QJEEAdv20Ma33e4: JEE Advanced 2020 Mathematics - Class 12 JEEAdv20Ma33e4 Solution: Let L 1 and L 2 be the following straight lines. and . Suppose…

Equation of given straight lines

${L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}$

and ${L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}$

having vector form respectively, are

and $$\begin{array}{rl}& r=(\hat{i}+\hat{k})+\lambda (\hat{i}-\hat{j}+3\hat{k})\\ & r=(\hat{i}+\hat{k})+v(-3\hat{i}-\hat{j}+\hat{k})\end{array}$$

$\because$ $(\widehat i - \widehat j + 3\widehat k)\,.\,( - 3\widehat i - \widehat j + \widehat k)$

= $-$ 3 + 1 + 3 = 1 is positive,

$\therefore$ Angle between supporting line vectors of lines L₁ and L₂ is acute, and point of intersection of given lines L₁ and L₂ is (1, 0, 1)

Now, vector along the acute angle bisector of vectors $(\widehat i - \widehat j + 3\widehat k)$ and $( - 3\widehat i - \widehat j + \widehat k)$ is $(\widehat i - \widehat j + 2\widehat k)$ or $(\widehat i + \widehat j - 2\widehat k)$

It is given that line

$L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}$ is the bisector of the acute angle between the lines L₁ and L₂, so

l = 1 and m = 1

and ${{I - \alpha } \over 1} = {{0 - 1} \over 1} = {{1 - \gamma } \over { - 2}}$

$\Rightarrow \alpha = 2,\,\gamma = - 1$

$\therefore$ $\alpha - \gamma = 3,\,l + m = 2$