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 cos θ=C^2=A^2 ∴ cos θ=-1/(√2) This gives θ=(3 π)/4 rad

Q89 / 180 · JIPMER 2016 · Physics

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

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

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

Hence, the correct answer is option: 'C'.

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

Solution

More Mathematics in Physics Questions