Question

In \(\triangle ABC\), \(P\) is a point on side \(BC\) such that \(BP:PC=4:1\) and \(Q\) is a point on the segment \(AP\) such that \(AQ:QP=3:2\). If \(\vec{BQ}\) is expressed as \(\vec{BQ}=x\vec{BA}+y\vec{BC}\), then the value of \(x+y\) is:

• A. \(\frac{22}{25}\)
• B. \(\frac{3}{5}\)
• C. \(\frac{17}{25}\)
• D. \(1\)

Hint:

Take B as origin and apply the section formula twice, first for P and then for Q.

Answer 1

The Correct Option is A

Step 1: Write Q Along AP
$\vec{BQ}=\vec{BA}+\dfrac{3}{5}\vec{AP}$ since $AQ=\dfrac{3}{5}AP$.

Step 2: Find AP
$\vec{AP}=\vec{BP}-\vec{BA}=\dfrac{4}{5}\vec{BC}-\vec{BA}$.

Step 3: Combine
$\vec{BQ}=\vec{BA}\left(1-\dfrac{3}{5}\right)+\dfrac{12}{25}\vec{BC}=\dfrac{2}{5}\vec{BA}+\dfrac{12}{25}\vec{BC}$, so $x+y=\dfrac{22}{25}$.

Final Answer:
The coefficients add up to 22/25. \[ \boxed{\text{(A) }\dfrac{22}{25}} \]