JEE MAIN - Mathematics (2024 - 27th January Evening Shift - No. 23)
The lines $$\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$$ and $$\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$$ intersect at the point $$P$$. If the distance of $$\mathrm{P}$$ from the line $$\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$$ is $$l$$, then $$14 l^2$$ is equal to __________.
Answer
108
Explanation
$$\begin{aligned} & \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \\ & \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \\ & \Rightarrow \lambda+2=4 \mathrm{k}-3 \\ & -\lambda=3 \mathrm{k}-2 \\ & \Rightarrow \mathrm{k}=1, \lambda=-1 \\ & 8 \lambda+7=\mathrm{k}-2 \\ & \therefore \mathrm{P}=(1,1,-1) \end{aligned}$$
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Projection of $$2 \hat{i}-2 \hat{k}$$ on $$2 \hat{i}+3 \hat{j}+\hat{k}$$ is
$$\begin{aligned} & =\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}} \\ & \therefore l^2=8-\frac{4}{14}=\frac{108}{14} \\ & \Rightarrow 14 l^2=108 \end{aligned}$$
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