Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of inte…

Question

Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x - 3y + 5z = 8. If the mirror image of the point $$\left( {2, - {1 \over 2},2} \right)$$ in the rotated plane is B(a, b, c), then :

Accepted Answer

Consider the equation of plane,

$$P:(2x + 3y + z + 20) + \lambda (x - 3y + 5z - 8) = 0$$

$$P:(2 + \lambda )x + 3(3 - 3\lambda )y + 1(1 + 5\lambda )z + (20 - 8\lambda ) = 0$$

$$\because$$ Plane P is perpendicular to $2x + 3y + z + 20 = 0$

So, $$4 + 2\lambda  + 9 - 9\lambda  + 1 + 5\lambda  = 0$$

$$ \Rightarrow \lambda  = 7$$

$P:9x - 18y + 36z - 36 = 0$

or $P:x - 2y + 4z = 4$

If image of $$\left( {2, - {1 \over 2},2} \right)$$ in plane P is (a, b, c) then

$${{a - 2} \over 1} = {{b + {1 \over 2}} \over { - 2}} = {{c - 2} \over 4}$$

and $$\left( {{{a + 2} \over 2}} \right) - 2\left( {{{b - {1 \over 2}} \over 2}} \right) + 4\left( {{{c + 2} \over 2}} \right) = 4$$

Clearly $$a = {4 \over 3}$$, $$b = {5 \over 6}$$ and $$c =  - {2 \over 3}$$

So, $a:b:c = 8:5: - 4$

Solutions · 1