Question

Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\].

 

Answer 1

Let \[\vec{a}\] be the position vector of the point (3, 2, 0). \[\therefore \vec{a} = 3 \hat{i} + 2 \hat{j}\] The line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\] passes through the point (3, 6, 4) and is parallel to the vector \[\hat{i} + 5 \hat{j} + 4 \hat{k}\] .

Suppose

\[\vec{b} = 3 \hat{i} + 6 \hat{j} + 4 \hat{k}\]

\[\vec{c} = \hat{i} + 5 \hat{j} + 4 \hat{k}\]

Let \[\vec{N}\]

be the vector normal to the required plane.

\[\therefore \vec{N} = \left( \vec{b} - \vec{a} \right) \times \vec{c} \]

\[ = \left[ \left( 3 \hat{i} + 6 \hat{j} + 4 {k} \right) - \left( 3 \hat {i} + 2 \hat{j} \right) \right] \times \left( \hat{i} + 5 \hat{j} + 4 \hat{k} \right)\]

\[ = \left( 4 \hat{j} + 4 \hat{k} \right) \times \left( \hat{i} + 5 \hat{j} + 4 \hat{k} \right)\]

\[ = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\ 0 & 4 & 4 \\ 1 & 5 & 4\end{vmatrix}\]

\[ = - 4 \hat{i} + 4 \hat{j}- 4 \hat{k}\]

So, the required plane passes through the point 

\[\vec{a} = 3 \hat{i} + 2 \hat{j}\] and is perpendicular to the vector

\[\vec{N} = - 4 \hat{i} + 4 \hat{j} - 4 \hat{k}\] .

∴ Equation of the required plane is given by

\[\left( \vec{r} - \vec{a} \right) \cdot \vec{N} = 0\]

\[ \Rightarrow \left[ \left( x \hat{i}+ y \hat{j} + z\hat{k} \right) - \left( 3 \hat{i} + 2\hat{ j}\right) \right] \cdot \left( - 4 \hat{i} + 4 {j} - 4 \hat{k} \right) = 0\]

\[ \Rightarrow \left[ \left( x - 3 \right) {i}+ \left( y - 2 \right) \hat{j} + z \hat{k} \right] \cdot \left( - 4 \hat{i} + 4 \hat{j} - 4 \hat{k} \right) = 0\]

\[ \Rightarrow - 4\left( x - 3 \right) + 4\left( y - 2 \right) - 4z = 0\]

\[ \Rightarrow x - 3 - y + 2 + z = 0\]

\[ \Rightarrow x - y + z = 1\]

Thus, the equation of the required plane is x − y + z = 1.