If $|\vec{a}|=\sqrt{27},|\bar{b}|=7$ and $|\vec{a} \times \bar{b}|=35$, then $\bar{a} \cdot \bar{b}$ is…

If $|\vec{a}|=\sqrt{27},|\bar{b}|=7$ and $|\vec{a} \times \bar{b}|=35$, then $\bar{a} \cdot \bar{b}$ is equal to
  1. $\sqrt{\frac{35}{2}}$
  2. $\frac{\sqrt{35}}{2}$
  3. $7 \sqrt{2}$
  4. $\sqrt{35}$

Solution

$|\vec{a}|=\sqrt{27},|\vec{b}|=7 \text { and }|\vec{a} \times \bar{b}|=35$
We know that $\begin{array}{rlrl} & |\overline{\mathrm{a}} \times \overline{\mathrm{b}}| & =|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \sin \theta \\ & \therefore \quad \sin \theta & =\frac{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}{|\overline{\mathrm{a}}||\overline{\mathrm{b}}|}=\frac{35}{\sqrt{27} \times 7}=\frac{5}{\sqrt{27}} \\ \therefore & \cos \theta=\sqrt{1-\frac{25}{27}}=\sqrt{\frac{2}{27}} \end{array}$ Now, $\begin{aligned} \bar{a} \cdot \bar{b} & =|\bar{a}||\bar{b}| \cos \theta \\ & =\sqrt{27} \times 7 \times \sqrt{\frac{2}{27}}=7 \sqrt{2}\end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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