Let $A B C$ be an equilateral triangle of side $a . M$ and $N$ are two points on the sides $A B$ and $A C$ respectively such that $\overrightarrow{\mathrm{AN}}=\mathrm{K} \overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{AB}}=3 \overrightarrow{\mathrm{AM}}$. If the vectors…

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Let $A B C$ be an equilateral triangle of side $a . M$ and $N$ are two points on the sides $A B$ and $A C$ respectively such that $\overrightarrow{\mathrm{AN}}=\mathrm{K} \overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{AB}}=3 \overrightarrow{\mathrm{AM}}$. If the vectors $\overrightarrow{\mathrm{BN}}$ and $\overrightarrow{\mathrm{CM}}$ are perpendicular, then $\mathrm{K}=$

Aicharya · Accepted Answer

Since, $\overrightarrow{A B}=3 \overrightarrow{A M}, \overrightarrow{A N}=K \overrightarrow{A C}$ $\begin{aligned} & \text {Now, } \overrightarrow{C M} \cdot \overrightarrow{B N}=0 \\ & \Rightarrow(\overrightarrow{A M}-\overrightarrow{A C}) \cdot(\overrightarrow{A N}-\overrightarrow{A B})=0 \\ & \Rightarrow\left(\frac{1}{3} \overrightarrow{A B}-\overrightarrow{A C}\right)(K \overrightarrow{A C}-\overrightarrow{A B})=0 \\ & \Rightarrow \frac{K}{3} \overrightarrow{A B} \cdot \overrightarrow{A C}--|\overrightarrow{A B}|^2-K|\overrightarrow{A C}|^2+\overrightarrow{A C} \cdot \overrightarrow{A B}=0 \\ & \Rightarrow \frac{K a^2}{6}-\frac{a^2}{3}-K a^2+\frac{a^2}{2}=0 \Rightarrow K=\frac{1}{5}\end{aligned}$

Let $A B C$ be an equilateral triangle of side $a . M$ and $N$ are two points on the sides $A B$ and $A C$…

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