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प्रश्न
Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersects the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.
उत्तर
\[\text{ Let } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} = \lambda (\text{ say } )\]
\[ \Rightarrow x = 3\lambda + 2; y = 4\lambda - 1; z = 2\lambda + 2 . . . \left( 1 \right)\]
\[\text{ Since (x,y,z) intersects the plane x - y + z - 5 = 0 } , \]
\[3\lambda + 2 - \left( 4\lambda - 1 \right) + 2\lambda + 2 - 5 = 0\]
\[ \Rightarrow 3\lambda + 2 - 4\lambda + 1 + 2\lambda + 2 - 5 = 0\]
\[ \Rightarrow \lambda = 0\]
\[\text{ Substituting this in (1), we get } \]
\[x = 2; y = - 1; z = 2\]
\[\text{ So } ,\left( x, y, z \right) = \left( 2, - 1, 2 \right)\]
\[\text{ Finding the angle }\]
\[\text{ The given line is parallel to the vector } \vec{b} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \text{ and the given plane is normal to the vector } \vec{n} = \hat{i} - \hat{j} + \hat{k} . \]
\[\text{ We know that the angle } \theta \text{ between a line and a plane is given by} \]
\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]
\[ = \frac{\left( 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right) . \left( \hat{i} - \hat{j} + \hat{k} \right)}{\left| 3 \hat{i} + 4 \hat{j} + 2 \hat{k} \right| \left| \hat{i} - \hat{j} + \hat{k} \right|} = \frac{3 - 4 + 2}{\sqrt{9 + 16 + 4} \sqrt{1 + 1 + 1}} = \frac{1}{\sqrt{87}}\]
\[ \Rightarrow \theta = \sin^{- 1} \left( \frac{1}{\sqrt{87}} \right)\]
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