Question

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

 

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 \[\frac{x - 1}{2} = \frac{y - \frac{3}{2}}{- 2} = \frac{z - 2}{4} = \lambda\left( \text{ Say } \right)\]

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 \[2x - 2y + 4z + 5 = 0\] \[\therefore 2\left( 2\lambda + 1 \right) - 2\left( - 2\lambda + \frac{3}{2} \right) + 4\left( 4\lambda + 2 \right) + 5 = 0\]

\[ \Rightarrow 24\lambda + 12 = 0\]

\[ \Rightarrow \lambda = - \frac{1}{2}\]

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,

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

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