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The length of the perpendicular drawn from the point $(3,-1,11)$ to the line $\frac{x}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ is :
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Verified Answer
The correct answer is:
$\sqrt{53}$
Let feet of perpendicular is
$(2 \alpha, 3 \alpha+2,4 \alpha+3)$
$\Rightarrow$ Direction ratio of the $\perp$ line is
$2 \alpha-3,3 \alpha+3,4 \alpha-8$. and
$\Rightarrow$ Direction ratio of the line are $2,3,4$
$\Rightarrow 2(2 \alpha-3)+3(3 \alpha+3)+4(4 \alpha-8)=0$
$\Rightarrow \alpha=1$
$\Rightarrow$ Feet of $\perp$ is $(2,5,7)$
$\Rightarrow$ Length $\perp$ is $\sqrt{1^{2}+6^{2}+4^{2}}=\sqrt{53}$
$(2 \alpha, 3 \alpha+2,4 \alpha+3)$
$\Rightarrow$ Direction ratio of the $\perp$ line is
$2 \alpha-3,3 \alpha+3,4 \alpha-8$. and
$\Rightarrow$ Direction ratio of the line are $2,3,4$
$\Rightarrow 2(2 \alpha-3)+3(3 \alpha+3)+4(4 \alpha-8)=0$
$\Rightarrow \alpha=1$
$\Rightarrow$ Feet of $\perp$ is $(2,5,7)$
$\Rightarrow$ Length $\perp$ is $\sqrt{1^{2}+6^{2}+4^{2}}=\sqrt{53}$
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