3.
What is the value of $$n$$ so that the angle between the lines having direction ratios $$\left( {1,\,1,\,1} \right)$$ and $$\left( {1,\, - 1,\,n} \right)$$ is $${60^ \circ }\,?$$
The point $$A\left( {6,\,7,\,7} \right)$$ is on the line . Let the perpendicular from $$P$$ meet the line in $$L$$. Then
$$A{P^2} = {\left( {6 - 1} \right)^2} + {\left( {7 - 2} \right)^2} + {\left( {7 - 3} \right)^2} = 66$$
5.
A line makes $${45^ \circ }$$ with positive $$x$$-axis and makes equal angles with positive $$y,\, z$$ axes, respectively. What is the sum of the three angles which the line makes with positive $$x,\,y$$ and $$z$$ axes ?
We know that sum of square of direction cosines $$ = 1$$
$$\eqalign{
& {\text{i}}{\text{.e}}{\text{., }}{\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1 \cr
& \Rightarrow {\cos ^2}{45^ \circ } + {\cos ^2}\beta + {\cos ^2}\beta = 1 \cr
& \left( {{\text{As given }}\alpha = {{45}^ \circ }{\text{ and }}\beta = \gamma } \right) \cr
& \Rightarrow \frac{1}{2} + 2{\cos ^2}\beta = 1 \cr
& \Rightarrow {\cos ^2}\beta = \frac{1}{4} \cr
& \Rightarrow \cos \,\beta = \pm \frac{1}{2}, \cr} $$
Negative value is discarded,
Since the line makes angle with positive axes.
Hence, $$\cos \,\beta = \frac{1}{2} \Rightarrow \cos \,\beta = \cos \,{60^ \circ } \Rightarrow \beta = {60^ \circ }$$
$$\therefore $$ Required sum $$ = \alpha + \beta + \gamma = {45^ \circ } + {60^ \circ } + {60^ \circ } = {165^ \circ }$$
6.
Given the line $$L:\frac{{x - 1}}{3} = \frac{{y + 1}}{2} = \frac{{z - 3}}{{ - 1}}$$ and the plane $$\pi \,:x - 2y = z.$$ Of the following assertions, the only one that is always true is :
Since $$3\left( 1 \right) + 2\left( { - 2} \right) + \left( { - 1} \right)\left( { - 1} \right) = 3 - 4 + 1 = 0$$
$$\therefore $$ given line is $$ \bot $$ to the normal to the plane i.e., given line is parallel to the given plane.
Also $$\left( {1,\, - 1,\,3} \right)$$ lies on the plane $$x - 2y - z = 0$$ if $$1 - 2\left( { - 1} \right) - 3 = 0$$ i.e. $$1 + 2 - 3 = 0$$ which is true.
$$\therefore \,L$$ lies in plane $$\pi .$$
7.
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
The equation of plane through the point $$\left( {1,\, - 2,\,1} \right)$$ and perpendicular to the planes $$2x-2y+z=0$$ and $$x-y+2z=4$$ is given by
\[\left| \begin{array}{l}
x - 1\,\,\,\,\,y + 2\,\,\,\,\,z - 1\\
\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 2\,\,\,\,\,\,\,\,\,\,\,\,\,1\\
\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 1\,\,\,\,\,\,\,\,\,\,\,\,\,2
\end{array} \right| = 0\,\, \Rightarrow x + y + 1 = 0\]
It’s distance from the point (1, 2, 2) is
$$\left| {\frac{{1 + 2 + 1}}{{\sqrt 2 }}} \right| = 2\sqrt 2 .$$
8.
The two lines $$x=ay+b,\,z=cy+d\,;$$ and $$x=a'y+b',\,z=c'y+d'$$ are perpendicular to each other if :
9.
The point $$P$$ is the intersection of the straight line joining the points $$Q\left( {2,\,3,\,5} \right)$$ and $$R\left( {1,\, - 1,\,4} \right)$$ with the plane $$5x-4y-z=1.$$ If $$S$$ is the foot of the perpendicular drawn from the
point $$T\left( {2,\,1,\,4} \right)$$ to $$QR,$$ then the length of the line segment $$PS$$ is :