Motion in a Plane - Result Question 22
24. The magnitudes of vectors $\vec{A}, \vec{B}$ and $\vec{C}$ are 3,4 and 5 units respectively. If $\vec{A}+\vec{B}=\vec{C}$, then the angle between $\vec{A}$ and $\vec{B}$ is
[1988]
(a) $\pi / 2$
(b) $\cos ^{-1} 0.6$
(c) $\tan ^{-1} 7 / 5$
(d) $\pi / 4$
Topic 2: Motion in a Plane with Constant acceleration
Show Answer
Answer:
Correct Answer: 24. (a)
Solution:
$ \begin{aligned} \text{ (a) } & (\vec{A}+\vec{B})^{2}=(\vec{C})^{2} \\ \Rightarrow & A^{2}+B^{2}+2 \vec{A} \cdot \vec{B}=C^{2} \\ \Rightarrow & 3^{2}+4^{2}+2 \vec{A} \cdot \vec{B}=5^{2} \\ \Rightarrow & 2 \vec{A} \cdot \vec{B}=0 \quad \text{ or } \Rightarrow \vec{A} \cdot \vec{B}=0 \quad \therefore \vec{A} \perp \vec{B} \end{aligned} $
Here $A^{2}+B^{2}=C^{2}$. Hence, $\vec{A} \perp \vec{B}$
