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If a point $\mathrm{R}(4, \mathrm{y}, \mathrm{z})$ lies on the line joining the points $P(2,-3,4)$ and $Q(8,0,10)$, then distance of $R$ from origin is
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The correct answer is:
$2 \sqrt{14}$
Equation of line joining point $\mathrm{P}$ and $\mathrm{Q}$ :
$\frac{x^{\prime}-2}{8-2}=\frac{y^{\prime}+3}{0+3}=\frac{z^{\prime}-4}{10-4}=\lambda($ say $)$
$\Rightarrow \quad x^{\prime}=6 \lambda+2, y^{\prime}=3 \lambda-3, z^{\prime}=6 \lambda+4$
$\because \quad$ Point $\mathrm{R}(4, \mathrm{y}, \mathrm{z})$ lies on line $\mathrm{PQ}$
$\because \quad 4=6 \lambda+2 \Rightarrow \lambda=\frac{1}{3}$
$\begin{aligned} & \text { and } y=3 \lambda-3=3 \times \frac{1}{3}-3=-2 \\ & z=6 \lambda+4=2+4=6\end{aligned}$
$\therefore \quad$ Point $\mathrm{R} \equiv(4,-2,6)$
Now, distance of $\mathrm{R}$ from origin is $=\sqrt{16+4+36}$ $=\sqrt{56}=2 \sqrt{14}$
$\frac{x^{\prime}-2}{8-2}=\frac{y^{\prime}+3}{0+3}=\frac{z^{\prime}-4}{10-4}=\lambda($ say $)$
$\Rightarrow \quad x^{\prime}=6 \lambda+2, y^{\prime}=3 \lambda-3, z^{\prime}=6 \lambda+4$
$\because \quad$ Point $\mathrm{R}(4, \mathrm{y}, \mathrm{z})$ lies on line $\mathrm{PQ}$
$\because \quad 4=6 \lambda+2 \Rightarrow \lambda=\frac{1}{3}$
$\begin{aligned} & \text { and } y=3 \lambda-3=3 \times \frac{1}{3}-3=-2 \\ & z=6 \lambda+4=2+4=6\end{aligned}$
$\therefore \quad$ Point $\mathrm{R} \equiv(4,-2,6)$
Now, distance of $\mathrm{R}$ from origin is $=\sqrt{16+4+36}$ $=\sqrt{56}=2 \sqrt{14}$
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