Question

If the image of point P(1, 2, 3) about the plane 2x - y + 3z = 2 is Q, then the area of triangle PQR, where coordinates of R is (4, 10, 12)

• A. $\sqrt{\frac{1531}{2}}$
• B. $\sqrt{\frac{1675}{2}}$
• C. $\sqrt{\frac{2443}{2}}$
• D. $\sqrt{\frac{1784}{2}}$

Answer 1

Answer: (A)

$\sqrt{\frac{1531}{2}}$

Answer 2

The correct option is (A): $\sqrt{\frac{1531}{2}}$
Image formula w.r.t P
$\frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{3}=-2\frac{(2\times 1-2+3\times 3-2)}{1^{2}+2^{2}+3^{2}}$
$\Rightarrow Q(-1,3,0)$

$Area=\frac{1}{2}\begin{Vmatrix} \widehat{i} & \widehat{j}& \widehat{k}\\\ 5& 7 & 12\\\ 2 & -1 & 3 \end{Vmatrix}$
$=\frac{1}{2}|33\widehat{i} +9\widehat{j}-19\widehat{k}|$
$Area=\frac{1}{2}\sqrt{(33)^{2}+9^{2}+(19)^{2}}$
$=\frac{1}{2}\sqrt{1089+81+361}=\frac{1}{2}\sqrt{1531}$