The direction ratios of the normal to the pla | vedclass

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Question

(B) Let the points be $A(1, 2, -3)$ and $B(-1, -2, 1)$. The vector $\vec{AB}$ lies in the plane, where $\vec{AB} = (-1-1, -2-2, 1-(-3)) = (-2, -4, 4)$

Vedclass · Accepted Answer

$(14, -8, -1)$

Vedclass · Answer

(B) Let the points be $A(1, 2, -3)$ and $B(-1, -2, 1)$. The vector $\vec{AB}$ lies in the plane, where $\vec{AB} = (-1-1, -2-2, 1-(-3)) = (-2, -4, 4)$.The plane is parallel to the line with direction ratios $(2, 3, 4)$. Thus, the vector $\vec{v} = (2, 3, 4)$ is parallel to the plane.The normal vector $\vec{n} = (a, b, c)$ to the plane is perpendicular to both $\vec{AB}$ and $\vec{v}$.Therefore, $\vec{n} = \vec{AB} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & -4 & 4 \ 2 & 3 & 4 \end{vmatrix}$.Calculating the cross product: $\vec{n} = \hat{i}(-16 - 12) - \hat{j}(-8 - 8) + \hat{k}(-6 - (-8)) = -28\hat{i} + 16\hat{j} + 2\hat{k}$.The direction ratios are proportional to $(-28, 16, 2)$, which simplifies to $(14, -8, -1)$ by dividing by $-2$.

The direction ratios of the normal to the plane passing through the points $(1, 2, -3)$, $(-1, -2, 1)$ and parallel to $\frac{x-2}{2}=\frac{y+1}{3}=\frac{z}{4}$ are:

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