Question

Given $ \mathbf{a} = 2\hat{i} + \hat{j} - \hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j}, \quad \mathbf{c} = 5\hat{i} - \hat{j} + \hat{k},$ then the unit vector parallel to $\mathbf{a} + \mathbf{b} - \mathbf{c} $ but in the opposite direction is

• A. \( \frac{1}{3} \left( 2\hat{i} - \hat{j} + 2\hat{k} \right) \)
• B. \( \frac{1}{2} \left( 2\hat{i} - \hat{j} + 2\hat{k} \right) \)
• C. \( \frac{1}{3} \left( 2\hat{i} - \hat{j} - 2\hat{k} \right) \)
• D. None of these

Hint:

To find a unit vector in the opposite direction, simply negate the vector and then divide it by its magnitude. This ensures that the resulting vector has a magnitude of 1.

Answer 1

The Correct Option is A

We are given three vectors: \[ \mathbf{a} = 2\hat{i} + \hat{j} - \hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j}, \quad \mathbf{c} = 5\hat{i} - \hat{j} + \hat{k} \] The task is to find the unit vector parallel to the vector \( \mathbf{a} + \mathbf{b} - \mathbf{c} \), but in the opposite direction.
Step 1: Find \( \mathbf{a} + \mathbf{b} - \mathbf{c} \)
First, add \( \mathbf{a} \) and \( \mathbf{b} \), then subtract \( \mathbf{c} \): \[ \mathbf{a} + \mathbf{b} - \mathbf{c} = \left( 2\hat{i} + \hat{j} - \hat{k} \right) + \left( \hat{i} - \hat{j} \right) - \left( 5\hat{i} - \hat{j} + \hat{k} \right) \] Now combine the components of \( \hat{i}, \hat{j}, \hat{k} \): \[ \mathbf{a} + \mathbf{b} - \mathbf{c} = \left( 2\hat{i} + \hat{i} - 5\hat{i} \right) + \left( \hat{j} - \hat{j} - (-\hat{j}) \right) + \left( -\hat{k} - \hat{k} \right) \] \[ = -2\hat{i} + \hat{j} - 2\hat{k} \]
Step 2: Find the unit vector in the opposite direction
The unit vector in the opposite direction is the negative of the vector \( \mathbf{a} + \mathbf{b} - \mathbf{c} \), normalized by its magnitude. So we have: \[ \mathbf{v} = -\left( -2\hat{i} + \hat{j} - 2\hat{k} \right) = 2\hat{i} - \hat{j} + 2\hat{k} \] Now, find the magnitude of \( \mathbf{v} \): \[ |\mathbf{v}| = \sqrt{(2)^2 + (-1)^2 + (2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \] Thus, the unit vector is: \[ \hat{v} = \frac{1}{3} \left( 2\hat{i} - \hat{j} + 2\hat{k} \right) \] Thus, the unit vector parallel to \( \mathbf{a} + \mathbf{b} - \mathbf{c} \) but in the opposite direction is \( \frac{1}{3} \left( 2\hat{i} - \hat{j} + 2\hat{k} \right) \).

Answer 2

To find the unit vector parallel to \( \mathbf{a} + \mathbf{b} - \mathbf{c} \) in the opposite direction, we first compute the vector \( \mathbf{a} + \mathbf{b} - \mathbf{c} \).

Given: \[ \mathbf{a} = 2\hat{i} + \hat{j} - \hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j}, \quad \mathbf{c} = 5\hat{i} - \hat{j} + \hat{k} \]

Calculate \( \mathbf{a} + \mathbf{b} \):

\[ \mathbf{a} + \mathbf{b} = (2\hat{i} + \hat{j} - \hat{k}) + (\hat{i} - \hat{j}) = (2+1)\hat{i} + (1-1)\hat{j} + (-1)\hat{k} = 3\hat{i} + 0\hat{j} - \hat{k} \]

Now subtract \( \mathbf{c} \):

\[ \mathbf{a} + \mathbf{b} - \mathbf{c} = (3\hat{i} - \hat{k}) - (5\hat{i} - \hat{j} + \hat{k}) = (3-5)\hat{i} + (0+1)\hat{j} + (-1-1)\hat{k} = -2\hat{i} + \hat{j} - 2\hat{k} \]

Find magnitude of \( -2\hat{i} + \hat{j} - 2\hat{k} \):

\[ \|\mathbf{v}\| = \sqrt{(-2)^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

The unit vector in the opposite direction is obtained by scaling by -1/3:

\[-\frac{1}{3}(-2\hat{i} + \hat{j} - 2\hat{k}) = \frac{1}{3}(2\hat{i} - \hat{j} + 2\hat{k})\]

Therefore, the unit vector parallel to \( \mathbf{a} + \mathbf{b} - \mathbf{c} \) but in the opposite direction is \(\frac{1}{3} \left( 2\hat{i} - \hat{j} + 2\hat{k} \right)\).