AP EAMCET · Maths · Vector Algebra

$D, E, F$ are respectively the points on the sides $B C, C A$ and $A B$ of a $∆ A B C$ dividing them in the ratio $2 : 3, 1 : 2, 3 : 1$ internally. The lines $B E$ and $C F$ intersect on the line $A D$ at $P$ . If $\vec{A P} = x_{1} \vec{A B} + y_{1} \cdot \vec{A C}$ , then $x_{1} + y_{1} =$

A $\frac{5}{6}$
B $1$
C $\frac{3}{2}$
D $2$

Verified Solution

Answer & Solution

Correct Answer

(A) $\frac{5}{6}$

Step-by-step Solution

Detailed explanation

By using section formula we can obtain required result If vector d,e,f are position vector of points D, E & F respectively then, by section formula d→=2c→+3b→5 ...i e→=2c→+a→3 ...ii f→=3b→+a→4 ...iii Equation of line AD is…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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D,E,F are respectively the points on the sides BC,CA and AB of a ∆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 AP→=x1AB→+y1·AC→, then x1+y1=

Step-by-step Solution

$D, E, F$ are respectively the points on the sides $B C, C A$ and $A B$ of a $∆ A B C$ dividing them in the ratio $2 : 3, 1 : 2, 3 : 1$ internally. The lines $B E$ and $C F$ intersect on the line $A D$ at $P$ . If $\vec{A P} = x_{1} \vec{A B} + y_{1} \cdot \vec{A C}$ , then $x_{1} + y_{1} =$