The distance of the point $(7,10,11)$ from the line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is
JEE · Math · previous-year question
- A.16
- B.12
- C.18
- D.14correct
Answer
D. 14
Explanation
Equation of line passing through $P(7,10,11)$ along the line $\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}$ is $$\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda$$ Let the point on the line is $$Q(2 \lambda+7,3 \lambda+10,6 \lambda+11)$$ $Q$ lies on line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$ $$\begin{aligned} & 3 \lambda+10=4 \Rightarrow \lambda=-2 \\ & \therefore \quad Q(3,4,-1) \\ & P Q=\sqrt{16+36+144}=14 \end{aligned}$$
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