The plane,passing through the points (0,-1,2) | vedclass

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(C) Let the two points be $A(0,-1,2)$ and $B(-1,2,1)$. The vector $\vec{AB} = (-1-0)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \

Vedclass · Accepted Answer

$(-2,5,0)$

Vedclass · Answer

(C) Let the two points be $A(0,-1,2)$ and $B(-1,2,1)$. The vector $\vec{AB} = (-1-0)\hat{i} + (2-(-1))\hat{j} + (1-2)\hat{k} = -\hat{i} + 3\hat{j} - \hat{k}$.The line is parallel to the vector $\vec{v}$ passing through $P(5,1,-7)$ and $Q(1,-1,-1)$,so $\vec{v} = (1-5)\hat{i} + (-1-1)\hat{j} + (-1-(-7))\hat{k} = -4\hat{i} - 2\hat{j} + 6\hat{k}$.The normal vector $\vec{n}$ to the plane is $\vec{AB} 	imes \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -1 & 3 & -1 \ -4 & -2 & 6 \end{vmatrix} = \hat{i}(18-2) - \hat{j}(-6-4) + \hat{k}(2+12) = 16\hat{i} + 10\hat{j} + 14\hat{k}$.Dividing by $2$,we get the normal vector $\vec{n}' = 8\hat{i} + 5\hat{j} + 7\hat{k}$.The equation of the plane is $8x + 5y + 7z = d$. Substituting point $(0,-1,2)$,we get $8(0) + 5(-1) + 7(2) = -5 + 14 = 9$. So,$d = 9$.The equation of the plane is $8x + 5y + 7z = 9$.Checking the options: For $(0,5,-2)$,$8(0) + 5(5) + 7(-2) = 25 - 14 = 11 
eq 9$. For $(-2,5,0)$,$8(-2) + 5(5) + 7(0) = -16 + 25 = 9$. Thus,the plane passes through $(-2,5,0)$.

The plane,passing through the points $(0,-1,2)$ and $(-1,2,1)$ and parallel to the line passing through $(5,1,-7)$ and $(1,-1,-1)$,also passes through the point.

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