If vec{a}=hat{i}-hat{j}+hat{k} and vec{b}=hat | vedclass

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(C) Given $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-3\hat{k}$.First,calculate $\vec{p}=\vec{a}-\vec{b} = (1-1)\hat{i} + (-1-2)\h

Vedclass · Accepted Answer

$\frac{1}{\sqrt{26}}\hat{i}+\frac{4}{\sqrt{26}}\hat{j}+\frac{3}{\sqrt{26}}\hat{k}$

Vedclass · Answer

(C) Given $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-3\hat{k}$.First,calculate $\vec{p}=\vec{a}-\vec{b} = (1-1)\hat{i} + (-1-2)\hat{j} + (1-(-3))\hat{k} = 0\hat{i}-3\hat{j}+4\hat{k}$.Next,calculate $\vec{q}=\vec{a}+\vec{b} = (1+1)\hat{i} + (-1+2)\hat{j} + (1-3)\hat{k} = 2\hat{i}+1\hat{j}-2\hat{k}$.The vector perpendicular to both $\vec{p}$ and $\vec{q}$ is given by $\vec{n} = \vec{p} 	imes \vec{q}$.$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & -3 & 4 \ 2 & 1 & -2 \end{vmatrix} = \hat{i}(6-4) - \hat{j}(0-8) + \hat{k}(0-(-6)) = 2\hat{i}+8\hat{j}+6\hat{k}$.The unit vector is $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{\sqrt{2^2+8^2+6^2}} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{\sqrt{4+64+36}} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{\sqrt{104}} = \frac{2\hat{i}+8\hat{j}+6\hat{k}}{2\sqrt{26}} = \frac{1}{\sqrt{26}}\hat{i}+\frac{4}{\sqrt{26}}\hat{j}+\frac{3}{\sqrt{26}}\hat{k}$.Thus,the correct option is $C$.

If $\vec{a}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-3\hat{k}$,then the unit vector perpendicular to both $\vec{p}=\vec{a}-\vec{b}$ and $\vec{q}=\vec{a}+\vec{b}$ is . . . . . . .

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