Question

If \( \overline{a} = 2\overline{i} - 3\overline{j} + 5\overline{k} \) and \( \overline{b} = -\overline{i} + 3\overline{j} + 3\overline{k} \) are two vectors, then the vector of magnitude 28 units in the direction of the vector \( \overline{a} - \overline{b} \) is:

• A. $ 3\overline{i} + 6\overline{j} - 2\overline{k} $
• B. $ 12\overline{i} - 24\overline{j} + 8\overline{k} $
• C. $ 3\overline{i} - 6\overline{j} - 2\overline{k} $
• D. $ 12\overline{i} + 24\overline{j} - 8\overline{k} $

Hint:

To find a vector of a specific magnitude in the direction of another vector, first compute the unit vector in that direction, then scale it by the desired magnitude.

Answer 1

The Correct Option is B

Step 1: Compute $ \overline{a} - \overline{b} $.
Given: $$ \overline{a} = 2\overline{i} - 3\overline{j} + 5\overline{k}, \quad \overline{b} = -\overline{i} + 3\overline{j} + 3\overline{k}. $$ Subtract $ \overline{b} $ from $ \overline{a} $: $$ \overline{a} - \overline{b} = (2\overline{i} - 3\overline{j} + 5\overline{k}) - (-\overline{i} + 3\overline{j} + 3\overline{k}). $$ Simplify: $$ \overline{a} - \overline{b} = (2 + 1)\overline{i} + (-3 - 3)\overline{j} + (5 - 3)\overline{k} = 3\overline{i} - 6\overline{j} + 2\overline{k}. $$ Step 2: Find the unit vector in the direction of $ \overline{a} - \overline{b} $.
The magnitude of $ \overline{a} - \overline{b} $ is: $$ |\overline{a} - \overline{b}| = \sqrt{(3)^2 + (-6)^2 + (2)^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7. $$ The unit vector in the direction of $ \overline{a} - \overline{b} $ is: $$ \frac{\overline{a} - \overline{b}}{|\overline{a} - \overline{b}|} = \frac{3\overline{i} - 6\overline{j} + 2\overline{k}}{7}. $$ Step 3: Scale the unit vector to magnitude 28.
To get a vector of magnitude 28 in the same direction, multiply the unit vector by 28: $$ 28 \cdot \frac{3\overline{i} - 6\overline{j} + 2\overline{k}}{7} = 4(3\overline{i} - 6\overline{j} + 2\overline{k}) = 12\overline{i} - 24\overline{j} + 8\overline{k}. $$ Step 4: Final Answer.
$$ \boxed{12\overline{i} - 24\overline{j} + 8\overline{k}} $$