Question

(i) 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.$

(ii) Let $\overrightarrow{a}=4\hat{i}+5\hat{j}-\hat{k},\overrightarrow{b}=\hat{i}-4\hat{j}+5\hat{k}\mathrm{and}\overrightarrow{c}=3\hat{i}+\hat{j}-\hat{k}$. Find a vector $\overrightarrow{d}$ which is perpendicular to both $\overrightarrow{c}\mathrm{and}\overrightarrow{b}\mathrm{and}\overrightarrow{d}.\overrightarrow{a}=21$.

Answer 1

(i)
$$\begin{gathered} \mathrm{Given}: \\ \overrightarrow{a}=\hat{i}+4\hat{j}+2\hat{k} \\ \overrightarrow{b}=3\hat{i}-2\hat{j}+7\hat{k} \\ \overrightarrow{c}=2\hat{i}-\hat{j}+4\hat{k} \\ \mathrm{Since}d\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{both}a\mathrm{and}b,\mathrm{it}\mathrm{is}\mathrm{parallel}\mathrm{to}\overrightarrow{a}\times \overrightarrow{b}. \\ \mathrm{Suppose}d=\lambda \left(\overrightarrow{a}\times \overrightarrow{b}\right)\mathrm{for}\mathrm{some}\mathrm{scalar}\lambda . \\ d=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 4& 2\\ 3& -2& 7\end{array}\right| \\ =\lambda \left[\left(28+4\right)\hat{i}-\left(7-6\right)\hat{j}+\left(-2-12\right)\hat{k}\right] \\ =\lambda \left[32\hat{i}-\hat{j}-14\hat{k}\right] \\ \overrightarrow{c.}\overrightarrow{d}=15\left(\mathrm{Given}\right) \\ \Rightarrow \left(2\hat{i}-\hat{j}+4\hat{k}\right).\lambda \left(32\hat{i}-\hat{j}-14\hat{k}\right)=15 \\ \Rightarrow \lambda \left(64+1-56\right)=15 \\ \Rightarrow \lambda =\frac{5}{3} \\ \therefore \overrightarrow{d}=\frac{5}{3}\left(32\hat{i}-\hat{j}-14\hat{k}\right) \\ \Rightarrow \overrightarrow{d}=\frac{1}{3}\left(160\hat{i}-5\hat{j}-70\hat{k}\right) \end{gathered}$$

Disclaimer: The question should contain
$$\begin{gathered} \text{"which is perpendicular to both}\overrightarrow{a}\text{and}\overrightarrow{b}\text{"} \\ \text{instead of} \\ \text{"which is perpendicular to both}\overrightarrow{a}\text{and}\overrightarrow{d}\text{"} \end{gathered}$$

(ii)

$$\begin{gathered} \mathrm{Let}\overrightarrow{d}=x\hat{i}+y\hat{j}+z\hat{k} \\ \mathrm{Since}\overrightarrow{d}\mathrm{is}\mathrm{perpendicular}\mathrm{to}\mathrm{both}\overrightarrow{c}\mathrm{and}\overrightarrow{b},\mathrm{so} \\ \overrightarrow{d.}\overrightarrow{c}=0\mathrm{and}\overrightarrow{d.}\overrightarrow{b}=0 \\ 3x+y-z=0.....\left(1\right) \\ x-4y+5z=0.....\left(2\right) \\ \overrightarrow{d.}\overrightarrow{a}=21 \\ 4x+5y-z=21.....\left(3\right) \\ \mathrm{Solving}\left(1\right),\left(2\right)\mathrm{and}\left(3\right) \\ x=\left(-\frac{1}{3}\right),y=\frac{16}{3},z=\frac{13}{3} \\ \overrightarrow{d}=\frac{1}{3}\left(-\hat{i}+16\hat{j}+13\hat{k}\right) \\ \overrightarrow{d}=-\frac{1}{3}\left(\hat{i}-16\hat{j}-13\hat{k}\right) \end{gathered}$$