Question

\(D,E,F\) are respectively the points on the sides \(BC,CA\) and \(AB\) of a \(\triangle ABC\), dividing them in the ratio \(2:3,\;1:2,\;3:1\) internally. The lines \(BE\) and \(CF\) intersect on the line \(AD\) at \(P\). If \[ \overrightarrow{AP}=x_1\overrightarrow{AB}+y_1\overrightarrow{AC}, \] then \[ x_1+y_1= \]

• A. \(\dfrac{5}{6}\)
• B. \(1\)
• C. \(\dfrac{3}{2}\)
• D. \(2\)

Hint:

For vector geometry problems, take one vertex as origin and express all points as linear combinations of two independent vectors such as \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\).

Answer 1

The Correct Option is A

Step 1: Represent points using vectors.
Let \[ \overrightarrow{AB}=\vec{b}, \qquad \overrightarrow{AC}=\vec{c} \] Taking \(A\) as origin, we have \[ B=\vec{b}, \qquad C=\vec{c} \]

Step 2: Find position vectors of \(E\) and \(F\).
Since \(E\) divides \(CA\) in the ratio \(1:2\), we get \[ E=\frac{2\vec{c}+\vec{0}}{3} \] \[ E=\frac{2}{3}\vec{c} \] Since \(F\) divides \(AB\) in the ratio \(3:1\), we get \[ F=\frac{3}{4}\vec{b} \]

Step 3: Write point \(P\) on line \(BE\).
Any point on \(BE\) can be written as \[ P=(1-\lambda)\vec{b}+\lambda\left(\frac{2}{3}\vec{c}\right) \] So, \[ P=(1-\lambda)\vec{b}+\frac{2\lambda}{3}\vec{c} \]

Step 4: Write point \(P\) on line \(CF\).
Any point on \(CF\) can be written as \[ P=(1-\mu)\vec{c}+\mu\left(\frac{3}{4}\vec{b}\right) \] So, \[ P=\frac{3\mu}{4}\vec{b}+(1-\mu)\vec{c} \]

Step 5: Compare coefficients of \(\vec{b}\) and \(\vec{c}\).
Comparing the coefficients of \(\vec{b}\), \[ 1-\lambda=\frac{3\mu}{4} \] Comparing the coefficients of \(\vec{c}\), \[ \frac{2\lambda}{3}=1-\mu \] Solving these two equations gives \[ \lambda=\frac{1}{2}, \qquad \mu=\frac{2}{3} \]

Step 6: Find \(\overrightarrow{AP}\).
Substitute \[ \lambda=\frac{1}{2} \] in \[ P=(1-\lambda)\vec{b}+\frac{2\lambda}{3}\vec{c} \] Thus, \[ P=\frac{1}{2}\vec{b}+\frac{1}{3}\vec{c} \] Therefore, \[ \overrightarrow{AP} = \frac{1}{2}\overrightarrow{AB} + \frac{1}{3}\overrightarrow{AC} \] Hence, \[ x_1=\frac{1}{2}, \qquad y_1=\frac{1}{3} \]

Step 7: Find \(x_1+y_1\).
\[ x_1+y_1=\frac{1}{2}+\frac{1}{3} \] \[ =\frac{3+2}{6} \] \[ =\frac{5}{6} \]

Step 8: Final conclusion.
Hence, \[ \boxed{\frac{5}{6}} \] which corresponds to option (1).