Find the image P′ of the point P having position vector (i + 3j + 4k) in the plane

Question

Find the image P′ of the point P having position vector \(\hat{i}+3\hat{j}+4\hat{k}\) in the plane \(\vec{r}.(2\hat{i}-\hat{j}+\hat{k})+3=0.\) Hence find the length of PP′.

Answer 1

Let given point be P (1, 3, 4) and the equation of given plane in cartisan form be

2x - y + z +3 = 0 ....(i)

Let R \((x_1, y_1, z_1)\) of foot of perpendicular and Q \((\alpha, \beta, \gamma)\) be the image of P

Since, R \((x_1, y_1, z_1)\) lie on plane (i)

Also, normal vector \(\vec{n}\) of plane (i) is \(\vec{n}\) = \(2\hat{i}-\hat{j}+\hat{k}\)

Putting \(x_1, y_1, z_1\) in (ii) we get

Since R is the mid point of PQ

Hence, image Q = (-3, 5, 2)