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Question

The orthogonal projection $$A'$$ of the point $$A$$ with position vector $$(1, 2, 3)$$ on the plane $$3x - y + 4z = 0$$ is


A
(1,3,1)
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B
(12,52,1)
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C
(12,52,1)
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D
(6,7,5)
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Solution

The correct option is D $$\displaystyle \left ( -\frac{1}{2}, \frac{5}{2}, 1 \right )$$
Equation of line passes through $$(1,2,3)$$ and perpendicular to the given plane is given by,
$$\displaystyle \frac{x-1}{3}=\frac{y-2}{-1}=\frac{z-3}{4}=k$$ (say)
Let any point on this line is $$P(3k+1, -k+2, 4k+3)$$
For orthogonal projection point $$P$$ lie on the given plane.
$$\Rightarrow 3(3k+1)-(2-k)+4(4k+3)=0\Rightarrow k = -\cfrac{1}{2}$$
Hence, required orthogonal point is $$\left(\cfrac{-1}{2},\cfrac{5}{2}, 1\right)$$

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