Find the vector equation of the plane passing through (1,2,3)and perpendicular to the plane r→.(i^+2j^-5k^)+9=0.
Find the vector equation of the plane passing through (1,2,3)and perpendicular to the plane r→.(i^+2j^-5k^)+9=0.
Solution and Explanation
The position vector of the point (1,2,3) is \(\overrightarrow{r_1}\) =\(\hat i+2\hat j^+3\hat k^.\)
The direction ratios of the normal to the plane,
\(\overrightarrow{r}\).(\(\hat i + 2\hat {j}-5\hat k\))+9=0, are 1,2,and -5 and the normal vector is \(\overrightarrow{N_1}\)=\(\hat i + 2\hat {j}-5\hat k\)
The equation of a line passing through a point and perpendicular to the given plane is given by,
\(\overrightarrow{I}\)=\(\overrightarrow{r}\)+λN→, λ∈R
⇒\(\overrightarrow{I}\)=(\(\hat i + 2\hat {j}+3\hat k\))+λ(\(\hat i + 2\hat {j}-5\hat k\)).
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