If a and b Are Unit Vectors, Then the Greatest Value of √3|a+b| + |a−b| Is

JEE Main 2020. Previous Year Question.

Problem

If $\vec a$ and $\vec b$ are unit vectors, then the greatest value of $\sqrt3\,|\vec a+\vec b|+|\vec a-\vec b|$ is?

Key insight. Since $\vec a$ and $\vec b$ are unit vectors, both $|\vec a+\vec b|$ and $|\vec a-\vec b|$ depend only on the angle $\theta$ between them — and each can be rewritten cleanly using the half-angle identity $1\pm\cos\theta=2\cos^2\left(\frac{\theta}2\right)$ or $2\sin^2\left(\frac{\theta}2\right)$. That turns the whole expression into a single sinusoid in $\theta/2$, whose maximum is read off directly.

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Approach

Let $\theta$ be the angle between $\vec a$ and $\vec b$. Express $|\vec a+\vec b|^2$ and $|\vec a-\vec b|^2$ in terms of $\cos\theta$, then rewrite each using the half-angle identities to get $|\vec a+\vec b|=2\cos(\theta/2)$ and $|\vec a-\vec b|=2\sin(\theta/2)$. Substitute into the given expression, and combine the resulting sum of a sine and cosine term into a single cosine using the auxiliary-angle method.

Solution

Step 1 — Express |a + b|² and |a − b|² using the angle θ between a and b

Since $|\vec a|=|\vec b|=1$:

$$|\vec a+\vec b|^2 = 1+1+2\cos\theta = 2(1+\cos\theta) = 4\cos^2\frac{\theta}{2}$$

$$|\vec a-\vec b|^2 = 1+1-2\cos\theta = 2(1-\cos\theta) = 4\sin^2\frac{\theta}{2}$$

Step 2 — Take square roots

$$|\vec a+\vec b| = 2\cos\frac{\theta}{2}, \qquad |\vec a-\vec b| = 2\sin\frac{\theta}{2}$$

(taking $\theta/2\in[0,\pi/2]$, where both cosine and sine are non-negative)

Step 3 — Substitute into the given expression

$$\sqrt3\,|\vec a+\vec b|+|\vec a-\vec b| = 2\sqrt3\cos\frac{\theta}{2}+2\sin\frac{\theta}{2} = 2\left(\sqrt3\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right)$$

Step 4 — Combine into a single cosine using the auxiliary-angle method

$$\sqrt3\cos\frac{\theta}{2}+\sin\frac{\theta}{2} = 2\left(\cos30°\cos\frac{\theta}{2}+\sin30°\sin\frac{\theta}{2}\right) = 2\cos\left(\frac{\theta}{2}-30°\right)$$

Step 5 — Combine and find the maximum

$$\sqrt3\,|\vec a+\vec b|+|\vec a-\vec b| = 2\cdot2\cos\left(\frac{\theta}{2}-30°\right) = 4\cos\left(\frac{\theta}{2}-30°\right)$$

Since the maximum value of $\cos$ is $1$ (achieved when $\theta/2=30°$, i.e. $\theta=60°$, a valid angle between two vectors):

$$\text{Greatest value} = 4\times1 = 4$$

Answer

Common mistakes

• Forgetting that $\theta/2$ stays within $[0°,90°]$, so a direct square root is valid without a $\pm$ sign. Since $\theta$ (the angle between two vectors) always lies in $[0°,180°]$, $\theta/2$ lies in $[0°,90°]$ — a range where both $\cos(\theta/2)$ and $\sin(\theta/2)$ are automatically non-negative, so no absolute-value case-splitting is needed.
• Trying to maximise $|\vec a+\vec b|$ and $|\vec a-\vec b|$ separately. These two quantities move in opposite directions as $\theta$ changes (one grows while the other shrinks), so there’s a genuine trade-off — the combined expression must be optimised as a whole, not term by term.

Practise next

• If $\vec a$ and $\vec b$ are unit vectors, find the greatest value of $|\vec a+\vec b|+\sqrt3\,|\vec a-\vec b|$, using the same half-angle and auxiliary-angle approach.