Existence and Properties of Vector in Three Dimensional Space

To determine the correct statements, we analyze the given conditions step-by-step.

Step 1: Analyze the given vectors and conditions

We are given:

1. $\hat{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ is a unit vector, so $|\hat{u}| = 1$.
2. $\hat{w} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$ is also a unit vector, as $|\hat{w}| = \sqrt{\frac{1^2 + 1^2 + 2^2}{6}} = 1$.
3. $|\hat{u} \times \vec{v}| = 1$
4. $\hat{w} \cdot (\hat{u} \times \vec{v}) = 1$

Let $\vec{z} = \hat{u} \times \vec{v}$. From the given conditions, we have $|\vec{z}| = 1$ and: $\hat{w} \cdot \vec{z} = 1$

Using the definition of the dot product: $\hat{w} \cdot \vec{z} = |\hat{w}||\vec{z}|\cos\theta = 1 \cdot 1 \cdot \cos\theta = 1 \implies \cos\theta = 1$

This implies $\theta = 0$, meaning $\vec{z}$ is parallel to and in the same direction as $\hat{w}$. Since both are unit vectors, we obtain: $\hat{u} \times \vec{v} = \hat{w}$

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Step 2: Determine the number of choices for $\vec{v}$

Since $\hat{u} \times \vec{v} = \hat{w}$, taking the dot product with $\hat{u}$ on both sides yields: $\hat{u} \cdot (\hat{u} \times \vec{v}) = \hat{u} \cdot \hat{w} \implies \hat{u} \cdot \hat{w} = 0$

Now, taking the cross product of $\hat{u}$ with both sides of $\hat{u} \times \vec{v} = \hat{w}$: $\hat{u} \times (\hat{u} \times \vec{v}) = \hat{u} \times \hat{w}$

Using the vector triple product identity $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$: $(\hat{u} \cdot \vec{v})\hat{u} - (\hat{u} \cdot \hat{u})\vec{v} = \hat{u} \times \hat{w}$

Since $|\hat{u}| = 1$, this simplifies to: $c \hat{u} - \vec{v} = \hat{u} \times \hat{w} \implies \vec{v} = \hat{w} \times \hat{u} + c \hat{u}$ where $c = \hat{u} \cdot \vec{v}$ can be any arbitrary real scalar $k \in \mathbb{R}$.

For any $k \in \mathbb{R}$, if we set $\vec{v} = \hat{w} \times \hat{u} + k\hat{u}$, we check: $\hat{u} \times \vec{v} = \hat{u} \times (\hat{w} \times \hat{u} + k\hat{u}) = \hat{u} \times (\hat{w} \times \hat{u}) = (\hat{u} \cdot \hat{u})\hat{w} - (\hat{u} \cdot \hat{w})\hat{u} = 1\hat{w} - 0 = \hat{w}$

Thus, there are infinitely many choices for the vector $\vec{v}$.

• Option (A) is incorrect.
• Option (B) is correct.

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Step 3: Analyze the condition $\hat{u} \cdot \hat{w} = 0$

Using $\hat{u} = u_1\hat{i} + u_2\hat{j} + u_3\hat{k}$ and $\hat{w} = \frac{1}{\sqrt{6}}(\hat{i} + \hat{j} + 2\hat{k})$: $\hat{u} \cdot \hat{w} = \frac{1}{\sqrt{6}}(u_1 + u_2 + 2u_3) = 0 \implies u_1 + u_2 + 2u_3 = 0$

Checking Option (C):

If $\hat{u}$ lies in the $xy$-plane, then its $z$-component is zero, i.e., $u_3 = 0$. Substituting $u_3 = 0$ into the equation: $u_1 + u_2 + 2(0) = 0 \implies u_1 + u_2 = 0 \implies u_1 = -u_2$ Taking the absolute value on both sides gives: $|u_1| = |u_2|$

• Option (C) is correct.

Checking Option (D):

If $\hat{u}$ lies in the $xz$-plane, then its $y$-component is zero, i.e., $u_2 = 0$. Substituting $u_2 = 0$ into the equation: $u_1 + 0 + 2u_3 = 0 \implies u_1 = -2u_3$ Taking the absolute value on both sides gives: $|u_1| = 2|u_3|$ Option (D) incorrectly states $2|u_1| = |u_3|$.

• Option (D) is incorrect.

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Conclusion

The correct options are B and C.