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,…

Question

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 the area (in sq. units) of $	riangle PQR$ is:

Accepted Answer

Image of Q in plane

$${{\left( {x - 0} \right)} \over 3} = {{\left( {y + 1} \right)} \over { - 1}} = {{z + 3} \over { + 4}} = {{ - 2(1 - 12 - 2)} \over {9 + 1 + 16}} = 1$$

x = 3, y = --2, z = 1

P(3, --2, 1), Q(0, --1, --3), R(3, --1, --2)

Now area of $$\Delta $$PQR is

$${1 \over 2}\left| {\overline {PQ}  \times \overline {QR} } \right| = {1 \over 2}\left| {\left| {\begin{pmatrix} {\widehat i} & {\widehat j} & {\widehat k}  \
   3 & { - 1} & 4  \
   3 & 0 & 1  \

\end{pmatrix} } \right|} \right|$$

$$ \Rightarrow {1 \over 2}\left| {\left\{ {\widehat i( - 1) - \widehat j(3 - 12) + \widehat k(3)} \right\}} \right|$$

$$ \Rightarrow {1 \over 2}\sqrt {(1 + 81 + 9)} $$

$$ \Rightarrow {{\sqrt {91} } \over 2}$$

Solutions · 1