Question

(i) Find a vector of magnitude 49, which is perpendicular to both the vectors $2\hat{i}+3\hat{j}+6\hat{k}\mathrm{and}3\hat{i}-6\hat{j}+2\hat{k}.$

(ii) Find a vector whose length is 3 and which is perpendicular to the vector $\overrightarrow{a}=3\hat{i}+\hat{j}-4\hat{k}\mathrm{and}\overrightarrow{b}=6\hat{i}+5\hat{j}-2\hat{k}.$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right)\mathrm{Given}: \\ \overrightarrow{a}=2\hat{i}+3\hat{j}+6\hat{k} \\ \overrightarrow{b}=3\hat{i}-6\hat{j}+2\hat{k} \\ \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 6\\ 3& -6& 2\end{array}\right| \\ =\left(6+36\right)\hat{i}-\left(4-18\right)\hat{j}+\left(-12-9\right)\hat{k} \\ =42\hat{i}+14\hat{j}-21\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{{42}^2+{14}^2+\left(-{21}^2\right)} \\ =\sqrt{2401} \\ =49 \\ \mathrm{Required}\mathrm{vector}=49\times \left\{\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}\right\} \\ =49\times \frac{42\hat{i}+14\hat{j}-21\hat{k}}{49} \\ =42\hat{i}+14\hat{j}-21\hat{k} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)\mathrm{Given}: \\ \overrightarrow{a}=3\hat{i}+\hat{j}-4\hat{k} \\ \overrightarrow{b}=6\hat{i}+5\hat{j}-2\hat{k} \\ \therefore \overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -4\\ 6& 5& -2\end{array}\right| \\ =\left(-2+20\right)\hat{i}-\left(-6+24\right)\hat{j}+\left(15-6\right)\hat{k} \\ =18\hat{i}-18\hat{j}+9\hat{k} \\ \Rightarrow \left|\overrightarrow{a}\times \overrightarrow{b}\right|=\sqrt{{18}^2+\left(-{18}^2\right)+9^2} \\ =\sqrt{729} \\ =27 \\ \mathrm{Required}\mathrm{vector}=3\times \left\{\frac{\overrightarrow{a}\times \overrightarrow{b}}{\left|\overrightarrow{a}\times \overrightarrow{b}\right|}\right\} \\ =3\times \frac{18\hat{i}-18\hat{j}+9\hat{k}}{27} \\ =\frac{3\left(2\hat{i}-2\hat{j}+\hat{k}\right)}{3} \\ =2\hat{i}-2\hat{j}+\hat{k} \end{gathered}$$