If u and v are unit vectors and θ is the acute angle between them, then 2u × 3v is a unit vector for

Q: Wait, why does the scalar $2$ and $3$ multiply to $6$? Shouldn't it be different?

That is a fantastic question! It comes down to the linearity of the cross product. When you have a scalar $k$ multiplying a vector $\vec{a}$, the magnitude $|k\vec{a}|$ is simply $|k| |\vec{a}|$. So, $|2\vec{u} \times 3\vec{v}|$ becomes $|2| |3| |\vec{u} \times \vec{v}|$. Since $2$ and $3$ are positive, they just pull out as $2 \times 3 = 6$. It is a beautiful property that simplifies our lives immensely!

Q: Why is the answer 'exactly one' and not 'none'? How do we know $1/6$ is a valid value for $\sin\theta$?

Great catch! We must always check the range. The sine function, $\sin\theta$, maps the interval $(0, \frac{\pi}{2})$ to the interval $(0, 1)$. Since $\frac{1}{6}$ is clearly between $0$ and $1$, there must be a unique angle $\theta$ that satisfies this. If the value had been, say, $2$, we would have had 'no value of $\theta$' because $\sin\theta$ can never exceed $1$.

Q: What if the problem didn't say $\theta$ was acute? Would the answer change?

Absolutely! If $\theta$ were not restricted to the acute range, we would have to consider the entire domain of the sine function. For $\sin\theta = \frac{1}{6}$, there would be infinitely many solutions (like $\theta = \arcsin(1/6) + 2n\pi$ or $\pi - \arcsin(1/6) + 2n\pi$). The 'acute' constraint is the gatekeeper that restricts us to exactly one solution.

Question

If $\vec{u}$ and $\vec{v}$ are unit vectors and $	heta$ is the acute angle between them, then $2\vec{u} 	imes 3\vec{v}$ is a unit vector for

Options:

(A) no value of $	heta$

(B) exactly one value of $	heta$

(C) exactly two values of $	heta$

(D) more than two values of $	heta$

Sigma Percentile · Accepted Answer

### The Vector Dance: Unlocking the Geometry of Cross Products

Welcome, fellow traveler on the path of JEE mastery! Today, we are not just solving a problem; we are peeling back the layers of a geometric reality.

We are looking at two unit vectors, $\vec{u}$ and $\vec{v}$, dancing in space, and we are asked to find the conditions under which their cross product, scaled by $2$ and $3$, behaves like a unit vector. It sounds simple, but the beauty lies in the details.

### Phase 1: The Geometry of the Cross Product

Imagine you are standing in a 3D space. You have two vectors, $\vec{u}$ and $\vec{v}$, originating from the same point. Because they are unit vectors, their lengths are fixed at $1$.

The angle between them, $	heta$, is the variable that dictates the 'spread' of these vectors. When we talk about the cross product $\vec{u} 	imes \vec{v}$, we are talking about a vector whose magnitude represents the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$.

Mathematically, the magnitude of the cross product is given by:

$$|\vec{u} 	imes \vec{v}| = |\vec{u}| |\vec{v}| \sin	heta$$

Since $|\vec{u}| = 1$ and $|\vec{v}| = 1$, this simplifies beautifully to $|\vec{u} 	imes \vec{v}| = \sin	heta$. This is our geometric foundation.

### Phase 2: The Algebraic Journey

Now, let us introduce the scalars. We are dealing with the vector $2\vec{u} 	imes 3\vec{v}$. The problem tells us this resultant vector is a unit vector, meaning its magnitude is $1$.

So, we set up our equation:

$$|2\vec{u} 	imes 3\vec{v}| = 1$$

Using the properties of the cross product, we know that scalars can be pulled out of the magnitude operation. Specifically, $|k\vec{a} 	imes m\vec{b}| = |k| |m| |\vec{a} 	imes \vec{b}|$.

Applying this to our expression, we get:

$$|2\vec{u} 	imes 3\vec{v}| = |2| |3| |\vec{u} 	imes \vec{v}| = 6 |\vec{u} 	imes \vec{v}|$$

Substituting our geometric identity $|\vec{u} 	imes \vec{v}| = \sin	heta$, we arrive at the elegant equation:

$$6 \sin	heta = 1$$

This is the heart of the problem. We have reduced a complex vector expression down to a simple trigonometric equation: $\sin	heta = \frac{1}{6}$.

### Phase 3: The Constraint Check

This is where many students stumble, but you will not. We are given the constraint that $	heta$ is an acute angle, meaning $0 < 	heta < \frac{\pi}{2}$.

In this interval, the sine function is strictly increasing. It starts at $0$ when $	heta = 0$ and climbs to $1$ when $	heta = \frac{\pi}{2}$.

Since our value $\frac{1}{6}$ lies strictly between $0$ and $1$, there must be exactly one unique angle $	heta$ in the first quadrant that satisfies $\sin	heta = \frac{1}{6}$. If the value were outside this range, we would have no solution.

### Conclusion

Isn't it satisfying? We started with vectors in 3D space, applied the properties of cross products, and used the constraints of the domain to arrive at a definitive answer.

The answer is **exactly one value of $	heta$**. Remember, in JEE Advanced, the math is not just about calculation; it is about understanding the constraints and the behavior of functions. Keep visualizing, keep questioning, and keep falling in love with the physics behind the math!

Animated Solution for Mathematics - Vector Algebra: If $u$ and $v$ are unit vectors and $θ$ is the acute angle between them, then $2 u \times 3 v$ is a unit vector for

Select Answer:

Visualized Solution

Introduction to Vectors $u$ and $v$

Unit Vector Magnitudes

The Resultant Vector $2 u \times 3 v$

Cross Product Magnitude Formula

Applying the Formula

Simplifying Scalar Multiples

Scalar Multiplication

Substituting Unit Magnitudes

Setting the Unit Vector Condition

Solving for $sin θ$

Domain Analysis: Acute Angle

Final Conclusion

The Sigma Insight: Vector (Cross) Product

The Vector Dance

Phase 1

Phase 2

Phase 3

Conclusion

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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 u and v are unit vectors and θ is the acute angle between them, then 2u×3v is a unit vector for