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Question
If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.
Solution
\[\text{ The normal is passing through the pointsO(0, 0, 0) andP(2, 3 ,-1). So } ,\]
\[ \vec{n} = \vec{OP} =\left( \text{ 2 } \hat{i} + \text{ 3 }\hat{j} - \hat{k} \right) - \left( 0 \hat{i} + \text{ 0 } \hat{j} + \text{ 0 }\hat{k} \right) = \text{ 2 }\hat{i} + \text{ 3 }\hat{j} - \hat{k} \]
\[\text{ Since the plane passes through the point } (2, 3 ,-1), \vec{a} = \text{ 2 } \hat{i} + \text{ 3 } \hat{j} - \hat{k} \]
\[\text{ We know that the vector equation of the plane passing through a point} \vec { a } \text { and normal to } \vec {n} \text { is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[\text{ Substituting } \vec{a} = \text{ 2 }\hat{i} + \text{ 3 } \hat{j} - \hat{k} \text{ and } \vec{n} = \text{ 2 } \hat{i} + \text{ 3 }\hat{j} - \hat{k} , \text{ we get } \]
\[ \vec{r} . \left( \text{ 2 } \hat{i} + \text{ 3 }\hat{j} - \hat{k} \right) = \left( \text{ 2 }\hat{i} + \text{ 3 }\hat{j} - \hat{k} \right) . \left( \text{ 2 } \hat{i} + \text{ 3 } \hat{j} - \hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left( \text{ 2 } \hat{i} + \text{ 3 }\hat{j} - \hat{k} \right) = 4 + 9 + 1\]
\[ \Rightarrow \vec{r} . \left( \text{ 2 }\hat{i} + \text{ 3 } \hat{j} - \hat{k} \right) = 14\]
\[ \Rightarrow \vec{r} . \left( \text{ 2 }\hat{i} + \text{ 3 }\hat{j} - \hat{k} \right) = 14\]
\[\text{ Substituting } \vec{r} = \text{ x }\hat{i} + \text{ y }\hat{j} + \text{ z }\hat{k} \text{ in the vector equation, we get } \]
\[\left( \text{ x }\hat{i} + \text{ y }\hat{j} + \text{ z } \hat{k} \right) . \left( \text{ 2 } \hat{i} + \text{ 3 } \hat{j} - \hat{k} \right) = 14\]
\[ \Rightarrow 2x + 3y - z = 14\]
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