

3. a) If a, ẞ and y are the angles which a line makes with the coordinate axes, prove that sin²α + sin²β + sin²γ = 2

Solution:

Since, α,β and γ are the angles made by a line with the axes, so it’s dc’s are cosα, cosβ, cos γ.

We have, cos²α + cos²β + cos² γ = 1.

→1 – sin²α + 1 – sin²β + 1 – sin²γ = 1.

So, sin²α + sin²β + sin²γ = 2.

 

b. If a, ẞ and y are the direction angles of a line, prove that cos 2α + cos 2β + cos 2γ + = 0.

Solution:

Since, α, β and γ are the direction angles of a line so the dc’s are l = cos α, m = cos β and n = cos γ.

We have, cos²α + cos²β + cos²γ = 1   [l² + m² + n² = 1]

Or, $\frac{{1 + cos2\alpha }}{2} + \frac{{1 + cos2\beta }}{2} + \frac{{1 + cos2\gamma }}{2}$ = 1.

So, cos2α + cos2β + cos2γ = 2 – 3 = - 1.

Hence, cos 2α + cos 2β + cos 2γ + = 0.

 

4. Find the direction cosines of each of the lines whose direction ratios are

a. -1,2,2

Solution:

The dr’s are – 1, 2 , 2.

Let 1,m,n be the dc’s

So, $\frac{l}{{ - 1}}$ = $\frac{m}{2}$ = $\frac{n}{2}$ = $\frac{{\left( {\sqrt {{l^2} + {m^2} + {n^2}} } \right)}}{{\sqrt {1 + 4 + 4} }}$ = $\frac{1}{3}$

So, l = $ - \frac{1}{3}$, m = $\frac{2}{3}$, n = $\frac{2}{3}$.

 

b. 2,3,6.

Solution:

The dr’s are 2,3,6

Let l,m,n be the dc’s

So, $\frac{l}{2}$ = $\frac{m}{3}$ = $\frac{n}{6}$ = $\frac{{\sqrt {{l^2} + {m^2} + {n^2}} }}{{\sqrt {4 + 9 + 36} }}$ = $\frac{1}{7}$.

So, l = $\frac{2}{7}$, m = $\frac{3}{7}$, n = $\frac{6}{7}$.

 

5. Find the direction cosines of the line passing through the points

a. O(0,0,0) and P(2,3,4)

Solution:

Given points are O(0,0,0) and P(2,3,4).

So, OP = $\sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( {3 - 0} \right)}^2} + {{\left( {4 - 0} \right)}^2}} $ = $\sqrt {29} $.

So, the dc’s of OP are,

Or, $\frac{{{x_2} - {x_1}}}{{OP}}$, $\frac{{{y_2} - {y_1}}}{{OP}}$, $\frac{{{z_2} - {z_1}}}{{OP}}$.

i.e. $\frac{2}{{\sqrt {29} }},\frac{3}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }}$.

 

b. P(2,3,4) and Q(1,4,6)

Solution:

Given points are P(2,3,4) and Q(1,4,6).

So, PQ = $\sqrt {{{\left( {1 - 2} \right)}^2} + {{\left( {4 - 3} \right)}^2} + {{\left( {6 - 4} \right)}^2}} $ = $\sqrt {1 + 1 + 4} $ = $\sqrt 6 $.

So, the dc’s of PQ are,

Or, $\frac{{{x_2} - {x_1}}}{{PQ}},\frac{{{y_2} - {y_1}}}{{PQ}},\frac{{{z_2} - {z_1}}}{{PQ}}$

i.e. $ - \frac{1}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }},\frac{2}{{\sqrt 6 }}$.

 

c. M(-1,2,-3) and N(4, - 1, 1)

Solution:

Given points are M(-1,2,-3) and N(4,-1,1)

So, MN = $\sqrt {{{\left( {4 + 1} \right)}^2} + {{\left( { - 1 - 2} \right)}^2} + {{\left( {1 + 3} \right)}^2}} $ = $\sqrt {50} $ = 5$\sqrt 2 $.

