Three Dimensional Geometry — Class 12 Maths Solution

.: (a) The given planes are
$7x + 5y + 6z + 30 = 0$

…(1)
and $3x - y - 10z + 4 = 0$

….(2)
These are neither parallel
nor perpendicular
If $\theta$ be the angle between (1) and (2), then

$\cos \theta = \left| {\cfrac{{(7)(3) + (5)( - 1) + (6)( - 10)}}{{\sqrt {49 + 25 + 36} \sqrt {9 + 1 + 100} }}} \right| = \left| {\cfrac{{21 - 5 - 60}}{{\sqrt {110} \sqrt {110} }}} \right| = \left| {\cfrac{{ - 44}}{{110}}} \right| = \left| {\cfrac{{ - 2}}{5}} \right|$

Hence $\theta = {\cos ^{ - 1}}\left( {\cfrac{2}{5}} \right)$

(b) The given planes are $x2 + y + 3z - 2 = 0$ ...(1)
and$x - 3y + 5 = 0$ ...(2)
Since$(2)(1) + (1)( - 2) + (3)(0) = 0$

$\therefore$ The planes are perpendicular.

(c) The given planes are $2x - 2y + 4z + 5 = 0$ ...(1)
and

$3x - 3y + 6z - 1 = 0$

...(2)
Since $\cfrac{2}{3} = \cfrac{{ - 2}}{{ - 3}} = \cfrac{4}{6},$ $\therefore$ The planes are parallel.

(d) The given planes are$2x - y + 3z - 1 = 0$ ...(1)
and $2x - y + 3z + 3 = 0$

...(2)

Since $\cfrac{2}{2} = \cfrac{{ - 1}}{{ - 1}} = \cfrac{3}{3},$ $\therefore$ The planes are parallel.

(e) The given planes are $4x + 8y + z - 8 = 0$ ...(1)
and $y + z - 4 = 0$

...(2)
Now, $\cfrac{4}{0} \ne \cfrac{8}{1} \ne \cfrac{1}{1},$ $\therefore$ These planes are not parallel.

Also, $(4)(0) + (8)(1) + (1)(1) \ne 0$

$\therefore$ These planes are not perpendicular.

If $\theta$ be the angle between (1) and (2), then

$\cos \theta = \left| {\cfrac{{(4)(0) + (8)(1) + (1)(1)}}{{\sqrt {16 + 64 + 1} \sqrt {0 + 1 + 1} }}} \right| = \left| {\cfrac{{0 + 8 + 1}}{{9\sqrt 2 }}} \right|$

$\Rightarrow$ $\cos \theta = \cfrac{1}{{\sqrt 2 }}$ $\Rightarrow$ $\theta = 45^\circ$