Question

Let \( \vec{a} = 2\hat{i} + 2\hat{j} + \hat{k} \) and \( \vec{b} \) be another vector such that \( \vec{a} \cdot \vec{b} = 14 \) and \( \vec{a} \times \vec{b} = 3\hat{i} + \hat{j} - 8\hat{k} \). The vector \( \vec{b} \) is

• A. \( 5\hat{i} + \hat{j} + 2\hat{k} \)
• B. \( 5\hat{i} - \hat{j} - 2\hat{k} \)
• C. \( 5\hat{i} + \hat{j} - 2\hat{k} \)
• D. \( 3\hat{i} + \hat{j} + 4\hat{k} \)

Hint:

When given both the dot product and cross product of an unknown vector \( \vec{b} \) with a known vector \( \vec{a} \), set up \( \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \). The dot product gives one linear equation, and the cross product gives a system of two independent linear equations. Solve this system to find the components \( x, y, z \).

Answer 1

The Correct Option is A

We are given \( \vec{a} = 2\hat{i} + 2\hat{j} + \hat{k} \), \( \vec{a} \cdot \vec{b} = 14 \), and \( \vec{a} \times \vec{b} = 3\hat{i} + \hat{j} - 8\hat{k} \). Let the unknown vector be \( \vec{b} = x\hat{i} + y\hat{j} + z\hat{k} \). From the dot product: \[ \vec{a} \cdot \vec{b} = (2\hat{i} + 2\hat{j} + \hat{k}) \cdot (x\hat{i} + y\hat{j} + z\hat{k}) = 2x + 2y + z = 14 \cdots(1) \] From the cross product: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 2 & 1 \\ x & y & z \end{vmatrix} = (2z - y)\hat{i} - (2z - x)\hat{j} + (2y - 2x)\hat{k} \] \[ \vec{a} \times \vec{b} = (2z - y)\hat{i} + (x - 2z)\hat{j} + (2y - 2x)\hat{k} \] We are given \( \vec{a} \times \vec{b} = 3\hat{i} + \hat{j} - 8\hat{k} \). By comparing the components, we get a system of equations: \[ 2z - y = 3 \implies y = 2z - 3 \cdots(2) \] \[ x - 2z = 1 \implies x = 2z + 1 \cdots(3) \] \[ 2y - 2x = -8 \implies y - x = -4 \cdots(4) \] Let's check if equations (2) and (3) are consistent with equation (4): \[ (2z - 3) - (2z + 1) = 2z - 3 - 2z - 1 = -4 \]. This is consistent. Now, substitute the expressions for \( x \) and \( y \) from (2) and (3) into equation (1): \[ 2(2z + 1) + 2(2z - 3) + z = 14 \] \[ 4z + 2 + 4z - 6 + z = 14 \] \[ 9z - 4 = 14 \] \[ 9z = 18 \implies z = 2 \] Now find \( x \) and \( y \): \[ x = 2z + 1 = 2(2) + 1 = 5 \] \[ y = 2z - 3 = 2(2) - 3 = 1 \] So, the vector \( \vec{b} \) is \( 5\hat{i} + 1\hat{j} + 2\hat{k} \).