The direction ratios of the normal to the pla | vedclass

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Question

(A) Let the points be $A(0,0,1)$,$B(0,1,2)$,and $C(1,0,3)$.Two vectors lying on the plane are $\vec{AB} = (0-0)\hat{i} + (1-0)\hat{j} + (2-1)\hat{k} =

Vedclass · Accepted Answer

$(2,1,-1)$

Vedclass · Answer

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

The direction ratios of the normal to the plane passing through $(0,0,1)$,$(0,1,2)$,and $(1,0,3)$ are

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