Question

Foot of the perpendicular drawn from the origin to the plane $2x-3y+4z=29$ is

• A. $(5,-1,4)$
• B. $(7,-1,3)$
• C. $(5,-2,3)$
• D. $(2,-3,4)$
• E. $(1,-3,4)$

Answer 1

The correct option is D $(2,-3,4)$

Let the foot of the perpendicular in the $2x-3y+4z=29$ be $P(\alpha ,\beta ,\gamma )$.
So, the point $(\alpha ,\beta ,\gamma )$ satisfy the given plane.
$2\alpha -3\beta +4\gamma =29$ ........ (i)
Now, DR's of $PO$ is $(\alpha ,\beta ,\gamma )$, where $O$ is origin.
Since, $OF$ is perpendicular to the given plane.
Therefore, normal to the plane is parallel to $OF$.
$\therefore \frac{\alpha }{2}=\frac{\beta }{-3}=\frac{\gamma }{4}=k$
$\Rightarrow \alpha =2k,\beta =-3k$ and $\gamma =4k$
On putting the value of $\alpha ,\beta$ and $\gamma$ in Eq. (i), we get
$2\left(2k\right)-3(-3k)+4\left(4k\right)=29$
$\Rightarrow 4k+9k+16k=29$
$\Rightarrow 29k=29\Rightarrow k=1$
Therefore, $\alpha =2,\beta =-3$ and $\gamma =4$
Hence, foot of perpendicular is $(2,-3,4)$.