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:

• A. $(6,5,2)$
• B. $(6,5,-2)$
• C. $(4,3,2)$
• D. $(3,4,-2)$

Answer 1

The correct option is B

$(6,5,-2)$

Explanation for the correct option:

Finding the image of $R$ in the plane $P$

Let $A=(2,1,0),B=(4,1,1)$and $C=(5,0,1)$
Let $X$ be any point on the required plane

$X\equiv (x,y,z)$

Two lines are said to be coplanar when they both lie on the same plane in a three-dimensional space.

Apply CO-Planarity condition for i.e. If there are three vectors in a 3d-space and their scalar triple product is zero.

$\overrightarrow{AP},\overrightarrow{AB}\text{and}\overrightarrow{AC}$
$$\begin{gathered} \left[\begin{array}{ccc}x-2& y-1& z-0\\ 4-2& 1-1& 1-0\\ 5-2& 0-1& 1-0\end{array}\right]=0\left[R_1:P-A;R_2:B-A;R_3:C-A\right] \\ \left[\begin{array}{ccc}x-2& y-1& z\\ 2& 0& 1\\ 3& -1& 1\end{array}\right]=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow x-2+y-1+z(-2)=0 \\ \Rightarrow x+y-2z=3 \end{gathered}$$

Now, $R\text{'}(x,y,z)$is image of $R(2,1,6)$ w.r.t. to plane $x+y-2z=3$
$$\begin{gathered} \Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=\frac{-2(2+1-2\times 6-3)}{(1+1+4)} \\ \Rightarrow \frac{x-2}{1}=\frac{y-1}{1}=\frac{z-6}{-2}=4 \end{gathered}$$

$\begin{array}{rcl}x-2& =& 4\\ x& =& 4+2=6\\ & \Rightarrow & x=6,y=5,z=-2\\ \therefore R\text{'}& \equiv & (6,5,-2)\end{array}$

Hence the correct option is (B).