JEE MAIN - Mathematics (2023 - 25th January Morning Shift - No. 10)
Consider the lines $$L_1$$ and $$L_2$$ given by
$${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$$
$${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$$.
A line $$L_3$$ having direction ratios 1, $$-$$1, $$-$$2, intersects $$L_1$$ and $$L_2$$ at the points $$P$$ and $$Q$$ respectively. Then the length of line segment $$PQ$$ is
$$4\sqrt3$$
$$2\sqrt6$$
4
$$3\sqrt2$$
Explanation
Let,
$$ \begin{aligned} & P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text { and } Q(\mu+2,2 \mu+2 \text {, } 3 \mu+3) \\\\ & \text { d.r's of } P Q \equiv<2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1> \\\\ & \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-1}{-2} \\\\ & \therefore \quad-2 \lambda+\mu+1=\lambda-2 \mu+1 \text { and }-2 \lambda+4 \mu-2= \\\\ & -2 \lambda+3 \mu+1 \\\\ & \Rightarrow 3 \lambda-3 \mu=0 \text { and } \mu=3 \\\\ & \therefore \quad \lambda=\pm 3 \text { and } \mu=3 \\\\ & \therefore \quad P \equiv(7,6,8) \text { and } Q(5,8,12) \\\\ & \therefore|P O|=\sqrt{2^{2}+2^{2}+4^{2}}=\sqrt{24}=2 \sqrt{6} \end{aligned} $$
$$ \begin{aligned} & P \equiv(2 \lambda+1, \lambda+3,2 \lambda+2) \text { and } Q(\mu+2,2 \mu+2 \text {, } 3 \mu+3) \\\\ & \text { d.r's of } P Q \equiv<2 \lambda-\mu-1, \lambda-2 \mu+1,2 \lambda-3 \mu-1> \\\\ & \therefore \quad \frac{2 \lambda-\mu-1}{1}-\frac{\lambda-2 \mu-1}{-1}=\frac{2 \lambda-3 \mu-1}{-2} \\\\ & \therefore \quad-2 \lambda+\mu+1=\lambda-2 \mu+1 \text { and }-2 \lambda+4 \mu-2= \\\\ & -2 \lambda+3 \mu+1 \\\\ & \Rightarrow 3 \lambda-3 \mu=0 \text { and } \mu=3 \\\\ & \therefore \quad \lambda=\pm 3 \text { and } \mu=3 \\\\ & \therefore \quad P \equiv(7,6,8) \text { and } Q(5,8,12) \\\\ & \therefore|P O|=\sqrt{2^{2}+2^{2}+4^{2}}=\sqrt{24}=2 \sqrt{6} \end{aligned} $$
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