Question

Given the position vectors of points \(A\) and \(B\) as \( \mathbf{A} = 2\mathbf{i} - 3\mathbf{j} + \mathbf{k} \) and \( \mathbf{B} = \mathbf{i} + 2\mathbf{j} - 3\mathbf{k} \), and \(C\) divides \(AB\) in the ratio 3:2. If \( \mathbf{D} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k} \) is the position vector of point \(D\), find the unit vector in the direction of \( \mathbf{CD} \):

• A. \(\frac{1}{\sqrt{7}}(8\mathbf{i} - 5\mathbf{j} - 3\mathbf{k})\)
• B. \(\frac{1}{\sqrt{266}}(4\mathbf{i} - 13\mathbf{j} + 9\mathbf{k})\)
• C. \(\frac{1}{\sqrt{42}}(8\mathbf{i} - 5\mathbf{j} + 17\mathbf{k})\)
• D. \(\frac{1}{\sqrt{7}}(8\mathbf{i} - 5\mathbf{j} + 3\mathbf{k})\)

Hint:

When solving vector problems involving the section formula, be sure to correctly apply the formula for dividing a line segment and use appropriate vector operations to find the unit vector.

Answer 1

The Correct Option is C

Step 1: Find the position vector of \(C\) using the section formula. Since \(C\) divides \(AB\) in the ratio \(3:2\), we use the section formula to find the position vector of \(C\): \[ \mathbf{C} = \frac{2\mathbf{A} + 3\mathbf{B}}{5} \] Substitute the position vectors of \(A\) and \(B\): \[ \mathbf{C} = \frac{2(2\mathbf{i} - 3\mathbf{j} + \mathbf{k}) + 3(\mathbf{i} + 2\mathbf{j} - 3\mathbf{k})}{5} \] \[ \mathbf{C} = \frac{(4\mathbf{i} - 6\mathbf{j} + 2\mathbf{k}) + (3\mathbf{i} + 6\mathbf{j} - 9\mathbf{k})}{5} \] \[ \mathbf{C} = \frac{7\mathbf{i} - 7\mathbf{k}}{5} \] Thus, the position vector of \(C\) is: \[ \mathbf{C} = \frac{7}{5}\mathbf{i} - \frac{7}{5}\mathbf{k} \] 

Step 2: Find the vector \( \mathbf{CD} \). The vector \( \mathbf{CD} \) is found by subtracting the position vector of \(C\) from the position vector of \(D\): \[ \mathbf{CD} = \mathbf{D} - \mathbf{C} = (3\mathbf{i} - \mathbf{j} + 2\mathbf{k}) - \left(\frac{7}{5}\mathbf{i} - \frac{7}{5}\mathbf{k}\right) \] \[ \mathbf{CD} = \left(3 - \frac{7}{5}\right)\mathbf{i} - \mathbf{j} + \left(2 - \left(-\frac{7}{5}\right)\right)\mathbf{k} \] 

Step 3: Find the magnitude of \( \mathbf{CD} \). The magnitude of \( \mathbf{CD} \) is: \[ |\mathbf{CD}| = \sqrt{\left(\frac{8}{5}\right)^2 + (-1)^2 + \left(\frac{17}{5}\right)^2} \] \[ |\mathbf{CD}| = \sqrt{\frac{64}{25} + 1 + \frac{289}{25}} = \sqrt{\frac{64 + 25 + 289}{25}} = \sqrt{\frac{378}{25}} = \frac{\sqrt{378}}{5} = \frac{3\sqrt{42}}{5} \]

 Step 4: Find the unit vector in the direction of \( \mathbf{CD} \). The unit vector in the direction of \( \mathbf{CD} \) is: \[ \hat{\mathbf{CD}} = \frac{\mathbf{CD}}{|\mathbf{CD}|} = \frac{\frac{8}{5}\mathbf{i} - \mathbf{j} + \frac{17}{5}\mathbf{k}}{\frac{3\sqrt{42}}{5}} = \frac{8\mathbf{i} - 5\mathbf{j} + 17\mathbf{k}}{3\sqrt{42}} \] Thus, the unit vector in the direction of \( \mathbf{CD} \) is: \[ \boxed{\frac{1}{\sqrt{42}}(8\mathbf{i} - 5\mathbf{j} + 17\mathbf{k})} \]