Question

Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained

Answer 1

Let   \[\vec{a} =\hat{ i} -\hat{ j} + 2 \hat{k}\]

Also, let

\[\vec{N_1} = 2\hat{ i}+ 3 \hat{j} - 2 \hat{k } \text { and } \vec{N_2} = \hat{i } + 2 \hat{j} - 3 \hat{k}\]

Now, the normal vector of the required plane,

\[\vec{N} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & - 2 \\ 1 & 2 & - 3\end{vmatrix} = - 5 \hat{i} + 4 \hat{j} + \hat{k}\]

So, the equation of the required plane is

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

\[ \Rightarrow \left( \vec{r} - \left(\hat{ i } - \hat{j} + 2 \hat{k} \right) \right) . \left( - 5 \hat{i} + 4 \hat{j} + \hat{k} \right) = 0\]

\[ \Rightarrow \vec{r} . \left( - 5\hat{ i } + 4 \hat{j} + \hat{k} \right) + 7 = 0 \text { is the vector equation}\]

\[ \text { and }\] \[ 5x - 4y - z = 7\text {  is the Cartesian equation} .\]

Also, the distance of point P (-2,5,5) from the plane obtained =\[\left| \frac{5( - 2) - 4(5) - (5) - 7}{\sqrt{25 + 16 + 1}} \right| = \left| \frac{- 42}{\sqrt{42}} \right| = \sqrt{42} \text { units }\]