Question

If $\overline{a}=\overline{i}+2\overline{j}+3\overline{k},\overline{b}=-\overline{i}+2\overline{j}+\overline{k},\overline{c}=3\overline{i}+\overline{j}$ and $\overline{d}$ is normal to both $\overline{a}$ and $\overline{b}$, then $(\overline{c},\overline{d})=$

• A. ${\cos }^{-1}\left(\frac{4}{\sqrt{30}}\right)$
• B. ${\sin }^{-1}\left(\frac{4}{\sqrt{30}}\right)$
• C. ${\cos }^{-1}\left(\frac{2}{\sqrt{30}}\right)$
• D. ${\sin }^{-1}\left(\frac{2}{\sqrt{30}}\right)$

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{4}{\sqrt{30}}\right)$

$\overrightarrow{d}=k(\overrightarrow{a}\times \overrightarrow{b})$
$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 2& 3\\ -1& 2& 1\end{array}\right|=-4\hat{i}-4\hat{j}+4\hat{k}$
$\overrightarrow{d}=k(-4\hat{i}-4\hat{j}+14\hat{k})$
$\overrightarrow{c}-\overrightarrow{d}=|\overrightarrow{c}||\overrightarrow{d}|cos\theta$
$k(-16)=k4\sqrt{3}\sqrt{10}cos\theta$
$co{s}^{-1}\left(\frac{-4}{\sqrt{30}}\right)=\theta$