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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Vector Algebra

NDA 2024 · Mathematics · Vector Algebra

Let θ be the angle between two unit vectors a and b . If a +2 b is...

Let θ\theta be the angle between two unit vectors a⃗\vec{a} and b⃗\vec{b}. If a⃗+2b⃗\vec{a}+2 \vec{b} is perpendicular to 5a⃗−4b⃗5 \vec{a}-4 \vec{b}, then what is cos⁡θ+cos⁡2θ\cos \theta+\cos 2 \theta equal to?

Options

  1. A00
  2. B12\frac{1}{2}
  3. C11
  4. D3+12\frac{\sqrt{3}+1}{2}

Correct answer

A. 00

Step-by-step solution

Since, (\vec{a}+2 \vec{b}) \ \text{ and } (5 \vec{a}-4 \vec{b}) are perpendicular vectors. ⇒(a⃗+2b⃗)⋅(5a⃗−4b⃗)=0⇒5∣a⃗∣2−4a⃗⋅b⃗+10a⃗⋅b⃗−8∣b⃗∣2=0⇒5×1+6a⃗⋅b⃗−8=0 Since, ∣a⃗∣=∣b⃗∣=1⇒6∣a⃗∣∣b⃗∣cos⁡θ=3⇒1⋅1cos⁡θ=12⇒cos⁡θ=12\begin{aligned} & \Rightarrow(\vec{a}+2 \vec{b}) \cdot(5 \vec{a}-4 \vec{b})=0 \\ & \Rightarrow 5|\vec{a}|^2-4 \vec{a} \cdot \vec{b}+10 \vec{a} \cdot \vec{b}-8|\vec{b}|^2=0 \\ & \Rightarrow 5 \times 1+6 \vec{a} \cdot \vec{b}-8=0 \quad \text{ Since, }|\vec{a}|=|\vec{b}|=1 \\ & \Rightarrow 6|\vec{a}||\vec{b}| \cos \theta=3 \\ & \Rightarrow 1 \cdot 1 \cos \theta=\frac{1}{2} \Rightarrow \cos \theta=\frac{1}{2} \end{aligned} So, cos⁡2θ=2cos⁡2θ−1=2×14−1=−12\cos 2 \theta=2 \cos ^2 \theta-1=2 \times \frac{1}{4}-1=-\frac{1}{2} Now, cos⁡θ+cos⁡2θ=12−12=0\cos \theta+\cos 2 \theta=\frac{1}{2}-\frac{1}{2}=0
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