Question

If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{j} - \hat{k}$ and $\vec{a} \times \vec{c} = \vec{b}$, $\vec{a} \cdot \vec{c} = 3$, then $\vec{c}$ is

• A. $\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}$
• B. $\frac{5}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$
• C. $\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$
• D. $\frac{5}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}$

Hint:

When given $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = \text{scalar}$, crossing $\vec{a}$ with the first equation is the standard and fastest algorithm to isolate $\vec{c}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We are given two vector equations involving an unknown vector $\vec{c}$: a cross product equation and a dot product equation. We can find $\vec{c}$ by using the vector triple product property.

Step 2: Key Formula or Approach:
Take the cross product of vector $\vec{a}$ with both sides of the equation $\vec{a} \times \vec{c} = \vec{b}$: $\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b}$ Expand the left side using the vector triple product formula: $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$. So, $(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} = \vec{a} \times \vec{b}$. Substitute the known values ($\vec{a} \cdot \vec{c} = 3$, calculate $\vec{a} \cdot \vec{a}$ and $\vec{a} \times \vec{b}$) to solve for $\vec{c}$.

Step 3: Detailed Explanation:
Given: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ $\vec{b} = 0\hat{i} + \hat{j} - \hat{k}$ $\vec{a} \times \vec{c} = \vec{b}$ $\vec{a} \cdot \vec{c} = 3$ From the vector triple product expansion: $(\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c} = \vec{a} \times \vec{b}$ Let's calculate the required components: 1. $\vec{a} \cdot \vec{a} = |\vec{a}|^2 = (1)^2 + (1)^2 + (1)^2 = 3$ 2. Calculate $\vec{a} \times \vec{b}$ using the determinant method: \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} \] \[ = \hat{i}(-1 - 1) - \hat{j}(-1 - 0) + \hat{k}(1 - 0) \] \[ = -2\hat{i} + \hat{j} + \hat{k} \] Now substitute these into our expanded equation: $(3)(\hat{i} + \hat{j} + \hat{k}) - (3)\vec{c} = -2\hat{i} + \hat{j} + \hat{k}$ $3\hat{i} + 3\hat{j} + 3\hat{k} - 3\vec{c} = -2\hat{i} + \hat{j} + \hat{k}$ Rearrange to solve for $3\vec{c}$: $3\vec{c} = (3\hat{i} + 3\hat{j} + 3\hat{k}) - (-2\hat{i} + \hat{j} + \hat{k})$ $3\vec{c} = (3 - (-2))\hat{i} + (3 - 1)\hat{j} + (3 - 1)\hat{k}$ $3\vec{c} = 5\hat{i} + 2\hat{j} + 2\hat{k}$ Divide by 3: $\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$

Step 4: Final Answer:
The vector $\vec{c}$ is $\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.