Vector Relationships and Matrix Equations Involving Unit Vectors

Given the vectors: $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$ $\vec{b} = \hat{i} + b_2\hat{j} + b_3\hat{k}, \quad b_2, b_3 \in \mathbb{R}$ $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}, \quad c_1, c_2, c_3 \in \mathbb{R}$

We are given that $\vec{a} \cdot \vec{b} = 0$: $\vec{a} \cdot \vec{b} = 3(1) + 1(b_2) + (-1)(b_3) = 3 + b_2 - b_3 = 0 \implies b_3 - b_2 = 3$

We are also given the matrix equation: $\begin{pmatrix} 0 & -c_3 & c_2 \\ c_3 & 0 & -c_1 \\ -c_2 & c_1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ b_2 \\ b_3 \end{pmatrix} = \begin{pmatrix} 3 - c_1 \\ 1 - c_2 \\ -1 - c_3 \end{pmatrix}$

Performing matrix multiplication on the left-hand side gives: $\begin{pmatrix} c_2 b_3 - c_3 b_2 \\ c_3 - c_1 b_3 \\ c_1 b_2 - c_2 \end{pmatrix} = \begin{pmatrix} 3 - c_1 \\ 1 - c_2 \\ -1 - c_3 \end{pmatrix}$

Notice that the vector corresponding to the left-hand side is the cross product $\vec{c} \times \vec{b}$: $\vec{c} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ c_1 & c_2 & c_3 \\ 1 & b_2 & b_3 \end{vmatrix} = (c_2 b_3 - c_3 b_2)\hat{i} + (c_3 - c_1 b_3)\hat{j} + (c_1 b_2 - c_2)\hat{k}$

The right-hand side corresponds to the vector: $(3 - c_1)\hat{i} + (1 - c_2)\hat{j} + (-1 - c_3)\hat{k} = (3\hat{i} + \hat{j} - \hat{k}) - (c_1\hat{i} + c_2\hat{j} + c_3\hat{k}) = \vec{a} - \vec{c}$

Thus, the given matrix equation simplifies to the vector relationship: $\vec{c} \times \vec{b} = \vec{a} - \vec{c} \implies \vec{a} = \vec{c} + \vec{c} \times \vec{b}$

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Step-by-Step Analysis of the Options:

1. Checking Option B: $\vec{b} \cdot \vec{c} = 0$

Taking the dot product with $\vec{b}$ on both sides of $\vec{c} \times \vec{b} = \vec{a} - \vec{c}$: $(\vec{c} \times \vec{b}) \cdot \vec{b} = (\vec{a} - \vec{c}) \cdot \vec{b}$

Since $(\vec{c} \times \vec{b}) \perp \vec{b}$, the left side is $0$: $0 = \vec{a} \cdot \vec{b} - \vec{c} \cdot \vec{b}$

Given that $\vec{a} \cdot \vec{b} = 0$, we get: $\vec{b} \cdot \vec{c} = 0$ Hence, Option B is TRUE.

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2. Checking Option A: $\vec{a} \cdot \vec{c} = 0$

Taking the dot product with $\vec{c}$ on both sides of $\vec{c} \times \vec{b} = \vec{a} - \vec{c}$: $(\vec{c} \times \vec{b}) \cdot \vec{c} = (\vec{a} - \vec{c}) \cdot \vec{c}$ $0 = \vec{a} \cdot \vec{c} - |\vec{c}|^2 \implies \vec{a} \cdot \vec{c} = |\vec{c}|^2$

If $\vec{a} \cdot \vec{c} = 0$, then $|\vec{c}|^2 = 0 \implies \vec{c} = \vec{0}$. Substituting $\vec{c} = \vec{0}$ into $\vec{c} \times \vec{b} = \vec{a} - \vec{c}$ yields $\vec{a} = \vec{0}$, which contradicts $\vec{a} = 3\hat{i} + \hat{j} - \hat{k} \neq \vec{0}$. Therefore, $\vec{c} \neq \vec{0}$ and $\vec{a} \cdot \vec{c} = |\vec{c}|^2 > 0$. Hence, Option A is FALSE.

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3. Checking Option C: $|\vec{b}| > \sqrt{10}$

The magnitude squared of $\vec{b}$ is: $|\vec{b}|^2 = 1^2 + b_2^2 + b_3^2$

Since $b_3 = b_2 + 3$, we have: $|\vec{b}|^2 = 1 + b_2^2 + (b_2 + 3)^2 = 2b_2^2 + 6b_2 + 10$

We are given that $b_2 b_3 > 0 \implies b_2(b_2 + 3) > 0$, which means $b_2 \in (-\infty, -3) \cup (0, \infty)$.

• For $b_2 > 0$: $2b_2^2 + 6b_2 + 10 > 10$.
• For $b_2 < -3$: Let $b_2 = -3 - x$ where $x > 0$. Then, $|\vec{b}|^2 = 2(-3 - x)^2 + 6(-3 - x) + 10 = 2x^2 + 6x + 10 > 10$

In all valid regions of $b_2$, we have $|\vec{b}|^2 > 10$, which implies: $|\vec{b}| > \sqrt{10}$ Hence, Option C is TRUE.

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4. Checking Option D: $|\vec{c}| \le \sqrt{11}$

Starting from $\vec{a} = \vec{c} + \vec{c} \times \vec{b}$, take the square of the magnitude on both sides: $|\vec{a}|^2 = |\vec{c} + \vec{c} \times \vec{b}|^2 = |\vec{c}|^2 + |\vec{c} \times \vec{b}|^2 + 2\vec{c} \cdot (\vec{c} \times \vec{b})$

Since $\vec{c} \cdot (\vec{c} \times \vec{b}) = 0$, we have: $|\vec{a}|^2 = |\vec{c}|^2 + |\vec{c} \times \vec{b}|^2$

From Option B, we know that $\vec{b} \perp \vec{c}$, so $|\vec{c} \times \vec{b}| = |\vec{c}||\vec{b}|$. Thus: $|\vec{a}|^2 = |\vec{c}|^2 + |\vec{c}|^2 |\vec{b}|^2 = |\vec{c}|^2 (1 + |\vec{b}|^2)$

Given $\vec{a} = 3\hat{i} + \hat{j} - \hat{k}$, we calculate $|\vec{a}|^2 = 3^2 + 1^2 + (-1)^2 = 11$. Substituting this yields: $11 = |\vec{c}|^2 (1 + |\vec{b}|^2) \implies |\vec{c}|^2 = \frac{11}{1 + |\vec{b}|^2}$

Since $|\vec{b}|^2 > 10$, we have $1 + |\vec{b}|^2 > 11$. Therefore: $|\vec{c}|^2 < \frac{11}{11} = 1 \implies |\vec{c}| < 1$

Since $|\vec{c}| < 1$, it logically follows that $|\vec{c}| \le \sqrt{11}$. Hence, Option D is TRUE.

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Conclusion:

The correct options are B, C, and D.