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(C) The line passes through the point $P(1, -1, 3)$ and has direction ratios $\vec{v} = (2, -1, 4)$.The equation of a plane passing through $(1, -1, 3

Vedclass · Accepted Answer

$a = 9, b = -2, c = -5$

Vedclass · Answer

(C) The line passes through the point $P(1, -1, 3)$ and has direction ratios $\vec{v} = (2, -1, 4)$.The equation of a plane passing through $(1, -1, 3)$ is $a(x-1) + b(y+1) + c(z-3) = 0$.Since this plane contains the line,its normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the line's direction vector $\vec{v} = (2, -1, 4)$. Thus,$2a - b + 4c = 0$.Since the plane is perpendicular to the plane $x+2y+z=12$,the normal vector $\vec{n} = (a, b, c)$ must be perpendicular to the normal of the given plane $\vec{n_1} = (1, 2, 1)$. Thus,$a + 2b + c = 0$.Solving the system of equations:$2a - b + 4c = 0$$a + 2b + c = 0$Using cross product of $(2, -1, 4)$ and $(1, 2, 1)$ to find the normal vector $\vec{n}$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & 4 \ 1 & 2 & 1 \end{vmatrix} = \hat{i}(-1-8) - \hat{j}(2-4) + \hat{k}(4+1) = -9\hat{i} + 2\hat{j} + 5\hat{k}$.Thus,the direction ratios are proportional to $(-9, 2, 5)$.The equation of the plane is $-9(x-1) + 2(y+1) + 5(z-3) = 0$.$-9x + 9 + 2y + 2 + 5z - 15 = 0 \Rightarrow -9x + 2y + 5z - 4 = 0$.Multiplying by $-1$,we get $9x - 2y - 5z + 4 = 0$.Comparing with $ax+by+cz+4=0$,we get $a=9, b=-2, c=-5$.

The equation of the plane passing through the line $\frac{x-1}{2} = \frac{y+1}{-1} = \frac{z-3}{4}$ and perpendicular to the plane $x+2y+z=12$ is given by $ax+by+cz+4=0$. Then:

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