Let ΔABC is fixed triangle with area = 1 unit2 . let points M,N,P are taken on AB¯, BC¯, CA¯ respectively such that AMAB=BNBC=CPCA=λ λ>0 then which of the following options is/are correct?

AN=→λc→+(1−λ)b¯−a¯BP¯=λa→+(1−λ)c¯−b¯CM¯=λ.b¯+(1−λ)a¯−c¯Area form by above three vectors0=12|s¯1×s¯2|=12|(λc→+(1−λ)b¯−a¯)×(λa→+(1−λ)c¯−b¯)|=12|λ2−λ+1| (Area ΔABC)=12|λ2−λ+1|

Let ΔABC is fixed triangle with area = 1 unit2 . let points M,N,P are taken on AB¯, BC¯, CA¯ respectively such that AMAB=BNBC=CPCA=λ λ>0 then which of the following options is/are correct?

AN=→λc→+(1−λ)b¯−a¯BP¯=λa→+(1−λ)c¯−b¯CM¯=λ.b¯+(1−λ)a¯−c¯Area form by above three vectors0=12|s¯1×s¯2|=12|(λc→+(1−λ)b¯−a¯)×(λa→+(1−λ)c¯−b¯)|=12|λ2−λ+1| (Area ΔABC)=12|λ2−λ+1|

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Let $Δ A B C$ is fixed triangle with area = 1 ${(u n i t)}^{2}$ . let points M,N,P are taken on $\bar{A B}, \bar{B C}, \bar{C A}$ respectively such that $\frac{A M}{A B} = \frac{B N}{B C} = \frac{C P}{C A} = λ (λ > 0)$ then which of the following options is/are correct?

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$\vec{A N} + \vec{B P} + \vec{C M} = \vec{0}$

The area of the triangle $Δ M N P$ always equal to $\frac{1}{4}$

The minimum area of the triangle formed by the vectors $\vec{A N}, \vec{B P}, \vec{C M}$ is $\frac{1}{4}$

$0 \leq λ \leq 1$

answer is A, D.

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Detailed Solution

$\begin{array}{l} \vec{A N =} λ \vec{c} + (1 - λ) \bar{b} - \bar{a} \\ \bar{B P} = λ \vec{a} + (1 - λ) \bar{c} - \bar{b} \\ \bar{C M} = λ . \bar{b} + (1 - λ) \bar{a} - \bar{c} \end{array}$

Area form by above three vectors

$\begin{array}{l} 0 = \frac{1}{2} | {\bar{s}}_{1} \times {\bar{s}}_{2} | \\ = \frac{1}{2} | (λ \vec{c} + (1 - λ) \bar{b} - \bar{a}) \times (λ \vec{a} + (1 - λ) \bar{c} - \bar{b}) | \\ = \frac{1}{2} | λ^{2} - λ + 1 | (A r e a Δ A B C) \\ = \frac{1}{2} | λ^{2} - λ + 1 | \end{array}$