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 (integer)
84
Solution
Let $A(3 \lambda+7,-\lambda+1, \lambda-2)$ and $B(2 \mu, 3 \mu+7, \mu)$ <br/><br/>So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
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Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
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$\Rightarrow 5 \lambda-3 \mu=-16$
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And $\lambda-5 \mu=10$
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From (i) and (ii) we get $\lambda=-5, \mu=-3$
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So, $A$ is $(-8,6,-7)$ and $B$ is $(-6,-2,-3)$
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$A B=\sqrt{4+64+16} \Rightarrow(A B)^{2}=84$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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