Vector Algebra
Cross Product
Grade None
Question:
<p>Let \(\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}\), \(\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}\), \(\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}\) be three non-zero vectors. If \(\vec{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\), and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\pi/6\), then</p>
<p>\[\begin{vmatrix}a_1&a_2&a_3\\b_1&b_2&b_3\\c_1&c_2&c_3\end{vmatrix}^2 =\]</p>
<li>\(\dfrac{1}{4}|\vec{a}|^2|\vec{b}|^2\)</li>
<li>\(\dfrac{3}{4}|\vec{a}|^2|\vec{b}|^2\)</li>
<li>\(|\vec{a}|^2|\vec{b}|^2\)</li>
<li>\(0\)</li>
Step-by-Step Solution
Key Concept: The determinant = [a,b,c] = (a \times b) \cdot c = |a \times b| \cdot |c| \cdot cos0° = |a||b|sin(\pi/6) \cdot 1 = |a||b|/2. Square it.
The determinant $D = [\vec{a},\vec{b},\vec{c}] = (\vec{a}\times\vec{b})\cdot\vec{c}$.
Since $\vec{c}\perp\vec{a}$ and $\vec{c}\perp\vec{b}$, $\vec{c}$ is parallel to $\vec{a}\times\vec{b}$. With $|\vec{c}|=1$:
$D = |\vec{a}\times\vec{b}||\vec{c}|\cos0° = |\vec{a}||\vec{b}|\sin\frac{\pi}{6} = \frac{|\vec{a}||\vec{b}|}{2}$
$D^2 = \frac{|\vec{a}|^2|\vec{b}|^2}{4} = \frac{(a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)}{4}$
Answer: (A)
Correct Answer: A