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(A) Given that $\overrightarrow{AB} = 3 \overrightarrow{AM}$ and $\overrightarrow{AN} = K \overrightarrow{AC}$.Since $\overrightarrow{BN} \perp \overr

Vedclass · Accepted Answer

$\frac{1}{5}$

Vedclass · Answer

(A) Given that $\overrightarrow{AB} = 3 \overrightarrow{AM}$ and $\overrightarrow{AN} = K \overrightarrow{AC}$.Since $\overrightarrow{BN} \perp \overrightarrow{CM}$,their dot product is zero,i.e.,$\overrightarrow{BN} \cdot \overrightarrow{CM} = 0$.We can write $\overrightarrow{BN} = \overrightarrow{AN} - \overrightarrow{AB} = K \overrightarrow{AC} - \overrightarrow{AB}$ and $\overrightarrow{CM} = \overrightarrow{AM} - \overrightarrow{AC} = \frac{1}{3} \overrightarrow{AB} - \overrightarrow{AC}$.Now,$(K \overrightarrow{AC} - \overrightarrow{AB}) \cdot (\frac{1}{3} \overrightarrow{AB} - \overrightarrow{AC}) = 0$.Expanding the dot product:$\frac{K}{3} (\overrightarrow{AC} \cdot \overrightarrow{AB}) - K |\overrightarrow{AC}|^2 - \frac{1}{3} |\overrightarrow{AB}|^2 + (\overrightarrow{AB} \cdot \overrightarrow{AC}) = 0$.Since $ABC$ is an equilateral triangle of side $a$,$|\overrightarrow{AB}| = |\overrightarrow{AC}| = a$ and $\overrightarrow{AB} \cdot \overrightarrow{AC} = a^2 \cos(60^{\circ}) = \frac{a^2}{2}$.Substituting these values:$\frac{K}{3} (\frac{a^2}{2}) - K a^2 - \frac{1}{3} a^2 + \frac{a^2}{2} = 0$.$\frac{K a^2}{6} - K a^2 = \frac{a^2}{3} - \frac{a^2}{2}$.$-\frac{5K a^2}{6} = -\frac{a^2}{6}$.$K = \frac{1}{5}$.

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