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

Correct Answer: -2.2 - -2.3

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)