MHT CET · Maths · Vector Algebra
If \(|\vec{a}|=\sqrt{27},|\bar{b}|=7\) and \(|\vec{a} \times \bar{b}|=35\), then \(\bar{a} \cdot \bar{b}\) is equal to
- A \(\sqrt{\frac{35}{2}}\)
- B \(\frac{\sqrt{35}}{2}\)
- C \(7 \sqrt{2}\)
- D \(\sqrt{35}\)
Answer & Solution
Correct Answer
(C) \(7 \sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(|\vec{a}|=\sqrt{27},|\vec{b}|=7 \text { and }|\vec{a} \times \bar{b}|=35\)
We know that
\(|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| =|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \sin \theta \)
\( \therefore \sin \theta =\frac{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}{|\overline{\mathrm{a}}||\overline{\mathrm{b}}|}=\frac{35}{\sqrt{27} \times 7}=\frac{5}{\sqrt{27}} \)
\( \therefore \cos \theta=\sqrt{1-\frac{25}{27}}=\sqrt{\frac{2}{27}}\)
Now,
\(\bar{a} \cdot \bar{b} =|\bar{a}||\bar{b}| \cos \theta \)
\( =\sqrt{27} \times 7 \times \sqrt{\frac{2}{27}}=7 \sqrt{2}\)
We know that
\(|\overline{\mathrm{a}} \times \overline{\mathrm{b}}| =|\overline{\mathrm{a}}||\overline{\mathrm{b}}| \sin \theta \)
\( \therefore \sin \theta =\frac{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}{|\overline{\mathrm{a}}||\overline{\mathrm{b}}|}=\frac{35}{\sqrt{27} \times 7}=\frac{5}{\sqrt{27}} \)
\( \therefore \cos \theta=\sqrt{1-\frac{25}{27}}=\sqrt{\frac{2}{27}}\)
Now,
\(\bar{a} \cdot \bar{b} =|\bar{a}||\bar{b}| \cos \theta \)
\( =\sqrt{27} \times 7 \times \sqrt{\frac{2}{27}}=7 \sqrt{2}\)
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