a=3 hat{i}+hat{j}-hat{k}, b=hat{i}-4 hat{j}+5 | vedclass

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(A) Given vectors $a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}$ and $c=4 \hat{i}+5 \hat{j}-\hat{k}$.Since $r$ is perpendicular to both

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$3(\hat{i}-\hat{j}-\hat{k})$

Vedclass · Answer

(A) Given vectors $a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}$ and $c=4 \hat{i}+5 \hat{j}-\hat{k}$.Since $r$ is perpendicular to both $b$ and $c$,$r$ must be parallel to $b 	imes c$.$b 	imes c = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -4 & 5 \ 4 & 5 & -1 \end{vmatrix} = \hat{i}(4-25) - \hat{j}(-1-20) + \hat{k}(5+16) = -21 \hat{i} + 21 \hat{j} + 21 \hat{k} = 21(-\hat{i} + \hat{j} + \hat{k})$.Let $r = \lambda(-\hat{i} + \hat{j} + \hat{k})$ for some scalar $\lambda$.Given $r \cdot a = 9$,we have $\lambda(-\hat{i} + \hat{j} + \hat{k}) \cdot (3 \hat{i} + \hat{j} - \hat{k}) = 9$.$\lambda(-3 + 1 - 1) = 9 \Rightarrow -3\lambda = 9 \Rightarrow \lambda = -3$.Thus,$r = -3(-\hat{i} + \hat{j} + \hat{k}) = 3(\hat{i} - \hat{j} - \hat{k})$.Therefore,option $A$ is correct.

$a=3 \hat{i}+\hat{j}-\hat{k}, b=\hat{i}-4 \hat{j}+5 \hat{k}, c=4 \hat{i}+5 \hat{j}-\hat{k}$ are three vectors and a vector $r$ is perpendicular to both the vectors $b$ and $c$. If $r \cdot a=9$,then $r=$

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