If vec{a} and vec{b} makes an angle cos ^{-1} | vedclass

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Question

$\cos 	heta=\frac{5}{9}$

$\frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9}$

$\mid \vec{a}+\cdots(1)$

$a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$\cos 	heta=\frac{5}{9}$

$\frac{\vec{a} \cdot \vec{b}}{a b}=\frac{5}{9}$

$\mid \vec{a}+\cdots(1)$

$a^2+b^2+2 \vec{a} \cdot \vec{b}=2 a^2+2 b^2-4 \vec{a} \cdot \vec{b}$

$6 \vec{a} \cdot \vec{b}=a^2+b^2$

$6 	imes \frac{5}{9} a b=a^2+b^2$

$\frac{10}{3} a b=a^2+b^2 \quad \& \quad a=n b$

$\frac{10}{3} n b^2=n^2 b^2+b^2$

$3 n^2-10 n+3=0$

$n=\frac{1}{3} 	ext { and } n=3$

$	ext { integer value } n=3$

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 . . . . . . ..