Question

Let \(a=i+j+2k\) and \(b=i-2j+3k\) be two vectors. Then the unit vector in the direction of \(a-b\) is

• A. \(\dfrac{1}{√10}(2j-3k)\)
• B. \(\dfrac{1}{√10}(3j-k)\)
• C. \((3j-k)\)
• D. \(\dfrac{1}{√5}(2j-3k)\)
• E. \(\dfrac{-1}{√5}(2j-3k)\)

Answer 1

The Correct Option is B

Step 1: Compute the vector \(\vec{a} - \vec{b}\). Given: \[ \vec{a} = \vec{i} + \vec{j} + 2\vec{k} \] \[ \vec{b} = \vec{i} - 2\vec{j} + 3\vec{k} \] Subtract \(\vec{b}\) from \(\vec{a}\): \[ \vec{a} - \vec{b} = (\vec{i} - \vec{i}) + (\vec{j} - (-2\vec{j})) + (2\vec{k} - 3\vec{k}) = 0\vec{i} + 3\vec{j} - \vec{k} \] Simplified: \[ \vec{a} - \vec{b} = 3\vec{j} - \vec{k} \]

Step 2: Find the magnitude of \(\vec{a} - \vec{b}\). \[ |\vec{a} - \vec{b}| = \sqrt{0^2 + 3^2 + (-1)^2} = \sqrt{0 + 9 + 1} = \sqrt{10} \]

Step 3: Compute the unit vector in the direction of \(\vec{a} - \vec{b}\). Divide the vector by its magnitude: \[ \text{Unit vector} = \frac{\vec{a} - \vec{b}}{|\vec{a} - \vec{b}|} = \frac{3\vec{j} - \vec{k}}{\sqrt{10}} = \frac{1}{\sqrt{10}}(3\vec{j} - \vec{k}) \]

Conclusion: The correct answer is \(\boxed{B}\) \(\left( \frac{1}{\sqrt{10}}(3\vec{j} - \vec{k}) \right)\).

Answer 2

First, find the vector a - b:

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

Next, find the magnitude of a - b:

\[ \|\mathbf{a} - \mathbf{b}\| = \sqrt{(0^2 + 3^2 + (-1)^2)} = \sqrt{9 + 1} = \sqrt{10} \]

Finally, to find the unit vector in the direction of a - b, divide the vector by its magnitude:

\[ \text{Unit vector} = \frac{\mathbf{a} - \mathbf{b}}{\|\mathbf{a} - \mathbf{b}\|} = \frac{3\mathbf{j} - \mathbf{k}}{\sqrt{10}} = \frac{1}{\sqrt{10}}(3\mathbf{j} - \mathbf{k}) \]

Therefore, the unit vector in the direction of a - b is \( \frac{1}{\sqrt{10}}(3\mathbf{j} - \mathbf{k}) \).