So, the d.c’s of MN are,

Or, $\frac{{{x_2} - {x_1}}}{{PQ}},\frac{{{y_2} - {y_1}}}{{PQ}},\frac{{{z_2} - {z_1}}}{{PQ}}$

Ie. $\frac{5}{{5\sqrt 2 }}, - \frac{3}{{5\sqrt 2 }},\frac{4}{{5\sqrt 2 }}$.

 

6. Find the angle between the two lines whose direction ratios are

a.1,2,4 and -2,1,5

Solution:

The dr’s are 1,2,4 and -2,1,5.

We have,

Cos θ = $\frac{{{a_1}.{a_2} + {b_1}.{b_2} + {c_1}.{c_2}}}{{\sqrt {a_1^2 + b_2^2 + c_2^2} }}$ = $\frac{{1.\left( { - 2} \right) + 2.1 + 4.5}}{{\sqrt {\left( {1 + 4 + 16} \right)\left( {4 + 1 + 25} \right)} }}$

Or, cosθ = $\frac{{20}}{{\sqrt {21*30} }}$ = $\frac{{20}}{{3\sqrt {70} }}$.

So, θ = cos^-1$\left( {\frac{{20}}{{3\sqrt {70} }}} \right)$

 

b. 2, 3, 4 and 1, -2, 1

Solution:

cosθ = $\frac{{2*1 + 3x - 2 + 4*1}}{{\sqrt {\left( {4 + 9 + 16} \right)\left( {1 + 4 + 1} \right)} }}$ = $\frac{0}{{\sqrt {29*6} }}$ = 0.

So, θ= $\frac{\pi }{2}$.

 

c. 1, 2, 2 and 2, 3, 6

Solution:

cosθ = $\frac{{1.2 + 2.3 + 2.6}}{{\sqrt {\left( {1 + 4 + 4} \right)\left( {4 + 9 + 36} \right)} }}$ = $\frac{{20}}{{\sqrt {9*49} }}$ = $\frac{{20}}{{3*7}}$ = $\frac{{20}}{{21}}$.

So, θ= cos^-1$\left( {\frac{{20}}{{21}}} \right)$.

 

7. Show that the line joining the points (1, 2, 3) and (-1, -2, -3) is

Solution:

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

So, dr’s of AB are x₂ – x₁,y₂ – y₁, z₂ – z₁ i.e. 2,4,6

Now,

a. parallel to the line joining the points (2, 3, 4) and (5, 9, 13)

Given points are C(2,3,4) and D(5,9,13).

So, dr’s of CD are 5 – 2,9 – 3,13 – 4, i.e. 3,6,9.

Since, $\frac{2}{3}$$\frac{4}{6}$ = $\frac{6}{9}$$\left[ {i.e\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}} \right]$

So, AB is parallel to CD. 

b. perpendicular to the line joining the points (-2, 1, 5) and (3, 3, 2)

Given points are E(-2,1,5) and F(3,3,2).

So, dr’s of EF are 3 + 2, 3 – 1, 2 – 5 , i.e. 5,2 – 3.

Thus, dr’s of AB are a₁ = 2,b₁ = 4,c₁ = 6.

And the dr’s of EF are a₂ = 5,b₂ = 2,c₂ = -3.

Now, a₁.a₂ + b₁.b₂ + c₁.c₂ = 2.5 + 4.2 + 6.(-3) = 18 – 18 = 0.

So, a₁.a₂ + b₁.b₂ + c₁.c₂ = 0

So, AB is perpendicular to EF.

 

8. a) For what value of k makes the line joining the points (1, 2, k) and (5, 7, 15) perpendicular to the line joining the points (4, 7, 1) and (3, 5, 3)?

Solution:

Given points are A(1,2,k), B(5,7,15), C(4,7,1) and D(3,5,3).

So, the dr’s of AB are 5 – 1, 7 – 2 , 15 – k, i.e. 4,5,15 – k.

And the dr’s of CD are 3 – 4, 5 – 7, 3 – 1, i.e. – 1, - 2 ,2.

Since, AB is perpendicular to CD, so a₁a₂ + b₁.b₂ + c₁.c₂ = 0

Or, 4x – 1 + 5x – 2 + (15 – k)2 = 0

Or, - 14 + 30 – 2k = 0

So, k = 8.

