Question

The position vectors of points A, B and C are 8i + 4j – 3k, 6i + 3j - 4k and 7i + 5j – 5k, respectively. Find the area of ∆ ABC

Answer 1

find AB & AC
\[Then area of \triangle=\frac{ 1 }{2 }\left| Vector AB \times Vector AC \right|\]

Answer 2

Hmm, please solve

Answer 3

\[Vector AB=P.V of B- p.v. of A=\left( 6i+3j-4k \right)-\left( 8i+4j-3k \right)\]
\[=\left( 6-8 \right)i+\left( 3-4 \right)j+\left( -4+3 \right)k=-2i-j-k\]
\[similarly Vector AC=\left( 7-8 \right)i+\left( 5-4 \right)j+\left( -5+3 \right)k\]
\[=-i+j-2k\]

now you can find the absolute value.

Answer 4

Wow but no idea of the absolute value?

Answer 5

\[area=\frac{ 1 }{ 2}\sqrt{\left( 3 \right)^{2}+\left( -5 \right)^{2}+\left( -3 \right)^{2}}=\frac{ 1 }{2 }\sqrt{9+25+9}=\frac{ \sqrt{43} }{ 2 } units.\]

Answer 6

yw