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Question
Let \[\overline{a}\] and \[\overline{b}\] be two vectors such that |a| = 1, |b| = 4, \[\overline{a}\] · \[\overline{b}\] = 2. If \[\overline{c}\] = (2\[\overline{a}\] × \[\overline{b}\]) − 3\[\overline{b}\], then the angle between \[\overline{b}\] and \[\overline{c}\] is
Options
\[\frac{\pi}{3}\]
\[\frac{\pi}{6}\]
\[\frac{3\pi}{4}\]
\[\frac{5\pi}{6}\]
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Solution
\[\frac{5\pi}{6}\]
Explanation:
\[\mathbf{\overline{a}}\cdot\mathbf{\overline{b}}=\left|\mathbf{\overline{a}}\right|\left|\mathbf{\overline{b}}\right|\cos\theta\]
\[\mathbf{\overline{a}}\times\mathbf{\overline{b}}=\left|\mathbf{\overline{a}}\right|\left|\mathbf{\overline{b}}\right|\sin\theta\]
\[\left(\overline{\mathrm{a}}\cdot\overline{\mathrm{b}}\right)^2+\left(\overline{\mathrm{a}}\times\overline{\mathrm{b}}\right)^2=\left|\overline{\mathrm{a}}\right|^2\left|\overline{\mathrm{b}}\right|^2\left(\cos^2\theta+\sin^2\theta\right)\]
\[\therefore\quad\left(\overset{-}{\operatorname*{\operatorname*{a}}}\times\overset{-}{\operatorname*{\operatorname*{b}}}\right)^2\]\[=\left(1\right)^{2}\left(4\right)^{2}-\left(\overline{\mathrm{a}}\cdot\overline{\mathrm{b}}\right)^{2}=12\]
\[\overset{-}{\operatorname*{\mathbf{c}}}=\left(2\overset{-}{\operatorname*{\mathbf{a}}}\times\overset{-}{\operatorname*{\mathbf{b}}}\right)-3\overset{-}{\operatorname*{\mathbf{b}}}\]
\[\therefore\quad\overline{\mathrm{c}}\cdot\overline{\mathrm{c}}=\left(2\overline{\mathrm{a}}\times\overline{\mathrm{b}}-3\overline{\mathrm{b}}\right)\cdot\left(2\overline{\mathrm{a}}\times\overline{\mathrm{b}}-3\overline{\mathrm{b}}\right)\]
\[\therefore\quad\left|\overline{\mathrm{c}}\right|^2=\left|2\overline{\mathrm{a}}\times\overline{\mathrm{b}}\right|^2+\left|3\overline{\mathrm{b}}\right|^2-2\times3\overline{\mathrm{b}}\cdot\left(2\overline{\mathrm{a}}\times\overline{\mathrm{b}}\right)\]
\[\therefore\quad\left|\overline{\mathrm{c}}\right|^2=4\left|\overline{\mathrm{a}}\times\overline{\mathrm{b}}\right|^2+9\left|\overline{\mathrm{b}}\right|^2-0\quad\ldots\left[\because\overline{\mathrm{b}}\cdot\left(\overline{\mathrm{a}}\times\overline{\mathrm{b}}\right)=0\right]\]
\[=4\times12+9\times16\]
\[=44+144\]
= 192
\[\mathbf{\overline{b}}\cdot\mathbf{\overline{c}}=\left|\mathbf{\overline{b}}\right|\left|\mathbf{\overline{c}}\right|\cos\theta\]
\[\therefore\quad|\overline{\mathrm{b}}||\overline{\mathrm{c}}|\cos\theta=\overline{\mathrm{b}}\cdot\left(2\overline{\mathrm{a}}\times\overline{\mathrm{b}}-3\overline{\mathrm{b}}\right)\]
\[\therefore\quad4\sqrt{192}\cos\theta=2\overline{\mathrm{b}}\cdot\left(\overline{\mathrm{a}}\times\overline{\mathrm{b}}\right)-3\overline{\mathrm{b}}\cdot\overline{\mathrm{b}}\]
\[\therefore\quad4\times8\sqrt{3}\cos\theta=-3\times16\]
\[\therefore\quad\cos\theta=\frac{-3}{2\sqrt{3}}=\frac{-\sqrt{3}}{2}\]
\[\therefore\quad\theta=150^\circ=\frac{5\pi}{6}\]