 

b. For what value of k makes the line joining the points (1, 2, k) and (4, 5, 6) parallel the line joining the points (-4, 3, -6) and (2, 9, 2)?

Solution:

Given points are A(1,2,k), B(4,5,6), C(-4,3,-6) and D(2,9,2).

So, the dr’s of AB are 4 – 1, 5 – 2, 6 – k, i.e. 3,3,6 – k.

And the dr’s of CD are 2 + 9, 9 – 3 , 2 + 6, i.e. 6,6,8.

Since, AB is parallel to CD if,

Or, $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$

Ie. $\frac{3}{6} = \frac{3}{6} = \frac{{6 - k}}{8}$

Or, $\frac{{6 - k}}{8}$ = $\frac{1}{2}$→ 6 – k = 4.

So, k = 2.

9. If O is the origin and P(2, 3, 4) and Q(1, - 2, 1) be any two points, show that OP is perpendicular to OQ.

 Solution:

It is given that

Line joining O (0, 0, 0) and P (2, 3, 4) is written as

OP = 2i + 3j + 4k

Line joining O (0, 0, 0) and Q (1, -2, 1) is written as

OQ = i – 2j + k

In order prove that these two lines are perpendicular we must show that angle between these two lines is π/2

Dot product

|OP||OQ| Cos θ = O where

Cos θ =0

OP. OQ=0

Substituting the values (2i + 3j + 4k). (i – 2j + k) = 2 – 6 + 4 = 0 Therefore, it is proved that these two lines are perpendicular.

10. Find the direction cosines of the line which is perpendicular to the lines with direction cosines proportional to 3,-1, 1 and -3, 2, 4.

Solution:

Let l,m,n be the d.c’s of a line which is perpendicular to the lines with dr’s 3,-1,-1 and – 3,2,4/

So, 3l – m + n = 0    [l₁l₂ + m₁.m₂ + n₁.n₂ = 0]

And – 3l + 2m + 4n = 0

Or, $\frac{l}{{ - 4 - 2}} = \frac{m}{{ - 3 - 12}} = \frac{n}{{6 - 3}}$

Or, $\frac{l}{2} = \frac{m}{5} = \frac{n}{{ - 1}}$ = $\frac{{\sqrt {{l^2} + {m^2} + {n^2}} }}{{\sqrt {4 + 25 + 1} }}$ = $\frac{1}{{\sqrt {30} }}$.

So, the d.c’s l,m,n are $\frac{2}{{\sqrt {30} }}$, $\frac{5}{{\sqrt {30} }}$, $ - \frac{1}{{\sqrt {30} }}$.

 

11. Show that the angle between two diagonals of a cube is cos^-1$\left( {\frac{1}{3}} \right)$

Solution:

Let 0(0,0,0) be the origin taken as one of the vertex of a cube of side OA = OB = OC = a, whose two diagonals are OP and AR where A (a,0,0) , R(0,a,a) and P(a,a,a). The cube is shown aside.

Now,

OP = $\sqrt {{{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2}} $ = a$\sqrt 3 $.

And AR = $\sqrt {{{\left( {0 - a} \right)}^2} + {{\left( {a - 0} \right)}^2} + {{\left( {a - 0} \right)}^2}} $ = a$\sqrt 3 $.

So, the d.c’s of AR = $\frac{{a - 0}}{{a\sqrt 3 }}$, $\frac{{a - 0}}{{a\sqrt 3 }}$, $\frac{{a - 0}}{{a\sqrt 3 }}$.

i.e. $\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$.

And th d.c.’s of AR are, $\frac{{0 - a}}{{\sqrt 3 }},\frac{{a - 0}}{{\sqrt 3 }},\frac{{a - 0}}{{\sqrt 3 }}$.

i.e. $ - \frac{1}{{\sqrt 3 }}$, $\frac{1}{{\sqrt 3 }}$, $\frac{1}{{\sqrt 3 }}$.

Let θ be the angle between OP and AR.

