Question

Let $\overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k},\overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k}\mathrm{and}\overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k}.$ Find a vector $\overrightarrow{d}$ which is perpendicular to both $\overrightarrow{a}\mathrm{and}\overrightarrow{b}\mathrm{and}\overrightarrow{c}·\overrightarrow{d}=15.$

Answer 1

$$\begin{gathered} \mathrm{Given}: \\ \overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k} \\ \overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k} \\ \overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k} \\ \mathrm{Since}d\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{both}a\mathrm{and}b,\mathrm{it}\mathrm{is}\mathrm{parallel}\mathrm{to}\overrightarrow{a}\times \overrightarrow{b}. \\ \mathrm{Suppose}d=\lambda \left(\overrightarrow{a}\times \overrightarrow{b}\right)\mathrm{for}\mathrm{some}\mathrm{scalar}\lambda . \\ d=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 4& 2\\ 3& -2& 7\end{array}\right| \\ =\lambda \left[\left(28+4\right)\hat{i}-\left(7-6\right)\hat{j}+\left(-2-12\right)\hat{k}\right] \\ =\lambda \left[32\hat{i}-\hat{j}-14\hat{k}\right] \\ \overrightarrow{c.}\overrightarrow{d}=15\left(\mathrm{Given}\right) \\ \Rightarrow \left(2\hat{i}-\hat{j}+4\hat{k}\right).\lambda \left(32\hat{i}-\hat{j}-14\hat{k}\right)=15 \\ \Rightarrow \lambda \left(64+1-56\right)=15 \\ \Rightarrow \lambda =\frac{5}{3} \\ \therefore \overrightarrow{d}=\frac{5}{3}\left(32\hat{i}-\hat{j}-14\hat{k}\right) \end{gathered}$$