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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Vector Algebra

NDA 2024 · Mathematics · Vector Algebra

Let a and b are two vectors of magnitude 4 inclined at an angle...

Let a⃗\vec{a} and b⃗\vec{b} are two vectors of magnitude 4 inclined at an angle π3\frac{\pi}{3}, then what is the angle between a⃗\vec{a} and a⃗−b⃗\vec{a}-\vec{b} ?

Options

  1. Aπ2\frac{\pi}{2}
  2. Bπ3\frac{\pi}{3}
  3. Cπ4\frac{\pi}{4}
  4. Dπ6\frac{\pi}{6}

Correct answer

B. π3\frac{\pi}{3}

Step-by-step solution

The angle between a⃗\vec{a} and a⃗−b⃗\vec{a}-\vec{b} is given by cos⁡θ=a⃗⋅(a⃗−b⃗)∣a⃗∣∣a⃗−b⃗∣\cos \theta=\frac{\vec{a} \cdot(\vec{a}-\vec{b})}{|\vec{a}||\vec{a}-\vec{b}|} Now, a⃗⋅(a⃗−b⃗)=a⃗⋅a⃗−a⃗⋅b⃗∴a⃗⋅a⃗=∣a⃗∣2\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 a⃗andb⃗\vec{a} and \vec{b} is 4 ∣a⃗∣=∣b⃗∣=4∣a⃗∣2=42=16\begin{aligned} |\vec{a}|=|\vec{b}| & =4 \\ |\vec{a}|^2 & =4^2=16 \end{aligned} and a⃗b⃗=∣a⃗∥b⃗∣cos⁡π3=4⋅4⋅12=8\qquad \vec{a} \vec{b}=|\vec{a} \| \vec{b}| \cos \frac{\pi}{3}=4 \cdot 4 \cdot \frac{1}{2}=8∣a⃗−b⃗∣2=∣a⃗∣2+∣b⃗∣2−a⃗b⃗−b⃗a⃗=16+16−8−8=16cos⁡θ=16−8416=84⋅4=12=cos⁡π3\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}⇒θ=π3\Rightarrow \quad \theta=\frac{\pi}{3}
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