The point which lies on the plane passing thr | vedclass

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(C) Let the points be $A(1, -2, -3)$, $B(3, -1, 4)$, and $C(-3, 2, -5)$.First, we find two vectors in the plane: $\vec{AB} = (3-1)\hat{i} + (-1+2)\hat

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$-\hat{i}+9\hat{j}+14\hat{k}$

Vedclass · Answer

(C) Let the points be $A(1, -2, -3)$, $B(3, -1, 4)$, and $C(-3, 2, -5)$.First, we find two vectors in the plane: $\vec{AB} = (3-1)\hat{i} + (-1+2)\hat{j} + (4+3)\hat{k} = 2\hat{i} + \hat{j} + 7\hat{k}$ and $\vec{AC} = (-3-1)\hat{i} + (2+2)\hat{j} + (-5+3)\hat{k} = -4\hat{i} + 4\hat{j} - 2\hat{k}$.The normal vector $\vec{n} = \vec{AB} 	imes \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 7 \ -4 & 4 & -2 \end{vmatrix} = \hat{i}(-2-28) - \hat{j}(-4+28) + \hat{k}(8+4) = -30\hat{i} - 24\hat{j} + 12\hat{k}$.Dividing by $-6$, we get the normal vector $\vec{n}' = 5\hat{i} + 4\hat{j} - 2\hat{k}$.The equation of the plane is $5(x-1) + 4(y+2) - 2(z+3) = 0$, which simplifies to $5x + 4y - 2z - 5 + 8 - 6 = 0$, or $5x + 4y - 2z - 3 = 0$.Checking option $C$: $5(-1) + 4(9) - 2(14) - 3 = -5 + 36 - 28 - 3 = 0$. Thus, the point $-\hat{i} + 9\hat{j} + 14\hat{k}$ lies on the plane.

The point which lies on the plane passing through the points $A(\hat{i}-2\hat{j}-3\hat{k})$, $B(3\hat{i}-\hat{j}+4\hat{k})$, and $C(-3\hat{i}+2\hat{j}-5\hat{k})$ is:

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