If A+B=C and that C is perpendicular to A . What is the angle between A and B , if |A|=|C| ?

Correct answer: C. (3 π)/(4) rad Given, A+B=C and B=C-A Since, C A , therefore B^2=C^2+A^2 Also, |C|=|A| , therefore B^2=2 A^2 B= 2 A Now, A^2+B^2+2 A B θ=C^2=A^2 θ=- 1 2 This gives θ=(3 π)/(4) rad

JIPMER 2016

Exam: JIPMER
Year: 2016
Subject: Physics
Chapter: Mathematics in Physics

JIPMER 2016 · Physics · Mathematics in Physics

If A+B=C and that C is perpendicular to A . What is the angle...

\mathbf{A}+\mathbf{B}=\mathbf{C}

and that

\mathbf{C}

is perpendicular to

\mathbf{A}

. What is the angle between

\mathbf{A}

and

\mathbf{B}

, if

|\mathbf{A}|=|\mathbf{C}|

Options

A $\frac{\pi}{4} \mathrm{rad}$
B $\frac{\pi}{2} \mathrm{rad}$
C $\frac{3 \pi}{4} \mathrm{rad}$
D $\pi$ rad

Correct answer

C. $\frac{3 \pi}{4} \mathrm{rad}$

Step-by-step solution

Given,

A+B=C

and

\mathrm{B}=\mathrm{C}-\mathrm{A}

Since,

\mathbf{C} \perp \mathbf{A}

, therefore

B^2=C^2+A^2

Also,

|\mathbf{C}|=|\mathbf{A}|

, therefore

B^2=2 A^2 \Rightarrow B=\sqrt{2} A

Now,

A^2+B^2+2 A B \cos \theta=C^2=A^2

\therefore \quad \cos \theta=-\frac{1}{\sqrt{2}}

This gives

\quad \theta=\frac{3 \pi}{4} \mathrm{rad}

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