Q. Foot of the perpendicular drawn from the point (1, 3, 4) to the plane 2x - y + z + 3 = 0 is

Solution:

The dr’s of PA are $x_{1}-1, y_{1}-3, z_{1}-4$
The dr’s $\overrightarrow{n}$ are 2, -1, 1
The dr’s of PA and $\overrightarrow{n}$ are parallel
$\therefore\, \frac{x_{1}-1}{2}=\frac{y_{1}-3}{-1}=\frac{z_{1}-4}{1}=\lambda $
$x_{1}=2 \lambda+1, y_{1}=-\lambda+3, z_{1}=\lambda+4$
$\therefore\, A=\left(2\lambda+1, -\lambda+3, \lambda+4\right)lies\, on \,2x - y + z + 3 = 0$
$\Rightarrow\, -4\lambda+2-\lambda-3-\lambda+4+3=0$
$6\lambda=-6$
$\lambda=-1$
$\therefore\, P=\left(-1,4,3\right)$

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