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
Solution
<p>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</p>
<p>$\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda$</p>
<p>Let the point on the line is</p>
<p>$Q(2 \lambda+7,3 \lambda+10,6 \lambda+11)$</p>
<p>$Q$ lies on line $\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}$</p>
<p>$$\begin{aligned}
& 3 \lambda+10=4 \Rightarrow \lambda=-2 \\
& \therefore \quad Q(3,4,-1) \\
& P Q=\sqrt{16+36+144}=14
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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