Question

Find all vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+4\hat{k}$. [NCERT EXEMPLAR]

Answer 1

Let $\overrightarrow{a}=\hat{i}+2\hat{j}+\hat{k}$ and $\overrightarrow{b}=-\hat{i}+3\hat{j}+4\hat{k}$.

Unit vectors perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ = $\pm \frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}$

Now,
$$\begin{gathered} \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 1\\ -1& 3& 4\end{array}\right|=5\hat{i}-5\hat{j}+5\hat{k} \\ \therefore \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\left|5\hat{i}-5\hat{j}+5\hat{k}\right|=\sqrt{5^2+{\left(-5\right)}^2+5^2}=\sqrt{75}=5\sqrt{3} \end{gathered}$$

Unit vectors perpendicular to both $\overrightarrow{a}$ and $\overrightarrow{b}$ = $\pm \frac{5\hat{i}-5\hat{j}+5\hat{k}}{5\sqrt{3}}=\pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}$

∴ Required vectors = $10\sqrt{3}\left(\pm \frac{\hat{i}-\hat{j}+\hat{k}}{\sqrt{3}}\right)=\pm 10\left(\hat{i}-\hat{j}+\hat{k}\right)$

Thus, the vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane of $\hat{i}+2\hat{j}+\hat{k}$ and $-\hat{i}+3\hat{j}+4\hat{k}$ are $\pm 10\left(\hat{i}-\hat{j}+\hat{k}\right)$.