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(B) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$ a

Vedclass · Accepted Answer

$4$

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(B) The normal vector $\vec{n}$ of the required plane is perpendicular to the normals of the given planes $\vec{n}_1 = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{n}_2 = \hat{i} - \hat{j} - \hat{k}$.Thus,$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & -1 \ 1 & -1 & -1 \end{vmatrix} = \hat{i}(-1 - 1) - \hat{j}(-2 + 1) + \hat{k}(-2 - 1) = -2\hat{i} + \hat{j} - 3\hat{k}$.The equation of the plane passing through $(-1, 0, -2)$ with normal vector $\vec{n} = -2\hat{i} + \hat{j} - 3\hat{k}$ is:$-2(x + 1) + 1(y - 0) - 3(z + 2) = 0$$-2x - 2 + y - 3z - 6 = 0$$-2x + y - 3z - 8 = 0$Multiplying by $-1$,we get $2x - y + 3z + 8 = 0$.Comparing this with $ax + by + cz + 8 = 0$,we have $a = 2$,$b = -1$,and $c = 3$.Therefore,$a + b + c = 2 - 1 + 3 = 4$.

Let the plane passing through the point $(-1, 0, -2)$ and perpendicular to each of the planes $2x + y - z = 2$ and $x - y - z = 3$ be $ax + by + cz + 8 = 0$. Then the value of $a + b + c$ is equal to:

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