Or, cosθ = a₁.a₂ + b₁.b₂ + c₁.c₂ = $\frac{1}{{\sqrt 3 }}\left( { - \frac{1}{{\sqrt 3 }}} \right) + \frac{1}{{\sqrt 3 }}.\frac{1}{{\sqrt 3 }}$ + $\frac{1}{{\sqrt 3 }}$.$\frac{1}{{\sqrt 3 }}$ = $\frac{1}{3}$

So, θ= cos^-1$\left( {\frac{1}{3}} \right)$.

 

12. The projection of a line on the axes are 6, 2, 3. Find the length of the line and its direction cosines.

Solution:

Let PQ be a line joining P(x₁,y₁,z₁) and Q(x₂,y₂,z₂) whose d.c’s are $\frac{{{x_2} - {x_1}}}{{PQ}}$, $\frac{{{y_2} - {y_1}}}{{PQ}}$, $\frac{{{z_2} - {z_1}}}{{PQ}}$ and whose projection on the axes are 6,2,3.

So, projection of PQ on the x – axis is, x₂ – x₁ = 6.

projection of PQ on the y – axis is, y₂ – y₁ = 2.

projection of PQ on the z – axis is, z₂ – z₁ = 3.

Now, (x₂ – x₁)² + (y₂ – y₁)² + (z₂ – z₁)² = 6² + 2² + 3²

Or, PQ² = 49

So, PQ = 7.

So, the dc’s are $\frac{{{x_2} - {x_1}}}{{PQ}}$, $\frac{{{y_2} - {y_1}}}{{PQ}}$, $\frac{{{z_2} - {z_1}}}{{PQ}} = $

i.e. $\frac{6}{7},\frac{2}{7}.\frac{3}{7}$.

Hence, length = 7 and d.c’s are $\frac{6}{7}$,$\frac{2}{7}$.$\frac{3}{7}$.

 

13. Find the projection of the join of the pair of points (3,-1, 2) and (5, -7, 4)

a. on the coordinate axes

Solution:

Given points are P(x₁,y₁,z₁) = (3,-1,2) and P(x₂,y₂,z₂) = (5,-7,4)

Projection of PQ on the x-axis = x₂ – x₁ =  5 – 3 = 2

Projection of PQ on the y – axis = y₂ – y₁ = - 7 + 1 = - 6

And projection of PQ on z  - axis = z₂ – z₁ = 4 – 2 = 2.

 

b. on a line whose direction cosines are proportional to 1, -1, 2

Solution:

The dc’s of a line AB whose dr’s are 1,-1,2 are

l = $\frac{1}{{\sqrt 6 }}$, m = $ - \frac{1}{{\sqrt 6 }}\: $, and n = $\frac{2}{{\sqrt 6 }}$.

So, projection of PQ on AB is l(x₂ – x₁) + m(y₂ – y₁) + n(z₂ – z₁)

= $\frac{1}{{\sqrt 6 }}$(5 – 3) + $\left( { - \frac{1}{{\sqrt 6 }}} \right)$(-7 + 1) + $\frac{2}{{\sqrt 6 }}$(4 – 2).

= $\frac{1}{{\sqrt 6 }}$(2 + 6 + 4) = $\frac{{12}}{{\sqrt 6 }}$ = 2$\sqrt 6 $.

 

c. on a line joining the points (0, 1, 0) and (1, 3, 7)

Solution:

The given points are A(0,1,0) and B(1,3,7).

So, dc’s of AB are $\frac{{1 - 0}}{{AB}}$, $\frac{{3 - 1}}{{AB}}$, $\frac{{7 - 9}}{{AB}}$, i.e. $\frac{1}{{\sqrt {54} }},\frac{2}{{\sqrt {54} }},\frac{7}{{\sqrt {54} }}$.

So, projection of PQ on AB is = l(x₂ – x₁) + m(y₂ – y₁) + n(z₂ – z₁)

= $\frac{1}{{\sqrt {54} }}$ (5 – 3) + $\frac{2}{{\sqrt {54} }}$(- 7 + 1) + $\frac{7}{{\sqrt {54} }}$(4 – 2).

= $\frac{2}{{\sqrt {54} }} - \frac{{12}}{{\sqrt {54} }}$ + $\frac{{14}}{{\sqrt {54} }}$ = $\frac{4}{{\sqrt {54} }}$.

