The direction ratios of the two lines AB and | vedclass

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Question

(A) Let the direction ratios of the lines $AB$ and $AC$ be $\vec{u} = \langle 1, -1, -1 \rangle$ and $\vec{v} = \langle 2, -1, 1 \rangle$ respectively

Vedclass · Accepted Answer

$2, 3, -1$

Vedclass · Answer

(A) Let the direction ratios of the lines $AB$ and $AC$ be $\vec{u} = \langle 1, -1, -1 angle$ and $\vec{v} = \langle 2, -1, 1 angle$ respectively. Since the lines $AB$ and $AC$ lie in the plane $ABC$,their cross product $\vec{n} = \vec{u} 	imes \vec{v}$ will give the direction ratios of the normal to the plane. $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & -1 \ 2 & -1 & 1 \end{vmatrix}$ $= \hat{i}(-1 - 1) - \hat{j}(1 - (-2)) + \hat{k}(-1 - (-2))$ $= \hat{i}(-2) - \hat{j}(3) + \hat{k}(1)$ $= -2\hat{i} - 3\hat{j} + \hat{k}$ Thus,the direction ratios are $\langle -2, -3, 1 angle$. Multiplying by $-1$,we get the equivalent direction ratios $\langle 2, 3, -1 angle$. 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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