Let a and b be two unit vectors such that the angle between them is π/3 If λa + 2b and 3a - λb are p

Q: Wait, why is the dot product zero? I thought it was just for finding angles.

The dot product is the ultimate test for orthogonality! If two non-zero vectors are perpendicular, their projection onto each other is zero, hence the dot product is zero. It is the most elegant way to enforce a $90^\circ$ angle without needing to calculate coordinates.

Q: Can I just multiply these like normal binomials? Like $(x+y)(z-w)$?

Almost! You can use the distributive property, but you must respect the vector nature. You cannot simply write $\vec{a} \cdot \vec{b}$ as $ab$. You must keep the dot product operator. And remember, $\vec{a} \cdot \vec{a}$ is the square of the magnitude, not just $a^2$.

Q: Why did we reject the roots? They are valid numbers!

They are valid numbers, yes, but are they valid solutions to the constraint? The problem specifically asked for values of $\lambda$ in the interval $[-1, 3]$. Always check your final answers against the domain constraints—it is the final gatekeeper of a correct JEE solution.

Question

Let $\vec{a}$ and $\vec{b}$ be two unit vectors such that the angle between them is $\frac{\pi}{3}$ If $\lambda\vec{a}+2\vec{b}$ and $3\vec{a}-\lambda\vec{b}$ are perpendicular to each other, then the number of values of $\lambda$ in $[-1,3]$ is:

Options:

(A) 2

(B) 1

(C) 0

(D) 3

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a 3D space, holding two unit vectors, $\vec{a}$ and $\vec{b}$. They are like two arrows of length $1$, starting from the same origin, locked in a dance at an angle of $\frac{\pi}{3}$ (or $60^\circ$).

This is the foundation of our problem. We are finding the specific conditions under which two complex linear combinations of these vectors become perfectly perpendicular.

### The Dot Product: Our Mathematical Compass

When we are asked to enforce perpendicularity, the dot product is our most powerful tool. The dot product of two vectors $\vec{u}$ and $\vec{v}$ is defined as $|\vec{u}||\vec{v}| \cos 	heta$.

If the angle $	heta$ is $90^\circ$, then $\cos(90^\circ) = 0$, and the entire product vanishes. We are given two new vectors:
$$\vec{u} = \lambda\vec{a} + 2\vec{b}$$
$$\vec{v} = 3\vec{a} - \lambda\vec{b}$$

For these to be perpendicular, their dot product must be zero:
$$(\lambda\vec{a} + 2\vec{b}) \cdot (3\vec{a} - \lambda\vec{b}) = 0$$

### Expanding the Horizon

We expand this expression using the distributive property:
$$\lambda\vec{a} \cdot (3\vec{a}) + \lambda\vec{a} \cdot (-\lambda\vec{b}) + 2\vec{b} \cdot (3\vec{a}) + 2\vec{b} \cdot (-\lambda\vec{b}) = 0$$

As we simplify, remember that $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = 1$ and $\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1$. Furthermore, the dot product of the unit vectors is:
$$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\left(\frac{\pi}{3}ight) = (1)(1)\left(\frac{1}{2}ight) = \frac{1}{2}$$

Substituting these values, the equation transforms into:
$$3\lambda - \lambda^2\left(\frac{1}{2}ight) + 6\left(\frac{1}{2}ight) - 2\lambda = 0$$
$$3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda = 0$$

### The Quadratic Resolution

Simplifying further, we obtain:
$$\lambda - \frac{\lambda^2}{2} + 3 = 0$$

Multiplying by $-2$ to clear the fraction, we arrive at the standard quadratic form:
$$\lambda^2 - 2\lambda - 6 = 0$$

Using the quadratic formula $\lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, we find:
$$\lambda = \frac{2 \pm \sqrt{4 - 4(1)(-6)}}{2} = \frac{2 \pm \sqrt{28}}{2} = 1 \pm \sqrt{7}$$

### The Final Gatekeeper

We evaluate the roots: $\lambda_1 = 1 + \sqrt{7} \approx 3.64$ and $\lambda_2 = 1 - \sqrt{7} \approx -1.64$.

The problem asks for the number of values of $\lambda$ in the interval $[-1, 3]$. We check our roots against these constraints:
1. $3.64 > 3$ (Outside the interval)
2. $-1.64 < -1$ (Outside the interval)

Neither root fits within the specified range. Thus, the number of valid values is exactly **0**.

Animated Solution for Mathematics - Vector Algebra: Let $a$ and $b$ be two unit vectors such that the angle between them is $\frac{π}{3}$ If $λ a + 2 b$ and $3 a - λ b$ are perpendicular to each other, then the number of values of $λ$ in $[- 1, 3]$ is:

Select Answer:

Visualized Solution

Identify Unit Vectors

Calculate Dot Product $a \cdot b$

State Perpendicularity Condition

Expand the Dot Product

Substitute Known Values

Simplify the Equation

Form the Quadratic Equation

Solve Using Quadratic Formula

Simplify the Roots

Approximate the Values

Check the Interval $[- 1, 3]$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Dot Product

Expanding the Horizon

The Quadratic Resolution

The Final Gatekeeper

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