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CBSE Sample Paper 2021-22 (Term 2) · Section A · 2 marks

Q3: Prove ∣a^+b^∣=2cos⁡θ2|\hat a + \hat b| = 2\cos\frac{\theta}{2} for unit vectors a^\hat a⁠, b^\hat b

Sample 2021-22 T2Ch 10 · Vector Algebra
The question

If a^\hat a and b^\hat b are unit vectors, then prove that ∣a^+b^∣=2cos⁡θ2|\hat a + \hat b| = 2\cos\dfrac{\theta}{2}⁠, where θ\theta is the angle between them.

Exam answer
∣a^+b^∣2=(a^+b^)⋅(a^+b^)=∣a^∣2+∣b^∣2+2(a^⋅b^)\begin{aligned} &|\hat a + \hat b|^2 = (\hat a + \hat b)\cdot(\hat a + \hat b) \\ &= |\hat a|^2 + |\hat b|^2 + 2(\hat a\cdot\hat b) \end{aligned}
Why?

The square of a length is a dot product: ∣v⃗∣2=v⃗⋅v⃗|\vec v|^2 = \vec v\cdot\vec v⁠. Expand it like brackets:

(a^+b^)⋅(a^+b^)=a^⋅a^+a^⋅b^+b^⋅a^+b^⋅b^\begin{aligned} &(\hat a + \hat b)\cdot(\hat a + \hat b) \\ &= \hat a\cdot\hat a + \hat a\cdot\hat b \\ &\quad + \hat b\cdot\hat a + \hat b\cdot\hat b \end{aligned}

Since a^⋅a^=∣a^∣2\hat a\cdot\hat a = |\hat a|^2 and b^⋅a^=a^⋅b^\hat b\cdot\hat a = \hat a\cdot\hat b⁠, the two middle terms add to 2(a^⋅b^)2(\hat a\cdot\hat b)⁠.

As ∣a^∣=∣b^∣=1|\hat a| = |\hat b| = 1 and a^⋅b^=∣a^∣∣b^∣cos⁡θ=cos⁡θ\hat a\cdot\hat b = |\hat a||\hat b|\cos\theta = \cos\theta⁠,

Why?

Unit vectors have length 1. The dot product of two vectors at angle θ\theta is

a^⋅b^=∣a^∣∣b^∣cos⁡θ\hat a\cdot\hat b = |\hat a||\hat b|\cos\theta

so here a^⋅b^=1×1×cos⁡θ=cos⁡θ\hat a\cdot\hat b = 1 \times 1 \times \cos\theta = \cos\theta⁠.

∣a^∣2=(1)2=1∣b^∣2=(1)2=12(a^⋅b^)=2cos⁡θ\begin{aligned} |\hat a|^2 &= (1)^2 = 1 \\ |\hat b|^2 &= (1)^2 = 1 \\ 2(\hat a\cdot\hat b) &= 2\cos\theta \end{aligned}
∣a^+b^∣2=1+1+2cos⁡θ=2(1+cos⁡θ)=4cos⁡2θ2\begin{aligned} |\hat a + \hat b|^2 &= 1 + 1 + 2\cos\theta \\ &= 2(1 + \cos\theta) \\ &= 4\cos^2\frac{\theta}{2} \end{aligned}
Why?

Half-angle identity:

1+cos⁡θ=2cos⁡2θ21 + \cos\theta = 2\cos^2\frac{\theta}{2}

It comes from cos⁡2A=2cos⁡2A−1\cos 2A = 2\cos^2A - 1 with A=θ2A = \frac{\theta}{2}⁠. Then 2×2cos⁡2θ2=4cos⁡2θ22 \times 2\cos^2\frac{\theta}{2} = 4\cos^2\frac{\theta}{2}⁠.

1+1+2cos⁡θ=2+2cos⁡θ=2(1+cos⁡θ)=2(2cos⁡2θ2)=4cos⁡2θ2\begin{aligned} 1 + 1 + 2\cos\theta &= 2 + 2\cos\theta \\ &= 2(1 + \cos\theta) \\ &= 2\left(2\cos^2\frac{\theta}{2}\right) \\ &= 4\cos^2\frac{\theta}{2} \end{aligned}

As 0≤θ≤π0 \le \theta \le \pi⁠, cos⁡θ2≥0\cos\frac{\theta}{2} \ge 0⁠. Taking the positive square root,

Why?

The angle between two vectors always lies between 00 and π\pi⁠. So θ2\frac{\theta}{2} lies between 00 and π2\frac{\pi}{2}⁠, where cosine is never negative.

Watch out: 4cos⁡2θ2=2∣cos⁡θ2∣\sqrt{4\cos^2\frac{\theta}{2}} = 2\left|\cos\frac{\theta}{2}\right|⁠. The modulus can be dropped only because cos⁡θ2≥0\cos\frac{\theta}{2} \ge 0⁠.

∣a^+b^∣=4cos⁡2θ2=4×cos⁡2θ2=2∣cos⁡θ2∣=2cos⁡θ2\begin{aligned} |\hat a + \hat b| &= \sqrt{4\cos^2\frac{\theta}{2}} \\ &= \sqrt4 \times \sqrt{\cos^2\frac{\theta}{2}} \\ &= 2\left|\cos\frac{\theta}{2}\right| = 2\cos\frac{\theta}{2} \end{aligned}

∴ ∣a^+b^∣=2cos⁡θ2\therefore\ |\hat a + \hat b| = 2\cos\dfrac{\theta}{2}⁠. Hence proved.

Where this comes from

Chapter 10, Vector Algebra: Dot product: magnitude of a sum of unit vectors.

Question type: Dot products with letters

How to spot it: click to see

Type 7. Dot products with letters: expand like brackets

Spot it: no i^,j^,k^\hat i, \hat j, \hat k at all, just a⃗\vec a⁠, b⃗\vec b⁠, their lengths and a dot product.

Here: Write ∣a^+b^∣2|\hat a + \hat b|^2 as (a^+b^)⋅(a^+b^)(\hat a + \hat b)\cdot(\hat a + \hat b) and expand, with ∣a^∣=∣b^∣=1|\hat a| = |\hat b| = 1 and a^⋅b^=cos⁡θ\hat a\cdot\hat b = \cos\theta⁠. That gives 2+2cos⁡θ=4cos⁡2θ22 + 2\cos\theta = 4\cos^2\frac{\theta}{2}⁠.

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