Let $\vec{a}$ and $\vec{b}$ be two vectors 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 ___________.
Answer (integer)
36
Solution
$|\vec{a}|=\sqrt{14},|\vec{b}|=\sqrt{6}$ and $|\vec{a} \times \vec{b}|=\sqrt{48}$
<br/><br/>$$
\begin{aligned}
& \Rightarrow |\vec{a} \times \vec{b}|^{2}+(\vec{a} \cdot \vec{b})^{2}=|\vec{a}|^{2}|\vec{b}|^{2} \\\\
& \Rightarrow 48+(\vec{a} \cdot \vec{b})^{2}=6 \times 14 \\\\
& \Rightarrow (\vec{a} \cdot \vec{b})^{2}=84-48 \\\\
&=36
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
This question is part of PrepWiser's free JEE Main question bank. 169 more solved questions on Vector Algebra are available — start with the harder ones if your accuracy is >70%.