Question

Find the length and the foot of perpendicular from the point $\left(1,\frac{3}{2},2\right)$ to the plane $2x-2y+4z+5=0$. [NCERT EXEMPLAR]

Answer 1

Let M be the foot of the perpendicular from P$\left(1,\frac{3}{2},2\right)$ on the plane $2x-2y+4z+5=0$.

Then, PM is the normal to the plane. So, its direction ratios are proportional to 2, −2, 4.

Since PM passes through P$\left(1,\frac{3}{2},2\right)$, therefore, its equation is

$\frac{x-1}{2}=\frac{y-{\displaystyle \frac{3}{2}}}{-2}=\frac{z-2}{4}=\lambda \left(\mathrm{Say}\right)$

Let the coordinates of M be $\left(2\lambda +1,-2\lambda +\frac{3}{2},4\lambda +2\right)$.

Now, M lies on the plane $2x-2y+4z+5=0$.

$$\begin{gathered} \therefore 2\left(2\lambda +1\right)-2\left(-2\lambda +\frac{3}{2}\right)+4\left(4\lambda +2\right)+5=0 \\ \Rightarrow 24\lambda +12=0 \\ \Rightarrow \lambda =-\frac{1}{2} \end{gathered}$$

So, the coordinates of M are $\left(2\times \left(-\frac{1}{2}\right)+1,-2\times \left(-\frac{1}{2}\right)+\frac{3}{2},4\times \left(-\frac{1}{2}\right)+2\right)$ or $\left(0,\frac{5}{2},0\right)$.

Thus, the coordinates of the foot of the perpendicular are $\left(0,\frac{5}{2},0\right)$.

Now,

$\mathrm{PM}=\sqrt{{\left(1-0\right)}^2+{\left(\frac{3}{2}-\frac{5}{2}\right)}^2+{\left(2-0\right)}^2}=\sqrt{1+1+4}=\sqrt{6}$

Thus, the length of the perpendicular from the given point to the plane is $\sqrt{6}$ units.