The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$, where $\bar{a}=\hat{i}+\hat{j}+\hat{k}$ and $\overline{\mathrm{b}}=3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ is

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Aicharya · Accepted Answer

$\begin{aligned} \bar{a}+\bar{b} & =(\hat{i}+\hat{j}+\hat{k})+(3 \hat{i}-2 \hat{j}+5 \hat{k}) \ & =4 \hat{i}-\hat{j}+6 \hat{k} \ \bar{a}-\bar{b} & =(\hat{i}+\hat{j}+\hat{k})-(3 \hat{i}-2 \hat{j}+5 \hat{k}) \ & =-2 \hat{i}+3 \hat{j}-4 \hat{k}\end{aligned}$ $	herefore \quad$ Vector perpendicular to $(\bar{a}+\bar{b})$ and $(\bar{a}-\bar{b})$ is $\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \ 4 & -1 & 6 \ -2 & 3 & -4 \end{array}ight|=-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}$ $	herefore \quad$ Required unit vector is $\frac{-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{(-14)^2+4^2+(10)^2}}=\frac{-14 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}}{\sqrt{312}}$

The unit vector perpendicular to each of the vectors $\bar{a}+\bar{b}$ and $\bar{a}-\bar{b}$, where…

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