The direction ratios of the two lines AB and | vedclass

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Question

(A) Let the direction ratios of the two lines $AB$ and $AC$ be $\vec{v_1} = \langle 1, -1, -1 \rangle$ and $\vec{v_2} = \langle 2, -1, 1 \rangle$ resp

Vedclass · Accepted Answer

$2, 3, -1$

Vedclass · Answer

(A) Let the direction ratios of the two lines $AB$ and $AC$ be $\vec{v_1} = \langle 1, -1, -1 angle$ and $\vec{v_2} = \langle 2, -1, 1 angle$ respectively.Since the lines $AB$ and $AC$ lie in the plane $ABC$,the normal vector $\vec{n}$ to the plane is given by the cross product of the vectors along these lines:$\vec{n} = \vec{v_1} 	imes \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -1 \ 2 & -1 & 1 \end{vmatrix}$$\vec{n} = \hat{i}(-1 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - (-2))$$\vec{n} = \hat{i}(-2) - \hat{j}(3) + \hat{k}(1)$$\vec{n} = -2\hat{i} - 3\hat{j} + \hat{k}$Thus,the direction ratios of the normal are $\langle -2, -3, 1 angle$. Multiplying by $-1$,we get $\langle 2, 3, -1 angle$.Therefore,the correct option is $A$.

The direction ratios of the two lines $AB$ and $AC$ are $1, -1, -1$ and $2, -1, 1$. The direction ratios of the normal to the plane $ABC$ are

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