Question

Find length of foot of perpendicular from point (1,1,2) to plane r.(i-2j+4k)+5=0

Answer 1

Solution:

The plane in vector form is

r. (i - 2 j + 4 k) + 5=0

r = x i + y j + z k

(x i + y j + z k) . (i - 2 j + 4 k) + 5=0

x - 2 y + 4 z+ 5=0 →→The plane in Scalar form.

Distance of plane A x + B y + C z +D=0 from P (p,q,r) is = $\frac{Ap+B q + C r+D}{\sqrt{A^2+B^2+C^2}}$

Length of Perpendicular form point (1,1,2) to plane x - 2 y + 4 z +5=0 is given by

=   $\frac{1-2 \times 1+ 4 \times 2 +5}{\sqrt{1^2+(-2)^2+(4)^2}}=\frac{12}{\sqrt{21}}$