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

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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, $\vec{A} \perp \vec{B}$