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: Understanding the Concept
The determinant of the three rows equals \(\vec a\cdot(\vec b\times\vec c)\).

Step 2: Key Formula or Approach
\(\vec a\) is perpendicular to both \(\vec b\) and \(\vec c\), so it is parallel to \(\vec b\times\vec c\).

Step 3: Detailed Explanation
\(|\vec b\times\vec c|=|\vec b||\vec c|\sin\dfrac\pi3=\dfrac{\sqrt3}{2}|\vec b||\vec c|\).
\(\vec a\) is a unit vector along \(\vec b\times\vec c\), so \(\vec a\cdot(\vec b\times\vec c)=\pm\dfrac{\sqrt3}{2}|\vec b||\vec c|\).
Squaring: \[ \text{det}^2=\frac34|\vec b|^2|\vec c|^2 \]

Final Answer:
The square of the determinant is \(\frac34|\vec b|^2|\vec c|^2\), option (A). \[ \boxed{\dfrac34|\vec b|^2|\vec c|^2\ \text{(A)}} \]