Question

vector A*B is grater then 73% of vector A.B then what is the angle between them ​

Answer 1

Correct Question:-

If two vectors $\vec{\sf{A}}$ and $\vec{\sf{B}}$ are such that magnitude of $\vec{\sf{A}}\times\vec{\sf{B}}$ is $\sf{73\%}$ greater than $\vec{\sf{A}}\cdot\vec{\sf{B}},$ find angle between them.

Solution:-

The dot product of vectors $\vec{\sf{A}}$ and $\vec{\sf{B}}$ is given by,

$\longrightarrow\vec{\sf{A}}\cdot\vec{\sf{B}}=\sf{AB\cos\theta}$

And the cross product is given by,

$\longrightarrow\vec{\sf{A}}\times\vec{\sf{B}}=\sf{AB\sin\theta}$

According to the question,

$\longrightarrow\vec{\sf{A}}\times\vec{\sf{B}}=\left(\vec{\sf{A}}\cdot\vec{\sf{B}}\right)\sf{\left[1+\dfrac{73}{100}\right]}$

$\longrightarrow\sf{AB\sin\theta=AB\cos\theta\times\dfrac{173}{100}}$

$\longrightarrow\sf{\sin\theta=\cos\theta\times1.73}$

$\longrightarrow\sf{\dfrac{\sin\theta}{\cos\theta}=1.73}$

$\longrightarrow\sf{\tan\theta=\sqrt3}$

$\longrightarrow\sf{\underline{\underline{\theta=60^o}}}$

Hence the angle between them is 60°.