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(D) The plane is perpendicular to the line of intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$. The direction vector of this line

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(D) The plane is perpendicular to the line of intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$. The direction vector of this line is given by the cross product of the normals of the two planes: $\vec{n}_1 = (1, -3, 2)$ and $\vec{n}_2 = (4, -1, 1)$.$\vec{n} = \vec{n}_1 	imes \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -3 & 2 \ 4 & -1 & 1 \end{vmatrix} = \hat{i}(-3 + 2) - \hat{j}(1 - 8) + \hat{k}(-1 + 12) = -\hat{i} + 7\hat{j} + 11\hat{k}$.Thus,the normal vector to the required plane is $\vec{n} = (-1, 7, 11)$.The equation of the plane passing through $(1, 1, 2)$ with normal vector $(-1, 7, 11)$ is:$-1(x - 1) + 7(y - 1) + 11(z - 2) = 0$$-x + 1 + 7y - 7 + 11z - 22 = 0$$-x + 7y + 11z = 28$.Dividing by $28$ to get the form $Ax + By + Cz = 1$:$-\frac{1}{28}x + \frac{7}{28}y + \frac{11}{28}z = 1$.Comparing with $Ax + By + Cz = 1$,we have $A = -\frac{1}{28}$,$B = \frac{7}{28}$,and $C = \frac{11}{28}$.Now,calculate $140(C - B + A)$:$140 \left( \frac{11}{28} - \frac{7}{28} - \frac{1}{28} ight) = 140 \left( \frac{3}{28} ight) = 5 	imes 3 = 15$.

If the equation of the plane passing through the point $(1, 1, 2)$ and perpendicular to the intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$ is $Ax + By + Cz = 1$,then $140(C - B + A)$ is equal to $.........$.

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