If a and b are unit vectors, then the greatest value of √3|a + b| + |a - b| is

Q: Wait, why can't we just assume $\vec{a}$ and $\vec{b}$ are along the x and y axes?

That is a common trap! If you fix the vectors, you lose the generality of the problem. The angle $\theta$ is a variable parameter here. To find the *greatest* value, we must treat $\theta$ as a variable and optimize the expression over all possible angles.

Q: Why do we use the half-angle identity instead of just keeping it as $\cos\theta$?

Great question. If we kept it as $\cos\theta$, we would have an expression involving $\sqrt{1+\cos\theta}$ and $\sqrt{1-\cos\theta}$. By using the half-angle identities, we transform the expression into a linear combination of $\sin$ and $\cos$ of the same angle, which is the standard form for finding maxima.

Q: Is there a geometric way to solve this without trigonometry?

Yes, you could rotate the coordinate system such that the bisector of the vectors lies on an axis, but the trigonometric approach is the most robust and standard for JEE. It directly leads to the harmonic form, which is a powerful tool in your arsenal.

Question

If $\vec{a}$ and $\vec{b}$ are unit vectors, then the greatest value of $\sqrt{3}|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a coordinate plane, holding two unit vectors, $\vec{a}$ and $\vec{b}$. They are the simplest building blocks of vector algebra, each with a magnitude of exactly one.

We aim to find the greatest value of the expression:
$$E = \sqrt{3}|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|$$

### The Parallelogram Law

Let the angle between our two unit vectors be $	heta$. When we add them, we form a parallelogram where the diagonal is the sum vector $\vec{a} + \vec{b}$.

According to the parallelogram law, the square of its magnitude is:
$$|\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2|\vec{a}||\vec{b}|\cos	heta$$

Since $\vec{a}$ and $\vec{b}$ are unit vectors, their magnitudes are $1$. Substituting these values, we obtain:
$$|\vec{a} + \vec{b}|^2 = 1 + 1 + 2(1)(1)\cos	heta = 2 + 2\cos	heta = 2(1 + \cos	heta)$$

### The Trigonometric Transformation

We utilize the trigonometric identity $1 + \cos	heta = 2\cos^2(\frac{	heta}{2})$. Substituting this into our expression, we get:
$$|\vec{a} + \vec{b}|^2 = 2 \left( 2\cos^2\left(\frac{	heta}{2}ight) ight) = 4\cos^2\left(\frac{	heta}{2}ight)$$

Taking the square root, we find:
$$|\vec{a} + \vec{b}| = 2\cos\left(\frac{	heta}{2}ight)$$

Similarly, for the difference vector $\vec{a} - \vec{b}$, we use the identity $1 - \cos	heta = 2\sin^2(\frac{	heta}{2})$:
$$|\vec{a} - \vec{b}|^2 = 2 - 2\cos	heta = 2(1 - \cos	heta) = 4\sin^2\left(\frac{	heta}{2}ight)$$
$$|\vec{a} - \vec{b}| = 2\sin\left(\frac{	heta}{2}ight)$$

### The Harmonic Maximization

Now, we substitute these results into our original expression $E$:
$$E = \sqrt{3} \left( 2\cos\left(\frac{	heta}{2}ight) ight) + 2\sin\left(\frac{	heta}{2}ight)$$

Factoring out the $2$, we have:
$$E = 2 \left[ \sqrt{3}\cos\left(\frac{	heta}{2}ight) + \sin\left(\frac{	heta}{2}ight) ight]$$

This expression follows the form $A\cos\alpha + B\sin\alpha$, where the maximum value is given by $\sqrt{A^2 + B^2}$. Here, $A = \sqrt{3}$ and $B = 1$.

The maximum value of the bracketed term is:
$$\sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$$

Finally, multiplying by the $2$ outside the bracket, we find the result:
$$E_{max} = 2 	imes 2 = 4$$

The **greatest value** of the expression is **4**.

Animated Solution for Mathematics - Vector Algebra: If $a$ and $b$ are unit vectors, then the greatest value of $3 ∣ a + b ∣ + ∣ a - b ∣$ is

Enter Numerical Value:

Visualized Solution

Visualizing Unit Vectors

Magnitude of Sum Vector

Substituting Unit Magnitudes

Applying Half-Angle Identity

Magnitude of Difference Vector

Difference Vector Simplification

Forming the Expression

Factoring the Expression

The Maximization Rule

Applying the Rule

Calculating the Final Answer

The Sigma Insight: Addition of Vectors

Analyzing the Setup

The Parallelogram Law

The Trigonometric Transformation

The Harmonic Maximization

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a and b are unit vectors, then the greatest value of 3​∣a+b∣+∣a−b∣ is