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Let \vec{a}$ and $\vec{b} be two vectors such that |2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}| and the angle between \vec{a} and \overrightarrow{\mathrm{b}} is 60^{\circ}. If \frac{1}{8} \overrightarrow{\mathrm{a}} is a unit vector, then |\overrightarrow{\mathrm{b}}| is equal to :
Option: 1 8
Option: 2 4
Option: 3 6
Option: 4 5

Answers (1)

best_answer

\left | \frac{\vec{a}}{8} \right |= 1\Rightarrow \left | \vec{a} \right |= 8
\left | 2\vec{a}+3\vec{b} \right |^{2}= \left | 3\vec{a}+\vec{b} \right |^{2}
\Rightarrow 4\left | \vec{a} \right |^{2}+9\left | \vec{b} \right |^{2}+ 12 \, \vec{a}\cdot \; \vec{b}= 9\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+6\, \vec{a}\,\;\cdot \vec{b}
\Rightarrow 5\left | \vec{a} \right |^{2}-8\left | \vec{b} \right |^{2}-6\; \left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}= 0
\Rightarrow 5\times 8^{2}-8\left | \vec{b} \right |^{2}-6\times 8\times \left | \vec{b} \right |\times \frac{1}{2}= 0
\Rightarrow \left | \vec{b} \right |^{2}+3 \left | \vec{b} \right |-40= 0
\Rightarrow \left | \vec{b} \right |^{2}+8 \left | \vec{b} \right |-5\left | \vec{b} \right |-40= 0
\Rightarrow \left | \vec{b} \right |^{2}= -8 \times \left ( Not\: possible \right )
OR \left | \vec{b} \right |= 5
option (4)

Posted by

Kuldeep Maurya

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