Vector Algebra Question 16
Question 16 - 31 January - Shift 1
Let $\vec{a}$ and $b$ be two vector such that $|\vec{a}|=\sqrt{ } 14$, $|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^{2}$ is equal to
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Answer: (36)
Solution:
Formula: Vector products of two vectors: Lagrange’s identity
$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6} \quad|\vec{a} \times \vec{b}|=\sqrt{48}$
$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2}$
$\Rightarrow(\vec{a} \cdot \vec{b})^{2}=84-48=36$
