Question

Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.

  

Answer 1

\[\text{ The coordinates of any point on this line are of the form} \]

\[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2} = \lambda\]

\[ \Rightarrow x = 3\lambda + 2; y = 4\lambda - 1; z = 2\lambda + 2\]

\[\text{ So, the coordinates of the point on the given line are } \left( 3\lambda + 2, 4\lambda - 1, 2\lambda + 2 \right). \text{ This point lies on the plane x - y + z - 5 = 0 . }\]

\[ \Rightarrow 3\lambda + 2 - 4\lambda + 1 + 2\lambda + 2 - 5 = 0\]

\[ \Rightarrow \lambda = 0\]

\[\text{ So, the coordinates of the point are} \]

\[\left( 3\lambda + 2, 4\lambda - 1, 2\lambda + 2 \right)\]

\[ = \left( 3 \left( 0 \right) + 2, 4 \left( 0 \right) - 1, 2 \left( 0 \right) + 2 \right)\]

\[ = \left( 2, - 1, 2 \right)\]

\[\text{ Finding the angle between the line and the plane } \]

\[\text{ The given line is parallel to the vector }  \vec{b} = 3 \hat{i}  + 4 \hat{j}  + 2 \hat{k}  \text{ and the given plane is normal to the vector }  \vec{n} = \hat{i} - \hat{j}  + \hat{k}  . \]

\[\text{ We know that the angle } \theta \text{ between the line and the plane is given by } \]

\[\sin \theta = \frac{\vec{b} . \vec{n}}{\left| \vec{b} \right| \left| \vec{n} \right|}\]

\[ = \frac{\left( 3 \hat{i} + 4 \hat{j}  + 2 \hat{k}  \right) . \left( \hat{i}  - \hat{j}  + \hat{k}  \right)}{\left| 3 \hat{i}  + 4 \hat{j}  + 2 \hat{k}  \right| \left| \hat{i}  - \hat{j} + \hat{k}  \right|} = \frac{3 - 4 + 2}{\sqrt{9 + 16 + 4} \sqrt{1 + 1 + 1}} = \frac{1}{\sqrt{87}}\]

\[ \Rightarrow \theta = \sin^{- 1} \left( \frac{1}{\sqrt{87}} \right)\]