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 ________.

Updated On: Sep 13, 2024
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Correct Answer: -2.2 - -2.3

Solution and Explanation

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)

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Concepts Used:

Shortest Distance Between Two Parallel Lines

Formula to find distance between two parallel line:

Consider two parallel lines are shown in the following form :

\(y = mx + c_1\) …(i)

\(y = mx + c_2\) ….(ii)

Here, m = slope of line

Then, the formula for shortest distance can be written as given below:

\(d= \frac{|c_2-c_1|}{\sqrt{1+m^2}}\)

If the equations of two parallel lines are demonstrated in the following way :

\(ax + by + d_1 = 0\)

\(ax + by + d_2 = 0\)

then there is a little change in the formula.

\(d= \frac{|d_2-d_1|}{\sqrt{a^2+b^2}}\)