Q220 / 270 · NDA 2024 (Phase 1) · Mathematics

Let $\vec{a}$ and $\vec{b}$ are two vectors of magnitude 4 inclined at an angle $\frac{\pi}{3}$, then what is the angle between $\vec{a}$ and $\vec{a}-\vec{b}$ ?

  • A
    $\frac{\pi}{2}$
  • Correct answer
    B
    $\frac{\pi}{3}$
  • C
    $\frac{\pi}{4}$
  • D
    $\frac{\pi}{6}$

Solution

The angle between $\vec{a}$ and $\vec{a}-\vec{b}$ is given by $\cos \theta=\frac{\vec{a} \cdot(\vec{a}-\vec{b})}{|\vec{a}||\vec{a}-\vec{b}|}$ Now, $\quad \vec{a} \cdot(\vec{a}-\vec{b})=\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}$ $\therefore \quad \vec{a} \cdot \vec{a}=|\vec{a}|^2$ Given that magnitude of $\vec{a}$ and $\vec{b}$ is 4 $\begin{aligned} |\vec{a}|=|\vec{b}| & =4 \\ |\vec{a}|^2 & =4^2=16 \end{aligned}$ and$\qquad \vec{a} \vec{b}=|\vec{a} \| \vec{b}| \cos \frac{\pi}{3}=4 \cdot 4 \cdot \frac{1}{2}=8$ $\begin{aligned} |\vec{a}-\vec{b}|^2 & =|\vec{a}|^2+|\vec{b}|^2-\vec{a} \vec{b}-\vec{b} \vec{a} \\ & =16+16-8-8=16 \\ \cos \theta & =\frac{16-8}{4 \sqrt{16}}=\frac{8}{4 \cdot 4}=\frac{1}{2}=\cos \frac{\pi}{3} \end{aligned}$ $\Rightarrow \quad \theta=\frac{\pi}{3}$

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

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