Question

Let \[ \overrightarrow{OA}=\hat{i}+2\hat{j}-4\hat{k} \] and \[ \overrightarrow{OB}=3\hat{i}-4\hat{j}-2\hat{k} \] be the position vectors of points \(A\) and \(B\). If a point \(C\) divides the line segment \(AB\) in the ratio \(1:3\) externally, then the position vector of a point which divides \(OC\) in the ratio \(4:1\) internally is

• A. \(5(\hat{i}-\hat{j})\)
• B. \(\hat{i}-4\hat{j}+2\hat{k}\)
• C. \(4\hat{i}-2\hat{j}+\hat{k}\)
• D. \(4(\hat{j}-\hat{k})\) \bigskip

Hint:

Remember the section formula: \[ \text{External division: } \frac{m\vec{b}-n\vec{a}}{m-n}, \qquad \text{Internal division: } \frac{m\vec{b}+n\vec{a}}{m+n}. \] Always find the external division point first and then apply the internal division formula if required.

Answer 1

The Correct Option is D

Concept: Use section formulas for external and internal division. Alternatively, use the parametric form where any point on line \(AB\) is \(\vec{A} + \lambda(\vec{B} - \vec{A})\), find \(C\) by external division condition, then find \(P\) on \(OC\).

Step 1: Find point \(C\) using the section formula for external division. Given \(\vec{A} = \hat{i} + 2\hat{j} - 4\hat{k}, \vec{B} = 3\hat{i} - 4\hat{j} - 2\hat{k}\). \(C\) divides \(AB\) externally in \(1:3\): \[ \vec{C} = \frac{1 \cdot \vec{B} - 3 \cdot \vec{A}}{1-3} = \frac{(3\hat{i} - 4\hat{j} - 2\hat{k}) - (3\hat{i} + 6\hat{j} - 12\hat{k})}{-2} = \frac{-10\hat{j} + 10\hat{k}}{-2} = 5\hat{j} - 5\hat{k}. \]

Step 2: Find point \(P\) on \(OC\) dividing it internally in \(4:1\). Origin \(O\) is \(\vec{0}\). \(P\) divides \(OC\) in \(4:1\): \[ \vec{P} = \frac{4 \cdot \vec{C} + 1 \cdot \vec{O}}{4+1} = \frac{4}{5}(5\hat{j} - 5\hat{k}) = 4\hat{j} - 4\hat{k} = 4(\hat{j} - \hat{k}). \]

Step 3: Write the final answer. \[ \boxed{4(\hat{j} - \hat{k})} \]