kukuhushlq3p

2023-02-19

If Q(0,−1,−3) is the image of the point P in the plane 3x−y+4z=2 and R is the point (3,−1,−2), then find the area (in sq. units) of triangle PQR

alihikaua8xxc

Beginner2023-02-20Added 9 answers

M = a point on the plane such that PQ passes through M.

The point R lies on the plane : 3x−y+4z−2=0

$PM=|\frac{1-12-2}{\sqrt{9+1+16}}|=\sqrt{\frac{13}{2}}\phantom{\rule{0ex}{0ex}}PR=\sqrt{9+1}=\sqrt{10}\phantom{\rule{0ex}{0ex}}\text{Using pythagorus theorem in}\Delta PMR,\phantom{\rule{0ex}{0ex}}RM=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta PQR=2\times \frac{1}{2}\times \sqrt{\frac{13}{2}}\times \sqrt{\frac{7}{2}}=\sqrt{\frac{91}{2}}$

The point R lies on the plane : 3x−y+4z−2=0

$PM=|\frac{1-12-2}{\sqrt{9+1+16}}|=\sqrt{\frac{13}{2}}\phantom{\rule{0ex}{0ex}}PR=\sqrt{9+1}=\sqrt{10}\phantom{\rule{0ex}{0ex}}\text{Using pythagorus theorem in}\Delta PMR,\phantom{\rule{0ex}{0ex}}RM=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}\phantom{\rule{0ex}{0ex}}\Rightarrow \Delta PQR=2\times \frac{1}{2}\times \sqrt{\frac{13}{2}}\times \sqrt{\frac{7}{2}}=\sqrt{\frac{91}{2}}$