Question

The position vectors of the vertices \( A, B, C \) of a triangle are given as: \[ \vec{A} = \hat{i} - 2\hat{j} + k, \quad \vec{B} = 2\hat{i} + \hat{j} - k, \quad \vec{C} = \hat{i} - \hat{j} - 2k \] If \( D \) and \( E \) are the midpoints of \( BC \) and \( CA \) respectively, then the unit vector along \( \overrightarrow{DE} \) is:

• A. \( \frac{1}{7} (3\hat{i} - 2\hat{j} + 6\hat{k}) \)
• B. \( \frac{1}{\sqrt{14}} (-\hat{i} - 3\hat{j} + 2\hat{k}) \)
• C. \( \frac{1}{\sqrt{3}} (\hat{i} - \hat{j} - \hat{k}) \)
• D. \( \frac{1}{13} (12\hat{i} + 3\hat{j} + 4\hat{k}) \)

Hint:

To find the unit vector along a segment, compute the midpoint vectors, determine the segment direction, and normalize it.

Answer 1

The Correct Option is B

Step 1: Finding midpoints of \( BC \) and \( CA \).
The midpoint \( D \) of \( BC \) is: \[ \vec{D} = \frac{\vec{B} + \vec{C}}{2} \] \[ = \frac{(2\hat{i} + \hat{j} - \hat{k}) + (\hat{i} - \hat{j} - 2\hat{k})}{2} \] \[ = \frac{(3\hat{i} + 0\hat{j} - 3\hat{k})}{2} = \frac{3}{2} \hat{i} - \frac{3}{2} \hat{k} \] Similarly, the midpoint \( E \) of \( CA \) is: \[ \vec{E} = \frac{\vec{C} + \vec{A}}{2} \] \[ = \frac{(\hat{i} - \hat{j} - 2\hat{k}) + (\hat{i} - 2\hat{j} + k)}{2} \] \[ = \frac{(2\hat{i} - 3\hat{j} - \hat{k})}{2} = \hat{i} - \frac{3}{2} \hat{j} - \frac{1}{2} \hat{k} \] Step 2: Finding \( \overrightarrow{DE} \).
\[ \overrightarrow{DE} = \vec{E} - \vec{D} \] \[ = \left( \hat{i} - \frac{3}{2} \hat{j} - \frac{1}{2} \hat{k} \right) - \left( \frac{3}{2} \hat{i} - \frac{3}{2} \hat{k} \right) \] \[ = \left( -\frac{1}{2} \hat{i} - \frac{3}{2} \hat{j} + \frac{2}{2} \hat{k} \right) \] Step 3: Finding the unit vector.
The magnitude of \( \overrightarrow{DE} \) is: \[ \left| \overrightarrow{DE} \right| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 + \left(\frac{2}{2}\right)^2} \] \[ = \sqrt{\frac{1}{4} + \frac{9}{4} + \frac{4}{4}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2} \] Thus, the unit vector is: \[ \frac{1}{\sqrt{14}} (-\hat{i} - 3\hat{j} + 2\hat{k}) \]