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

To determine the vector \(\vec{c}\) that satisfies the given conditions, we will solve the problem step-by-step:

Given:

• \(\vec{a} = \hat{i} + \hat{j} + \hat{k}\)
• \(\vec{b} = \hat{j} - \hat{k}\)
• \(\vec{a} \times \vec{c} = \vec{b}\)
• \(\vec{a} \cdot \vec{c} = 3\)

Objective: Find \(\vec{c}\).

Let's assume \(\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}\).

Step 1: Calculate \(\vec{a} \times \vec{c}\)

The cross product \(\vec{a} \times \vec{c}\) can be computed using the determinant:

\(\vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{vmatrix}\)

Expanding the determinant:

\(\vec{a} \times \vec{c} = \hat{i}(1 \cdot z - 1 \cdot y) - \hat{j}(1 \cdot z - 1 \cdot x) + \hat{k}(1 \cdot y - 1 \cdot x)\)

Simplifying, we get:

\(\vec{a} \times \vec{c} = (z - y)\hat{i} - (z - x)\hat{j} + (y - x)\hat{k}\)

Step 2: Set \(\vec{a} \times \vec{c} = \vec{b}\) to find equations

Equating to \(\vec{b} = \hat{j} - \hat{k}\), we have:

• \(z - y = 0\)
• \(-(z - x) = 1 \Rightarrow z - x = -1\)
• \(y - x = -1\)

Step 3: Solve the system of equations

From \(z - y = 0\), we have \(z = y\).

Substituting \(z = y\) into \(z - x = -1\) gives:

\(y - x = -1 \Rightarrow y = x - 1\)

Use \(y - x = -1\): \((x - 1) - x = -1\) (true for all \(x\))

Step 4: Use \(\vec{a} \cdot \vec{c} = 3\)

Substitute \(\vec{c} = x\hat{i} + (x - 1)\hat{j} + (x - 1)\hat{k}\) into the dot product:

\(\vec{a} \cdot \vec{c} = (1 \cdot x) + (1 \cdot (x - 1)) + (1 \cdot (x - 1)) = 3\)

Thus, \(x + x - 1 + x - 1 = 3\).

Then, \(3x - 2 = 3 \Rightarrow 3x = 5 \Rightarrow x = \frac{5}{3}\).

Step 5: Find \(\vec{c}\) components

So, \(x = \frac{5}{3}\), \(y = x - 1 = \frac{2}{3}\), \(z = \frac{2}{3}\).

Thus, \(\vec{c} = \frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\).

Conclusion: The correct option is \(\frac{5}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}\).