Practicing Success
If $a_m \hat{i}+b_m \hat{j}+c_m \hat{k}, m=1,2,3$, are pairwise perpendicular unit vectors, then $\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|$ is equal to |
0 1 or –1 3 or -3 4 or –4 |
1 or –1 |
$\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|^2=\left|\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right|=1 \Rightarrow\left|\begin{array}{lll}a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{array}\right|= \pm 1$ Hence (2) is correct answer. |