Question

Let \(\overset{̄}{a} = (a_1\hat{i}+a_2\hat{j}+a_3\hat{k}), \overset{̄}{b} = (b_1\hat{i}+b_2\hat{j}+b_3\hat{k}), \overset{̄}{c} = (c_1\hat{i}+c_2\hat{j}+c_3\hat{k})\) be three non-zero vectors such that \(\overset{̄}{a}\) is a unit vector perpendicular to both \(\overset{̄}{b}\) and \(\overset{̄}{c}\). If the angle between \(\overset{̄}{b}\) and \(\overset{̄}{c}\) is \(\frac{π}{3}\) then \(\begin{array}{ccc}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}^2 =\)

• A. \(\frac{3}{4}|\overset{̄}{b}|^2|\overset{̄}{c}|^2\)
• B. \(1\)
• C. \(0\)
• D. \(\frac{1}{4}|\overset{̄}{b}|^2|\overset{̄}{c}|^2\)

Hint:

The determinant is the scalar triple product \(\vec a\cdot(\vec b\times\vec c)\).

Answer 1

The Correct Option is A

Step 1: Use the Gram determinant
$\det^2=\begin{vmatrix}\vec a\cdot\vec a&\vec a\cdot\vec b&\vec a\cdot\vec c\\ \vec b\cdot\vec a&\vec b\cdot\vec b&\vec b\cdot\vec c\\ \vec c\cdot\vec a&\vec c\cdot\vec b&\vec c\cdot\vec c\end{vmatrix}$.

Step 2: Fill it
With $\vec a\cdot\vec b=\vec a\cdot\vec c=0$ and $\vec a\cdot\vec a=1$, this is $|\vec b|^2|\vec c|^2-(\vec b\cdot\vec c)^2=|\vec b|^2|\vec c|^2(1-\tfrac14)=\tfrac34|\vec b|^2|\vec c|^2$, option (A).

Final Answer:
The square equals $\frac34|\vec b|^2|\vec c|^2$, option (A). \[ \boxed{\dfrac34|\vec b|^2|\vec c|^2} \]