Question

# Find the vector equation of the plane passing through the points P (2, 5, âˆ’3), Q (âˆ’2, âˆ’3, 5) and R (5, 3, âˆ’3).

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Solution

## $\text{The required plane passes through the point}\mathit{\text{P}}\text{(2, 5, -3) whose position vector is}\stackrel{â†’}{a}\text{=2}\stackrel{^}{i}\text{+5}\stackrel{^}{j}\text{-3}\stackrel{^}{k}\text{and is normal to the vector}\stackrel{â†’}{n}\text{given by}\phantom{\rule{0ex}{0ex}}\stackrel{â†’}{n}=\stackrel{â†’}{PQ}Ã—\stackrel{â†’}{PR.}\phantom{\rule{0ex}{0ex}}\text{Clearly,}\stackrel{â†’}{PQ}=\stackrel{â†’}{OQ}-\stackrel{â†’}{OP}=\left(-2\stackrel{^}{i}-3\stackrel{^}{j}+5\stackrel{^}{k}\right)-\left(2\stackrel{^}{i}\text{+ 5}\stackrel{^}{j}\text{- 3}\stackrel{^}{k}\right)=-4\stackrel{^}{i}-8\stackrel{^}{j}+8\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{â†’}{PR}=\stackrel{â†’}{OR}-\stackrel{â†’}{OP}=\left(5\stackrel{^}{i}+3\stackrel{^}{j}-3\stackrel{^}{k}\right)-\left(2\stackrel{^}{i}\text{+ 5}\stackrel{^}{j}\text{- 3}\stackrel{^}{k}\right)=3\stackrel{^}{i}-2\stackrel{^}{j}-0\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\stackrel{â†’}{n}=\stackrel{â†’}{PQ}Ã—\stackrel{â†’}{PR}=\left|\begin{array}{ccc}\stackrel{^}{i}& \stackrel{^}{j}& \stackrel{^}{k}\\ -4& -8& 8\\ 3& -2& 0\end{array}\right|=16\stackrel{^}{i}+24\stackrel{^}{j}+32\stackrel{^}{k}\phantom{\rule{0ex}{0ex}}\text{The vector equation of the required plane is}\phantom{\rule{0ex}{0ex}}\stackrel{â†’}{r}.\stackrel{â†’}{n}=\stackrel{â†’}{a}.\stackrel{â†’}{n}\phantom{\rule{0ex}{0ex}}â‡’\stackrel{â†’}{r}.\left(16\stackrel{^}{i}+24\stackrel{^}{j}+32\stackrel{^}{k}\right)=\left(2\stackrel{^}{i}\text{+5}\stackrel{^}{j}\text{-3}\stackrel{^}{k}\right).\left(16\stackrel{^}{i}+24\stackrel{^}{j}+32\stackrel{^}{k}\right)\phantom{\rule{0ex}{0ex}}â‡’\stackrel{â†’}{r}.\left[8\left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)\right]=32+120-96\phantom{\rule{0ex}{0ex}}â‡’\stackrel{â†’}{r}.\left[8\left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)\right]=56\phantom{\rule{0ex}{0ex}}â‡’\stackrel{â†’}{r}.\left(2\stackrel{^}{i}+3\stackrel{^}{j}+4\stackrel{^}{k}\right)=7\phantom{\rule{0ex}{0ex}}$

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