Direction ratios of the normal to the plane p | vedclass

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(D) Let the points be $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$.Two vectors lying in the plane are $\vec{AB} = (1-0)\hat{i} + (1-1)\hat{j} + (2-1)\

Vedclass · Accepted Answer

$(1, -2, -1)$

Vedclass · Answer

(D) Let the points be $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$.Two vectors lying in the plane are $\vec{AB} = (1-0)\hat{i} + (1-1)\hat{j} + (2-1)\hat{k} = \hat{i} + 0\hat{j} + \hat{k}$ and $\vec{AC} = (-1-0)\hat{i} + (2-1)\hat{j} + (-2-1)\hat{k} = -\hat{i} + \hat{j} - 3\hat{k}$.The normal vector $\vec{n}$ to the plane is given by the cross product $\vec{AB} 	imes \vec{AC}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 0 & 1 \ -1 & 1 & -3 \end{vmatrix} = \hat{i}(0 - 1) - \hat{j}(-3 - (-1)) + \hat{k}(1 - 0) = -\hat{i} + 2\hat{j} + \hat{k}$.The direction ratios are $(-1, 2, 1)$,which is proportional to $(1, -2, -1)$.Thus,the correct option is $(D)$.

Direction ratios of the normal to the plane passing through the points $A(0, 1, 1)$,$B(1, 1, 2)$,and $C(-1, 2, -2)$ are

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