Question

Given: \[ \vec{a} = \hat{j} - \hat{k}, \quad \vec{c} = \hat{i} - \hat{j} - \hat{k} \] The vector \( \vec{b} \) satisfies: \[ \vec{a} \times \vec{b} + \vec{c} = \vec{0} \quad \text{and} \quad \vec{a} \cdot \vec{b} = 3 \] Find the vector \( \vec{b} \).

• A. \( \vec{b} = -\hat{i} + \hat{j} - 2\hat{k} \)
• B. \( \vec{b} = \hat{i} + \hat{j} + 2\hat{k} \)
• C. \( \vec{b} = \hat{i} - \hat{j} + 2\hat{k} \)
• D. \( \vec{b} = -\hat{i} + \hat{j} + \hat{k} \)

Hint:

When solving vector equations involving both cross and dot products, translate everything into component form and solve the system of equations step-by-step.

Answer 1

The Correct Option is A

We are given: \[ \vec{a} = \hat{j} - \hat{k} = \langle 0, 1, -1 \rangle, \quad \vec{c} = \hat{i} - \hat{j} - \hat{k} = \langle 1, -1, -1 \rangle \] Also: \[ \vec{a} \times \vec{b} = -\vec{c} \Rightarrow \vec{a} \times \vec{b} = \langle -1, 1, 1 \rangle \] Let: \[ \vec{b} = \langle x, y, z \rangle \] Compute: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
0 & 1 & -1
x & y & z \end{vmatrix} = \hat{i}(1 \cdot z - (-1) \cdot y) - \hat{j}(0 \cdot z - (-1) \cdot x) + \hat{k}(0 \cdot y - 1 \cdot x) = \langle z + y, -x, -x \rangle \] So: \[ \vec{a} \times \vec{b} = \langle z + y, -x, -x \rangle = \langle -1, 1, 1 \rangle \] Compare components: - \( z + y = -1 \) - \( -x = 1 \Rightarrow x = -1 \) - \( -x = 1 \Rightarrow x = -1 \) (consistent) From \( x = -1 \), and \( z + y = -1 \), we can write: \[ z = -1 - y \] Also given: \[ \vec{a} \cdot \vec{b} = 3 \Rightarrow (0)(x) + (1)(y) + (-1)(z) = y - z = 3 \] Substitute \( z = -1 - y \): \[ y - (-1 - y) = 3 \Rightarrow y + 1 + y = 3 \Rightarrow 2y = 2 \Rightarrow y = 1 \] Now: \[ z = -1 - y = -1 - 1 = -2 \] So: \[ \vec{b} = \langle x, y, z \rangle = \langle -1, 1, -2 \rangle \] Final Answer: \[ \boxed{\vec{b} = -\hat{i} + \hat{j} - 2\hat{k}} \]