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  4. Let veca and vecb be two vectors such that | veca|=√14,| vecb|=√6 and | veca × vecb|=√48. Then ( veca ⋅ vecb)2 is equal to

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Q. 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

JEE MainJEE Main 2023Vector Algebra

Solution:

$ |\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 \times|\vec{b}|^2$
$ \Rightarrow(\vec{a} \cdot \vec{b})^2=84-48=36$