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  • A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, -1, 1), then the projection of<img alt="\overrightarrow{OP}" src="https://entrancecorner.codecogs.com/gif.latex?%5Coverrightarrow%7BOP%7D

A plane passes through the points A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2). If O is the origin and P is (2, -1, 1), then the projection of\overrightarrow{OP} on this plane is of length :
Option: 1 \sqrt{\frac{2}{7}}
Option: 2 \sqrt{\frac{2}{3}}
Option: 3 \sqrt{\frac{2}{11}}
Option: 4 \sqrt{\frac{2}{5}}

Answers (1)

best_answer

Equation of the plane passing through the point A(1, 2, 3), B(2, 3, 1) and C(2, 4, 2) is

\\\begin{vmatrix} x-1 &y-2 &z-3 \\ -1 & -1 &2 \\ -1& -2 &1 \end{vmatrix}=0\\\Rightarrow 3x-y+z-4=0

\\\text{normal to the plane }\vec n=3\hat i-\hat j+\hat k\\

\overrightarrow{\mathrm{OP}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}

\cos \theta=\frac{6+1+1}{\sqrt{6} \sqrt{11}}=\frac{8}{\sqrt{66}} \Rightarrow \sin \theta=\sqrt{\frac{2}{66}}

\\\therefore \text { Projection of } \overrightarrow{\mathrm{OP}} \text { on plane }=|\overrightarrow{\mathrm{OP}}| \sin \theta\\ =\sqrt{\frac{2}{11}} \\|\vec{OP}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt6

Posted by

himanshu.meshram

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