Let u = u1i + u2j + u3k be a unit vector in R³ and w = (1/√6)(i + j + 2k). Given that there exists a

Q: Wait, why do we assume $\hat{u} \times \vec{v} = \hat{w}$? Couldn't the angle be something else?

Great question! We are given $|\hat{u} \times \vec{v}| = 1$ and $\hat{w} \cdot (\hat{u} \times \vec{v}) = 1$. Let $\vec{a} = \hat{u} \times \vec{v}$. We know $|\vec{a}| = 1$ and $|\hat{w}| = 1$. The dot product formula is $\hat{w} \cdot \vec{a} = |\hat{w}||\vec{a}| \cos \theta$. Substituting the values, we get $1 = (1)(1) \cos \theta$, which forces $\cos \theta = 1$. The only way for the cosine of an angle to be 1 is if the angle is $0^\circ$. Thus, $\vec{a}$ and $\hat{w}$ must point in the exact same direction and have the same magnitude, making them identical.

Q: Why are there infinitely many solutions for $\vec{v}$?

Think of the equation $\hat{u} \times \vec{v} = \hat{w}$ as a constraint on the direction and magnitude of $\vec{v}$ perpendicular to $\hat{u}$. The cross product only defines the component of $\vec{v}$ that is perpendicular to $\hat{u}$. The component of $\vec{v}$ that is parallel to $\hat{u}$ (which we represent as $k\hat{u}$) does not affect the cross product because $\hat{u} \times (k\hat{u}) = 0$. Since $k$ can be any real number, we can add any multiple of $\hat{u}$ to a particular solution and still satisfy the original equation.

Q: How do I know if $\hat{u}$ is in the $xy$-plane?

In 3D Cartesian coordinates, any vector $\hat{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ lies in the $xy$-plane if and only if it has no vertical component. Geometrically, this means the $z$-coordinate must be zero. So, setting $u_3 = 0$ is the correct algebraic way to enforce the condition that $\hat{u}$ lies in the $xy$-plane.

Question

Let $\hat{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ be a unit vector in $\mathbb{R}^3$ and $\hat{w} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$. Given that there exists a vector $\vec{v}$ in $\mathbb{R}^3$ such that $|\hat{u} 	imes \vec{v}| = 1$ and $\hat{w} \cdot (\hat{u} 	imes \vec{v}) = 1$. Which of the following statement(s) is (are) correct?

Options:

(A) There is exactly one choice for such $\vec{v}$

(B) There are infinitely many choices for such $\vec{v}$

(C) If $\hat{u}$ lies in the xy-plane then $|u_1| = |u_2|$

(D) If $\hat{u}$ lies in the xz-plane then $2|u_1| = |u_3|$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We begin with a unit vector $\hat{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ and a specific vector $\hat{w} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$.

The problem provides two primary conditions:
1. $|\hat{u} 	imes \vec{v}| = 1$
2. $\hat{w} \cdot (\hat{u} 	imes \vec{v}) = 1$

These conditions define a geometric constraint on the cross product of our vectors.

### The Dot Product Revelation

Let us define a new vector, $\vec{a} = \hat{u} 	imes \vec{v}$. From the first condition, we know the magnitude $|\vec{a}| = 1$.

Calculating the magnitude of $\hat{w}$:
$$|\hat{w}| = \sqrt{\left(\frac{1}{\sqrt{6}}ight)^2 + \left(\frac{1}{\sqrt{6}}ight)^2 + \left(\frac{2}{\sqrt{6}}ight)^2} = \sqrt{\frac{1+1+4}{6}} = 1$$

Using the second condition, $\hat{w} \cdot \vec{a} = 1$, and the definition of the dot product $\hat{w} \cdot \vec{a} = |\hat{w}||\vec{a}| \cos 	heta$, we find:
$$1 = (1)(1) \cos 	heta \Rightarrow \cos 	heta = 1$$

This implies that the angle $	heta$ between $\hat{w}$ and $\vec{a}$ is $0^\circ$. Therefore, the vectors are perfectly aligned, leading to the identity:
$$\hat{u} 	imes \vec{v} = \hat{w}$$

### The Orthogonality Constraint

By the properties of the cross product, the resulting vector $\hat{w}$ must be perpendicular to both $\hat{u}$ and $\vec{v}$. Specifically, $\hat{w} \cdot \hat{u} = 0$.

Substituting the components of $\hat{w}$ and $\hat{u}$:
$$\frac{1}{\sqrt{6}}(1 \cdot u_1 + 1 \cdot u_2 + 2 \cdot u_3) = 0$$

Multiplying by $\sqrt{6}$, we arrive at the fundamental relation:
$$u_1 + u_2 + 2u_3 = 0$$

### The Infinite Solutions

The equation $\hat{u} 	imes \vec{v} = \hat{w}$ is a classic vector equation of the form $\vec{a} 	imes \vec{x} = \vec{b}$. The general solution for $\vec{v}$ is given by:
$$\vec{v} = \frac{\hat{w} 	imes \hat{u}}{|\hat{u}|^2} + k\hat{u}$$

Since $\hat{u}$ is a unit vector ($|\hat{u}|^2 = 1$), this simplifies to:
$$\vec{v} = (\hat{w} 	imes \hat{u}) + k\hat{u}$$

Because $k$ can be any real number, there exists an infinite family of vectors $\vec{v}$ that satisfy the condition. This confirms that **Option (B) is correct**.

### Testing the Planes

If $\hat{u}$ lies in the $xy$-plane, then $u_3 = 0$. Plugging this into our fundamental relation $u_1 + u_2 + 2u_3 = 0$, we get $u_1 + u_2 = 0$, which implies $|u_1| = |u_2|$. This confirms **Option (C)**.

If $\hat{u}$ lies in the $xz$-plane, then $u_2 = 0$. Our relation becomes $u_1 + 2u_3 = 0$, or $|u_1| = 2|u_3|$.

Since Option (D) claims $2|u_1| = |u_3|$, it is mathematically incorrect. We have successfully navigated the geometry and verified the constraints.

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Visualized Solution

Given Vectors and Conditions

Magnitude of $\overset{w}{^}$

Analyzing the Dot Product

Deducing $\overset{u}{^} \times v$

Orthogonality of $\overset{u}{^}$ and $\overset{w}{^}$

Expanding $\overset{w}{^} \cdot \overset{u}{^} = 0$

Solving for $v$

Infinitely Many Choices for $v$

Case: $\overset{u}{^}$ in $x y$ -plane

Verifying Option C

Case: $\overset{u}{^}$ in $x z$ -plane

Verifying Option D

Final Conclusion

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Dot Product Revelation

The Orthogonality Constraint

The Infinite Solutions

Testing the Planes

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Visualized Solution

Given Vectors and Conditions

Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w^

Analyzing the Dot Product

Deducing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u^×v

Orthogonality of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u^ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w^

Expanding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } w^⋅u^=0

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Infinitely Many Choices for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v

Case: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u^ in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } xy-plane