Question

2) Find the cosines of angle between a×b if (3a-2b)×(a+2b)are perpendicular to each one and 5a_3b,(a_b)are twi mutually perpendicular vectors
اک
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ماسه (د که
(د دی
find the cosines of angle between (3a-2b)×(a+2b)perpendicular to each one and
(5á -3b),(a-b)are two
mutualy
perperpendiculas vectors​

Answer 1

Vector Analysis

Solution:

The vectors $(3\vec{a}-2\vec{b})$ and $\vec{a}+2\vec{b})$ are perpendicular to each other

$\Rightarrow (3\vec{a}-2\vec{b}).(\vec{a}+2\vec{b})=0$

$\Rightarrow 3a^{2}-4b^{2}+4\vec{a}.\vec{b}=0$

The vectors $(5\vec{a}-3\vec{b})$ and $(\vec{a}-\vec{b})$ are perpendicular to each other

$\Rightarrow (5\vec{a}-3\vec{b}).(\vec{a}-\vec{b})=0$

$\Rightarrow 5a^{2}+3b^{2}-8\vec{a}.\vec{b}=0$

We have two relations:

$\quad\quad 3a^{2}-4b^{2}+4\vec{a}.\vec{b}=0$

$\quad\quad 5a^{2}+3b^{2}-8\vec{a}.\vec{b}=0$

By cross-multiplication, we get

$\quad\frac{a^{2}}{32-12}=\frac{b^{2}}{20+24}=\frac{\vec{a}.\vec{b}}{9+20}$

$\Rightarrow \frac{a^{2}}{20}=\frac{b^{2}}{44}=\frac{\vec{a}.\vec{b}}{29}=k\:(say)$

$\Rightarrow a^{2}=20k,\:b^{2}=44k,\:\vec{a}.\vec{b}=29k$

$\Rightarrow a=\sqrt{20k},\:b=\sqrt{44k},\:\vec{a}.\vec{b}=29k$

If $\theta$ be the angle between the vectors $\vec{a}$ and $\vec{b}$, then

$\quad cos\theta=\frac{\vec{a}.\vec{b}}{ab}$

$\Rightarrow cos\theta=\frac{29k}{\sqrt{20k\times 44k}}$

$\Rightarrow cos\theta=\frac{29}{\sqrt{880}}$

$\Rightarrow \boxed{cos\theta=\frac{29}{4\sqrt{55}}}$

This is the required cosine of the angle between the vectors $\vec{a}$ and $\vec{b}$.

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