Question

If \(\vec{a}=\hat{i}+2\hat{j}+3\hat{k},\vec{b}=3\hat{i}+4\hat{j}+4\hat{k},\vec{c}=3\hat{i}+2\hat{j}+3\hat{k}\) are the position vectors of the vertices \(A\), \(B\) and \(C\) respectively of \(\triangle ABC\) and the bisector of \(\angle A\) meets side \(BC\) at \(D\) where \(B-D-C\), then \(|\vec{AD}|\) is...

• A. \(\frac{2\sqrt{30}}{5}\)
• B. \(\frac{2}{3}\)
• C. \(\frac{\sqrt{13}}{5}\)
• D. \(1\)

Hint:

Use the angle bisector theorem to find the ratio in which D divides BC, then find the position vector of D.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
The bisector of angle A divides BC in the ratio \(AB:AC\).

Step 2: Find the side lengths
\(\vec{AB}=(2,2,1)\), so \(AB=3\). \(\vec{AC}=(2,0,0)\), so \(AC=2\). Hence \(BD:DC=3:2\).

Step 3: Position of D
\(\vec{D}=\frac{2\vec{b}+3\vec{c}}{5}=\frac{(6,8,8)+(9,6,9)}{5}=\left(3,\frac{14}{5},\frac{17}{5}\right)\).

Step 4: Find AD
\(\vec{AD}=\left(2,\frac{4}{5},\frac{2}{5}\right)\), so \(|AD|^{2}=4+\frac{16}{25}+\frac{4}{25}=\frac{24}{5}\).
\(|AD|=\sqrt{\frac{24}{5}}=\frac{2\sqrt{30}}{5}\).

Final Answer:
The length of AD is 2√30/5. \[ \boxed{\text{(A) }\dfrac{2\sqrt{30}}{5}} \]