Question

The vector equation of plane containing the point (1, -1, 2) and perpendicular to planes 2x + 3y - 2z = 5 and x + 2y - 3z = 8 is

• A. $\overrightarrow{r}.(-5i + 4j - k) = -7$
• B. $\overrightarrow{r}.(-5i + 4j + k) = 7$
• C. $\overrightarrow{r}.(-5i + 4j + k) = -7$
• D. $\overrightarrow{r}.(-5i + 4j + k) = 7$

Answer 1

Answer: (C)

$\overrightarrow{r}.(-5i + 4j + k) = -7$

Answer 2

The normal vectors of the given planes are $\mathbf{n}_1 = (2, 3, -2)$ and $\mathbf{n}_2 = (1, 2, -3)$. The normal vector to the required plane is the cross product of these two vectors:

$\mathbf{n} = \mathbf{n}_1 \times \mathbf{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -2 \\ 1 & 2 & -3 \end{vmatrix} = (-9 + 4)\mathbf{i} - (-6 + 2)\mathbf{j} + (4 - 3)\mathbf{k} = -5\mathbf{i} + 4\mathbf{j} + \mathbf{k}$.

So, the normal vector is $(-5, 4, 1)$. The equation of the plane is given by $\overrightarrow{r} \cdot (-5\mathbf{i} + 4\mathbf{j} + \mathbf{k}) = d$. Since the plane contains the point (1, -1, 2), we have:

$d = -5(1) + 4(-1) + 1(2) = -5 - 4 + 2 = -7$.

Thus, the equation of the plane is $\overrightarrow{r} \cdot (-5\mathbf{i} + 4\mathbf{j} + \mathbf{k}) = -7$.