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CBSE Sample Paper 2020-21 · Part A: Section I · 1 mark

Q12: Angle between unit vectors a^\hat a⁠, b^\hat b when ∣a^+b^∣=1|\hat a + \hat b| = 1

Sample 2020-21Ch 10 · Vector Algebra
The question

Find the angle between the unit vectors a^\hat a and b^\hat b⁠, given that ∣a^+b^∣=1|\hat a + \hat b| = 1

Exam answer
∣a^+b^∣2=1∣a^∣2+∣b^∣2+2 a^⋅b^=11+1+2 a^⋅b^=1a^⋅b^=−12\begin{aligned} |\hat a + \hat b|^2 &= 1 \\ |\hat a|^2 + |\hat b|^2 + 2\,\hat a\cdot\hat b &= 1 \\ 1 + 1 + 2\,\hat a\cdot\hat b &= 1 \\ \hat a\cdot\hat b &= -\frac{1}{2} \end{aligned}
Why?

Squaring brings in the dot product, which holds the angle. The square of a length is the dot product of the vector with itself:

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

a^\hat a and b^\hat b are unit vectors, so ∣a^∣2=∣b^∣2=1|\hat a|^2 = |\hat b|^2 = 1⁠. Then 2 a^⋅b^=1−2=−12\,\hat a\cdot\hat b = 1 - 2 = -1⁠.

2+2 a^⋅b^=12 a^⋅b^=1−2=−1a^⋅b^=−12=−12\begin{aligned} 2 + 2\,\hat a\cdot\hat b &= 1 \\ 2\,\hat a\cdot\hat b &= 1 - 2 = -1 \\ \hat a\cdot\hat b &= \frac{-1}{2} = -\frac{1}{2} \end{aligned}
∣a^∣∣b^∣cos⁡θ=−12cos⁡θ=−12θ=π−π3\begin{aligned} |\hat a||\hat b|\cos\theta &= -\frac{1}{2} \\ \cos\theta &= -\frac{1}{2} \\ \theta &= \pi - \frac{\pi}{3} \end{aligned}
Why?

By definition a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta⁠, with 0≤θ≤π0 \le \theta \le \pi⁠. Here ∣a^∣∣b^∣=1|\hat a||\hat b| = 1⁠.

cos⁡π3=12\cos\frac{\pi}{3} = \frac{1}{2} and cos⁡(π−α)=−cos⁡α\cos(\pi - \alpha) = -\cos\alpha⁠, so cos⁡(π−π3)=−12\cos\left(\pi - \frac{\pi}{3}\right) = -\frac{1}{2}⁠.

Then work out the angle over a common denominator:

π−π3=3π3−π3=3π−π3=2π3\begin{aligned} \pi - \frac{\pi}{3} &= \frac{3\pi}{3} - \frac{\pi}{3} \\ &= \frac{3\pi - \pi}{3} = \frac{2\pi}{3} \end{aligned}

Watch out: π3\frac{\pi}{3} is wrong. A negative cosine means the angle is obtuse.

∴ θ=2π3\therefore\ \theta = \dfrac{2\pi}{3}

Where this comes from

Chapter 10, Vector Algebra: Angle between two vectors (dot product).

Question type: Dot products with letters + The angle between two vectors (combines 2 types)

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: First, there are only unit vectors and a length: square ∣a^+b^∣=1|\hat a + \hat b| = 1 and expand, so 1+1+2 a^⋅b^=11 + 1 + 2\,\hat a\cdot\hat b = 1 and a^⋅b^=−12\hat a\cdot\hat b = -\frac12⁠.

Type 4. The angle between two vectors

Spot it: "find the angle between the vectors".

Here: Then the angle: both lengths are 1, so cos⁡θ=−12\cos\theta = -\frac12 and θ=2π3\theta = \frac{2\pi}{3}⁠.

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