Question

Please ans this Vectors ques !
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Answer 1

Explanation:

We know the equation,

${a}^{2} + {b}^{2} + 2ab \cos( \alpha ) = {r}^{2}$

Also,

$tan \alpha = \frac{b \sin( \beta ) }{a + b \cos( \beta ) }$

Where beta is the angle btw a and b.

We are given,

$\alpha = 90$

So,

$\frac{1}{0} = \frac{b \sin( \beta ) }{a + b \cos( \beta ) }$

Equating the denominator,

$0 = a + b \cos( \beta ) \\ b \cos( \beta ) = - a \\ \cos( \beta ) = - \frac{a}{b} \\ \beta = { \cos }^{ - 1} \frac{ - a}{b}$

Answer 2

$\huge\bf\fbox\red{Answer:-}$

We know the equation,

${a}^{2} + {b}^{2} + 2ab \cos( \alpha ) = {r}^{2}$

Also,

$tan \alpha = \frac{b \sin( \beta ) }{a + b \cos( \beta ) }$

Where beta is the angle btw a and b.

We are given,

$\alpha = 90$

So,

$\begin{gathered} \frac{1}{0} = \frac{b \sin( \beta ) }{a + b \cos( \beta ) } \\ \end{gathered}$

Equating the denominator,

$\begin{gathered}0 = a + b \cos( \beta ) \\ b \cos( \beta ) = - a \\ \cos( \beta ) = - \frac{a}{b} \\ \beta = { \cos }^{ - 1} \frac{ - a}{b} \end{gathered}$