JEE MAIN - Mathematics (2022 - 24th June Morning Shift - No. 17)
Let a line having direction ratios, 1, $$-$$4, 2 intersect the lines $${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$$ and $${x \over 2} = {{y - 7} \over 3} = {z \over 1}$$ at the points A and B. Then (AB)2 is equal to ___________.
Answer
84
Explanation
Let $A(3 \lambda+7,-\lambda+1, \lambda-2)$ and $B(2 \mu, 3 \mu+7, \mu)$
So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
$\Rightarrow 5 \lambda-3 \mu=-16$
And $\lambda-5 \mu=10$
From (i) and (ii) we get $\lambda=-5, \mu=-3$
So, $A$ is $(-8,6,-7)$ and $B$ is $(-6,-2,-3)$
$$ A B=\sqrt{4+64+16} \Rightarrow(A B)^{2}=84 $$
So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
$\Rightarrow 5 \lambda-3 \mu=-16$
And $\lambda-5 \mu=10$
From (i) and (ii) we get $\lambda=-5, \mu=-3$
So, $A$ is $(-8,6,-7)$ and $B$ is $(-6,-2,-3)$
$$ A B=\sqrt{4+64+16} \Rightarrow(A B)^{2}=84 $$
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