 

14. a) A, B, C and D are four points with coordinates (2, 3, 1), (3, 2, 5), (-1, 2, 4) and (-1, 5, 7) respectively. Prove that the projection of AB on CD is equal to the projection of CD on AB. Also, show that the angle between them is $\frac{\pi }{3}$

Solution:

 

b) A(1, 2, 3) B(- 2, 2, 0) and C(3, 1, 1) are three points. Find the foot of the perpendicular drawn from A to the line BC.

Solution:

 

15. Find the direction cosines l, m, n of two lines which satisfy the equations

a. l + m + n = 0 and 2lm – mn + 2nl = 0

Solution:

Given relations are,

l + m + n = 0 → l = - (m + n) …(1)

and 2lm – mn + 2nl = 0 …(2)

or, -2m(m + n) – mn – 2n(m + n) = 0 [from(1)]

or,  2m² + 5mn + 2n² = 0

or, (2m + n)(m + 2n) = 0

So, 2m + n = 0 and  m + 2n = 0

So, 0,l + 2m + n = 0 …(3)

And 0.l + m + 2n = 0 …(4)

Now, from (1) and (2), we have,

So, $\frac{l}{1-2} = \frac{m}{0-1} = \frac{n}{2-0} \quad \rightarrow \quad \frac{l}{-1} = \frac{m}{-1} = \frac{n}{2} \quad \left[ \begin{array}{c} l+m+n=0 \\ 0.l+2m+n=0 \end{array} \right]$

Or, $\frac{l}{1} = \frac{m}{1} = \frac{n}{{ - 2}}$→ l = $\frac{1}{{\sqrt 6 }}$, m = $\frac{1}{{\sqrt 6 }}$, n = $ - \frac{2}{{\sqrt 6 }}$.

Again, from (1) and (4), we have,

Or, $\frac{l}{2-1} = \frac{m}{0-2} = \frac{n}{1-0} \quad \left[ \begin{array}{c} l+m+n=0 \\ 0.l+2m+n=0 \end{array} \right]$

Or, $\frac{l}{1} = \frac{m}{{ - 2}} = \frac{n}{1}$

So, l = $\frac{1}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }}, - \frac{2}{{\sqrt 6 }}$ and $\frac{1}{{\sqrt 6 }}, - \frac{2}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }}$.

Let θ be the angle between the lines. Then we have,

Cosθ = l₁.l₂ + m₁.m₂ + n₁.n₂ = $\frac{1}{{\sqrt 6 }}.\frac{1}{{\sqrt 6 }} - \frac{2}{{\sqrt 6 }}.\frac{1}{{\sqrt 6 }} + \frac{1}{{\sqrt 6 }}.\left( { - \frac{2}{{\sqrt 6 }}} \right)$

= $ - \frac{3}{6}$ = $ - \frac{1}{2}$

So, θ = 120°.

Note: The dc’s $\frac{1}{{\sqrt 6 }}, - \frac{2}{{\sqrt 6 }},\frac{1}{{\sqrt 6 }}$ can be writeen as $ - \frac{1}{{\sqrt 6 }},\frac{2}{{\sqrt 6 }}, - \frac{1}{{\sqrt 6 }}$ then, we θ= 60°

b. 4l + 3m – 2n = 0 and lm – mn + nl = 0

Solution:

The given relations are:

4l + 3m – 2n = 0 ..(1)

i.e. n = $\frac{{4l + 3m}}{2}$

And lm – mn + nl = 0 ..(2)

Or, lm – m $\frac{{\left( {4l + 3m} \right)}}{2}$ + $\frac{{l\left( {4l + 3m} \right)}}{2}$ = 0    [from (1)]

Or, 2lm – 4lm – 3m² + 4l² + 3lm= 0.

