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Question

Find perpendicular unit vector of vectors
$$\hat i-2\hat j+\hat k$$
and $$2\hat i+\hat j-3\hat k$$.


Solution

Let $$\vec{a}=\hat{i}-2\hat{j}+\hat{k}$$
and $$\vec{b}=2\hat{i}+\hat{j}-3\hat{k}$$
Then perpendicular unit vector of $$\vec{a}$$ and $$\vec{b}$$,
$$\hat{n}=\dfrac{\vec{a}\times \vec{b}}{|\vec{a}\times \vec{b}|}$$
Now $$\vec{a}\times \vec{b}=\begin{vmatrix} \hat { i }  & \hat { j }  & \hat { k }  \\ 1 & -2 & 1 \\ 2 & 1 & -3 \end{vmatrix}$$
$$\Rightarrow \vec{a}\times \vec{b}=\hat{i}(6-1)-\hat{j}(-3-2)+\hat{k}(1+4)$$
$$\Rightarrow \vec{a}\times \vec{b}=5\hat{i}+5\hat{j}+5\hat{k}$$
and $$|\vec{a}\times \vec{b}|=\sqrt{(5)^{2}+(5)^{2}+(5)^{2}}$$
$$\Rightarrow |\vec{a}\times \vec{b}|=\sqrt{25+25+25}$$
$$\Rightarrow |\vec{a}\times \vec{b}|=\sqrt{75}$$
$$\Rightarrow |\vec{a}\times \vec{b}|=5\sqrt{3}$$
$$\therefore$$ Perpendicular unit vector between $$\vec{a}$$ and $$\vec{b}$$,
$$\hat{n}=\dfrac{5\hat{i}+5\hat{j}+5\hat{k}}{5\sqrt{3}}$$
$$=\dfrac{5(\hat{i}+\hat{j}+\hat{k})}{5\sqrt{3}}$$
$$=\dfrac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$$
Hence the perpendicular unit vector between $$\hat{i}-2\hat{j}+\hat{k}$$ and $$2\hat{i}+\hat{j}-3\hat{k}$$ is $$\dfrac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{3}}$$

Physics

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