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Q.
The coordinates of the foot of perpendicular drawn from origin to the plane $2x - 3y + 4z - 6 = 0$ is
Three Dimensional Geometry
Solution:
Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane is $(x_1, y_1, z_1)$.
Then, the direction ratios of the line $OP$ are $x_1$, $y_1$, $z_1$.
Writing the equation of the plane in the normal form, we
have $\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{6}{\sqrt{29}}$
where, $\frac{2}{\sqrt{29}}$, $\frac{-3}{\sqrt{29}}$, $\frac{4}{\sqrt{29}}$ are the direction cosines of the lines $OP$.
Since, $d$.$c's$ and $d$.$r's$ of a line are proportional. Then, we
have $\frac{x_{1}}{\frac{2}{\sqrt{29}}}=\frac{y_{1}}{\frac{-3}{\sqrt{29}}}=\frac{z_{1}}{\frac{4}{\sqrt{29}}}=k$
i.e., $x_{1}=\frac{2k}{\sqrt{29}}$,
$y_{1}=\frac{-3k}{\sqrt{29}}$,
$z_{1}=\frac{4k}{\sqrt{29}}$
Substituting these values in the equation of the plane,
we get $\frac{4k}{\sqrt{29}}+\frac{9k}{\sqrt{29}}+\frac{16k}{\sqrt{29}}-6=0$
$\Rightarrow k=\frac{6}{\sqrt{29}}$
Hence, the foot of the perpendicular is $\left(\frac{12}{29}, \frac{-18}{29}, \frac{24}{29}\right)$.
