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: Understanding the Concept
Take \(B\) as the origin. Let \(\vec{BA}=\vec{a}\) and \(\vec{BC}=\vec{c}\), so \(A=\vec{a}\), \(C=\vec{c}\).

Step 2: Find P
\(BP:PC=4:1\), so \(\vec{BP}=\dfrac{4}{5}\vec{c}\).

Step 3: Find Q
\(AQ:QP=3:2\), so \(\vec{BQ}=\dfrac{2\vec{a}+3\vec{BP}}{5}=\dfrac{2}{5}\vec{a}+\dfrac{3}{5}\cdot\dfrac{4}{5}\vec{c}=\dfrac{2}{5}\vec{a}+\dfrac{12}{25}\vec{c}\).

Step 4: Add
\(x=\dfrac{2}{5}=\dfrac{10}{25}\) and \(y=\dfrac{12}{25}\), so \(x+y=\dfrac{22}{25}\), option (A).

Final Answer:
The value of x + y is 22/25. \[ \boxed{\text{(A) }\dfrac{22}{25}} \]