First slide

Planes in 3D

Question

Let P be a plane passing through the points (2, 1, 0), (4, 1, 1) and (5, 0,1) and R be any point (2, 1,6). Then the image of R in the plane P is:

Difficult

A

(6,5,-2)
B

(6,5,2)
C

(4,3,2)
D

(3,4,-2)

Solution

The equation of the plane passing through three points (2, 1, 0), (4, 1, 1) and (5, 0,1) is $|\begin{matrix} x - 2 & y - 1 & z \\ 2 & 0 & 1 \\ 3 & - 1 & 1 \end{matrix}| = 0$
Expand the determinant as below.
$\begin{matrix} (x - 2) (0 + 1) - (y - 1) (- 1) + z (- 2) = 0 \\ \Rightarrow x - 2 + y - 1 - 2 z = 0 \\ \Rightarrow x + y - 2 z = 3 \end{matrix}$
Suppose that the image of the point $(2, 1, 6)$ in the plane $x + y - 2 z = 3$ is $(h, k, l)$
If the image of a point $P (x_{1}, y_{1}, z_{1})$ in the plane $a x + b y + c z + d = 0$ is $(h, k, l)$ then $\frac{h - x_{1}}{a} = \frac{k - y_{1}}{b} = \frac{l - z_{1}}{c} = \frac{- 2 (a x_{1} + b y_{1} + c)}{a^{2} + b^{2}}$
Hence, $\frac{h - 2}{1} = \frac{k - 1}{1} = \frac{l - 6}{- 2} = - \frac{2 (- 12)}{6}$
$\Rightarrow (h, k, l) = (6, 5, - 2)$

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