Or, 4l² + lm – 3m² = 0

Or, 4l² + 4lm – 3lm – 3m² = 0

So, (4l – 3m)(l + m) = 0

So, 4l – 3m + 0.n = 0 …(3)

And l + m + 0.n = 0. …(4)

Now, from (1) and (3), we have,

Or, $\frac{l}{0-6} = \frac{m}{-8-0} = \frac{n}{-12-12} \quad \left[ \begin{array}{c} 4l+3m-2n=0 \\ 4l-3m+0n=0 \end{array} \right]$

So, $\frac{l}{{ - 3}} = \frac{m}{{ - 4}} = \frac{m}{{ - 12}}$→ l = $\frac{3}{{13}}$, m = $\frac{4}{{13}}$, n = $\frac{{12}}{{13}}$.

Again, from (1) and (4), we have,

Or, $\frac{l}{0+2} = \frac{m}{-2-0} = \frac{n}{4-3} \quad \left[ \begin{array}{c} 4l+3m-2n=0 \\ 4l+m+0.n=0 \end{array} \right]$

Or, $\frac{l}{2}$ = $\frac{m}{{ - 2}}$ = $\frac{n}{1}$.

So, l = $\frac{2}{3}$,m = $ - \frac{2}{3}$, n = $\frac{1}{3}$.

Hence, the dc’s $\frac{2}{3}$, $ - \frac{2}{3}$, $\frac{1}{3}$ and $\frac{3}{{13}}$, $\frac{4}{{13}}$, $\frac{{12}}{{13}}$.

Let θ be the angle between the lines. Then,

Cosθ = $\frac{2}{3}.\frac{3}{{13}}$ + $\left( { - \frac{2}{3}} \right)$.$\frac{4}{{13}}$ + $\frac{1}{3}$.$\frac{{12}}{{13}}$ = $\frac{{10}}{{39}}$.

So, θ= cos^-1 $\left( {\frac{{10}}{{39}}} \right)$.

 

c. l + m + n = 0 and l² + m² – n² = 0

Solution:

Given relations are:

l + m + n = 0….(i)

So, k = - m – n

And l² + m² – n² = 0 ..(2)

Or, (-m-n)² + m² – n² = 0   [from (1)]

Or, m² + 2mn + n² + m² – n² = 0

Or, 2m² + 2mn = 0

Or, 2m(m + n) = 0

So, m = 0 …(3)

And, m + n = 0 ..(4)

Now, from (1) and (3), we have,

Or, l + m + n = 0 …(1)

Or, 0.l + m + 0.n = 0 …(3)

So, $\frac{l}{{0 - 1}} = \frac{m}{{0 - 0}} = \frac{n}{{1 - 0}}$→$\frac{l}{{ - 1}} = \frac{m}{0} = \frac{n}{1}$.

So, l = $ - \frac{1}{{\sqrt 2 }}$, m = 0 , n = $\frac{1}{{\sqrt 2 }}$.

Again from(1) and (4), we have,

Or, $\frac{l}{1-1} = \frac{m}{0-1} = \frac{n}{1-0} \quad \left[ \begin{array}{c} l+m+n=0 \\ 0.l+m+n=0 \end{array} \right]$

So, $\frac{l}{0}$ = $\frac{m}{{ - 1}}$ = $\frac{n}{1}$→ l = 0, m = $ - \frac{1}{{\sqrt 2 }}$, n = $\frac{1}{{\sqrt 2 }}$.

So, the dc’s are $ - \frac{1}{{\sqrt 2 }},0,\frac{1}{{\sqrt 2 }}$ and 0, $ - \frac{1}{{\sqrt 2 }}$,$\frac{1}{{\sqrt 2 }}$.

Let θ be the angle between the lines, Then,

Or, cosθ = l₁l₂ + m₁.m₂ + n₁.n₂ = 0 + 0 + $\frac{1}{2}$ = $\frac{1}{2}$.

So, θ= 60°.

 

16. a) A line makes a, ẞ, y, 8 with the four diagonals of a cube, prove that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta = \frac{4}{3}$

Solution:

 

b) Prove that the lines whose direction cosines are given by the relations pl + qm + rn = 0 and amn + bnl+ clm = 0 are perpendicular if $\frac{a}{p} + \frac{b}{q} + \frac{c}{r} = 0$ and parallel if $\sqrt{ap} \pm \sqrt{bq} \pm \sqrt{cr} = 0$.

Solution: In the book Example Number 8