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If $\vec{a}$ and $\vec{b}$ makes an angle $\cos ^{-1}\left(\frac{5}{9}\right)$ with each other, then $|\vec{a}+\vec{b}|=\sqrt{2}|\vec{a}-\vec{b}|$ for $|\vec{a}|=n|\vec{b}|$ The integer value of $n$ is . . . . . . ..
$3$
$5$
$4$
$6$
Solution
$\cos \theta=\frac{5}{9}$
$\frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9}$
$\mid \vec{a}+\cdots(1)$
$a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \vec{a} \cdot \vec{b}$
$6 \vec{a} \cdot \vec{b}=a^2+b^2$
$6 \times \frac{5}{9} a b=a^2+b^2$
$\frac{10}{3} a b=a^2+b^2 \quad \& \quad a=n b$
$\frac{10}{3} n b^2=n^2 b^2+b^2$
$3 n^2-10 n+3=0$
$n=\frac{1}{3} \text { and } n=3$
$\text { integer value } n=3$
Similar Questions
If $\left| {\vec A } \right|\, = \,2$ and $\left| {\vec B } \right|\, = \,4$ then match the relation in Column $-I$ with the angle $\theta $ between $\vec A$ and $\vec B$ in Column $-II$.
| Column $-I$ | Column $-II$ |
| $(a)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,0$ | $(i)$ $\theta = \,{30^o}$ |
| $(b)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,8$ | $(ii)$ $\theta = \,{45^o}$ |
| $(c)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4$ | $(iii)$ $\theta = \,{90^o}$ |
| $(d)$ $\left| {\vec A \, \times \,\,\vec B } \right|\, = \,\,4\sqrt 2$ | $(iv)$ $\theta = \,{0^o}$ |
