Question

The line of the shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of cos−1⁡$\sqrt{\frac{2}{27}}$ with the plane P: ax – y – z = 0, (a> 0). If the image of the point (1, 1, –5) in the plane P is (α, β, γ), then α + β – γ is equal to ________.

Answer 1

The line of shortest distance will be along $\vec {b_1}$×$\vec{b_2}$
Where,
$\vec {b_1}$=$\hat j$+$\hat k$
and
$\vec {b_2}$=2$\hat i$+2$\hat j$+$\hat k$
$\vec {b_1}$×$\vec {b_2}$=$\begin{vmatrix} \hat i &\hat j &\hat k \\\ 0&0 &1 \\\ 2&2 &1 \end{vmatrix}$=−$\hat i$+2$\hat j$−2$\hat k$
Angle between $\vec {b_1}$×$\vec {b_2}$ and plane P,
sin⁡θ=|$\frac{-a-2+2}{3.\sqrt{a^2+2}}$|=$\frac{5}{\sqrt{27}}$
⇒$\frac{|a|}{\sqrt{a^2+2}}$=$\frac{5}{\sqrt3}$
⇒ a2=-$\frac{25}{11}$(not possible)