12th NCERT Three Dimensional Geometry Exercise 11.1 Questions 5
Do or do not
There is no try

Question (1)

If a line makes andgles 90°, 135°, 45° with x, y and z axes respectively, find its direction cosines.

Solution

The angle made by a line with x axis is denoted by α, with y axis is β, and with z axis is γ.
The direction cosines of a line is the value of ( cos α, cos β, cos γ )
Now α = 90°, , β = 135° , and γ = 45°
cos 135° = cos ( π - 45°) = - cos 45° = - 1/√2
The direction cosines of a line = cos α, cos β, cos γ
= cos 90°, cos 135°, cos 45°
= 0, -1/√2 , 1/ √2 .

Question (2)

Find the direction cosines of a line which makes equal angles with the coordinate axes.

Solution

The angle made by a line with x axis is denoted by α, with y axis is β, and with z axis is γ.
The direction cosines of a line is the value of ( cos α, cos β, cos γ )
the line makes an equal angles with axes.
So α = β = γ
Now cos2 α + cos2 β + cos2 γ = 1
∴ 3 cos2 α = 1
∴ cos2 α = 1/3
∴ cos α = ± 1/ √3.
The direction cosines of a line = cos α, cos β, cos γ
= ± 1/ √3, ± 1/ √3, ± 1/ √3.

Question (3)

If a line has direction ratios -18, 12, -4, then what are its direction cosines?

Solution

The direction ratios are -18, 12, -4.
Then its direction cosines are given by
\[\frac{{ - 18}}{{\sqrt { - 18{)^2} + {{( 12)}^2} + {{( - 4)}^2}} }},\frac{{ 12}}{{\sqrt { - 18{)^2} + {{( 12)}^2} + {{( - 4)}^2}} }},\frac{{ - 4}}{{\sqrt { - 18{)^2} + {{( 12)}^2} + {{( - 4)}^2}} }}\] \[\frac{{ - 18}}{{\sqrt {484} }},\frac{{ 12}}{{\sqrt {484} }},\frac{{ - 4}}{{\sqrt {484} }}\] \[\frac{{ - 18}}{{22}},\frac{{ 12}}{{22}},\frac{{ - 4}}{{22}}\] \[\frac{{ - 9}}{{11}},\frac{6}{{11}},\frac{{ - 2}}{{11}}\]

Question (4)

Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.

Solution

To prove the points collinear , first we will find the direction ratios.
Let A(2, 3, 4), B(-1, -2, 1), C(5, 8, 7)
Direction ratio of AB = (-1, -2, 1) - (2, 3, 4) = -3, -5, -5
Direction ratio of AC = (5, 8, 7) - (2, 3, 4) = 3, 5, 5
These direction ratios are in proportion. and they have A as a common point. So the points A, B and C are collinear.
Other method
We calculate the vector AB and vector AC. And find the cross product between them.
IF it is zero vector then we can say the vectors are collinear as they have one common point.
Vector AB =( - 1,- 2, 1) - (2, 3, 4) = (- 3, - 5, - 5}
vector AC = (5, 8, 7) - (2, 3, 4) = (3, 5, 5)
Now cross product between them
$$\left| \begin{matrix} i & j & k \\ -3 & -5 & -5 \\ 3 & 5 & 5 \\ \end{matrix} \right| $$ =i(-25 + 25) -j(-15 +15) +k(-15+15)
= 0
The points A, B and C are collinear.

Question (5)

Find the direction cosines of the sides of the triangle whose vertices are (3, 5, -4), (-1, 1, 2) and (-5, -5, -2).

Solution

Let A(3, 5, -4), B(-1, 1, 2) and C(-5, -5, -2).
Now Direction ratio of AB = (-1, 1, 2) - (3, 5, -4) = -4, -4, 6.
Direction cosine of AB =
\[\frac{{ - 4}}{{\sqrt {{{( - 4)}^2} + {{( - 4)}^2} + {{(6)}^2}} }},\frac{{ - 4}}{{\sqrt {{{( - 4)}^2} + {{( - 4)}^2} + {{(6)}^2}} }},\frac{6}{{\sqrt {{{( - 4)}^2} + {{( - 4)}^2} + {{(6)}^2}} }}\]
\[\frac{{ - 4}}{{\sqrt {68} }},\frac{{ - 4}}{{\sqrt {68} }},\frac{6}{{\sqrt {68} }}\] \[\frac{{ - 4}}{{2\sqrt {17} }},\frac{{ - 4}}{{2\sqrt {17} }},\frac{6}{{2\sqrt {17} }}\] \[\frac{{ - 2}}{{\sqrt {17} }},\frac{{ - 2}}{{\sqrt {17} }},\frac{3}{{\sqrt {17} }}\] Direction ratio of BC = (-5, -5, -2) - (-1, 1, 2) = -4, -6, -4.
\[\frac{-4}{{\sqrt {{4^2} + {{( - 6)}^2} + {{( - 4)}^2}} }},\frac{{ - 6}}{{\sqrt {{4^2} + {{( - 6)}^2} + {{( - 4)}^2}} }},\frac{{ - 4}}{{\sqrt {{4^2} + {{( - 6)}^2} + {{( - 4)}^2}} }}\] \[\frac{-4}{{\sqrt {68} }},\frac{{ - 6}}{{\sqrt {68} }},\frac{{ - 4}}{{\sqrt {68} }}\] \[\frac{-4}{{2\sqrt {17} }},\frac{{ - 6}}{{2\sqrt {17} }},\frac{{ - 4}}{{2\sqrt {17} }}\] \[\frac{{ - 2}}{{\sqrt {17} }},\frac{{ - 3}}{{\sqrt {17} }},\frac{{ - 2}}{{\sqrt {17} }}\]
Direction ratio of AC = (-5, -5, -2) - (3, 5, -4)) = -8, -10, 2.
\[\frac{{ - 8}}{{\sqrt {{{( - 8)}^2} + {{( - 10)}^2} + {{(2)}^2}} }},\frac{{ - 10}}{{\sqrt {{{( - 8)}^2} + {{( - 10)}^2} + {{(2)}^2}} }},\frac{2}{{\sqrt {{{( - 8)}^2} + {{( - 10)}^2} + {{(2)}^2}} }}\] \[\frac{{ - 8}}{{\sqrt {168} }},\frac{{ - 10}}{{\sqrt {168} }},\frac{2}{{\sqrt {168} }}\] \[\frac{{ - 8}}{{2\sqrt {42} }},\frac{{ - 10}}{{2\sqrt {42} }},\frac{2}{{2\sqrt {42} }}\] \[\frac{{ - 4}}{{\sqrt {42} }},\frac{{ - 5}}{{\sqrt {42} }},\frac{1}{{\sqrt {42} }}\]
#
⇒ Exercise 11.2