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Question

(NONE) The normal vector $\vec{n_1}$ of the plane $3x - 2y + z = 8$ is $\langle 3, -2, 1 \rangle$.
The normal vector $\vec{n_2}$ of the plane $x + y

Vedclass · Accepted Answer

$\frac{-339}{11}$

Vedclass · Answer

(NONE) The normal vector $\vec{n_1}$ of the plane $3x - 2y + z = 8$ is $\langle 3, -2, 1 angle$.
The normal vector $\vec{n_2}$ of the plane $x + y + z = 6$ is $\langle 1, 1, 1 angle$.
The normal vector $\vec{n}$ of the required plane is perpendicular to both $\vec{n_1}$ and $\vec{n_2}$,so $\vec{n} = \vec{n_1} 	imes \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 1 \ 1 & 1 & 1 \end{vmatrix} = \hat{i}(-2-1) - \hat{j}(3-1) + \hat{k}(3+2) = -3\hat{i} - 2\hat{j} + 5\hat{k}$.
The equation of the plane passing through $(2, -1, 3)$ with normal $\vec{n} = \langle -3, -2, 5 angle$ is $-3(x - 2) - 2(y + 1) + 5(z - 3) = 0$.
Simplifying this,we get $-3x + 6 - 2y - 2 + 5z - 15 = 0$,which is $-3x - 2y + 5z = 11$.
Dividing by $11$,we get $-\frac{3}{11}x - \frac{2}{11}y + \frac{5}{11}z = 1$.
Comparing this with $lx + my + nz = 1$,we have $l = -3/11$,$m = -2/11$,and $n = 5/11$.
Now,calculate $4m + 2n - 31 = 4(-2/11) + 2(5/11) - 31 = -8/11 + 10/11 - 31 = 2/11 - 31 = (2 - 341)/11 = -339/11$.

If the equation of the plane passing through the point $(2, -1, 3)$ and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$,then $4m + 2n - 31 =$

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