Question

Let $\overrightarrow{a}=\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k}),\overrightarrow{b}=\frac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k}),\overrightarrow{c}=c_1\hat{i}+c_2\hat{j}+c-3\hat{k}$ and matrix $A=\left[\begin{array}{ccc}\frac{2}{7}& \frac{3}{7}& \frac{6}{7}\\ \frac{6}{7}& \frac{2}{7}& -\frac{3}{7}\\ c_1& c_2& c_3\end{array}\right]$
If $A{A}^T=I,then\overrightarrow{c}$ is equal to

• A. $\frac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})$
• B. $-\frac{1}{7}(3\hat{i}+6\hat{j}+2\hat{k})$
• C. $\frac{1}{7}(-3\hat{i}+6\hat{j}-2\hat{k})$
• D. $-\frac{1}{7}(-3\hat{i}+6\hat{j}-2\hat{k})$

Answer 1

Answer: (C), (D)

The correct options are
C

$\frac{1}{7}(-3\hat{i}+6\hat{j}-2\hat{k})$

D

$-\frac{1}{7}(-3\hat{i}+6\hat{j}-2\hat{k})$

$\because A{A}^T=I\Rightarrow \overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are orthogonal unit vectors
$\therefore \overrightarrow{C}=\pm (\overrightarrow{a}\times \overrightarrow{b})=\pm \frac{1}{49}\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 3& 6\\ 6& 2& -3\end{array}\right|\Rightarrow \overrightarrow{C}=\pm \frac{(-3\hat{i}+6\hat{j}-2\hat{k})}{7}\because |\overrightarrow{a}\times \overrightarrow{b}|=1$