Question

Given an equilateral triangle ABC with side length equal to ‘a’. Let M and N be two points respectively on the side AB and AC such that $\overrightarrow {AN} = K\overrightarrow {AC} $ and $\overrightarrow {AM} = \dfrac{{\overrightarrow {AB} }}{3}$. If $\overrightarrow {BN} $ and$\overrightarrow {CM} $ are orthogonal then the value of K is equal to
(A) $\dfrac{1}{5}$
(B) $\dfrac{1}{4}$
(C) $\dfrac{1}{3}$
(D) $\dfrac{1}{2}$

Answer 1

Hint: In equilateral$\Delta $all the sides are equal and all the angles are equal to each other. All the angle of an equilateral triangle are ${60^0}$

Complete Step by Step Solution:
Let us consider an equilateral$\Delta $ABC

$\overrightarrow {AM} = \dfrac{{\overrightarrow {AD} }}{3}$
Therefore,$BM = AB - AM$
$ = AB - \dfrac{{AB}}{3}$
$\overrightarrow {BM} = \dfrac{{2AB}}{3}$
$\overrightarrow {AN} = K\overrightarrow {AC} $ . . . . (given)
$\overrightarrow {CN} = \overrightarrow {AC} - \overrightarrow A .$
$ = \overrightarrow {AC} - K\overrightarrow {AC} $
$\overrightarrow {CN} = \overrightarrow {AC} (1 - k)$
If $\overrightarrow {BN} $and$\overrightarrow {CN} $are orthogonal, then
$\overrightarrow {BN} \times \overrightarrow {CN} = 0$
How, We can find the value of $\overrightarrow {BN} $ and $\overrightarrow {CN} $
So,
\[\overrightarrow {BN} = \overrightarrow c - \dfrac{{\overrightarrow b }}{3}\]
$\overrightarrow {CN} = \lambda \overrightarrow c - \overrightarrow b $
Therefore,
$\left( {\overrightarrow c - \dfrac{{\overrightarrow b }}{3}} \right)(\lambda \overrightarrow c - \overrightarrow b ) = 0$
$\lambda {\left| {\overrightarrow c } \right|^2} - \dfrac{\lambda }{3}\overrightarrow c .\overrightarrow b - \overrightarrow c .\overrightarrow b + \dfrac{{{{\left| {\overrightarrow b } \right|}^2}}}{3} = 0$
$\lambda {a^2} - \dfrac{\lambda }{3}{a^2} \times \dfrac{1}{2} - \dfrac{{{a^2}}}{2} + \dfrac{{{a^2}}}{3} = 0$
Then common$'\lambda '$from this equation.
$\lambda \left( {{a^2} - \dfrac{{{a^2}}}{3}} \right) - \dfrac{{{a^2}}}{2} + \dfrac{{{a^2}}}{3} = 0$
$\lambda \left( {{a^2} - \dfrac{{{a^2}}}{6}} \right) = \dfrac{{{a^2}}}{2} - \dfrac{{{a^2}}}{3}$
$\lambda \left( {\dfrac{{6{a^2} - {a^2}}}{6}} \right) = \dfrac{{3{a^2} - 2{a^2}}}{6}$
$\lambda \left( {\dfrac{{5{a^2}}}{6}} \right) = \dfrac{{{a^2}}}{6}$
$\lambda = \dfrac{{{a^2}}}{{5{a^2}}}$
$\lambda = \dfrac{1}{5}$
If the value of $\lambda = \dfrac{1}{5}$, then the side of $C$ is $AM = 1$and $NC = 4$ and $AC = 5$
Then $\overrightarrow {AN} = K\overrightarrow {AC} $
Then $K = \dfrac{{\overrightarrow {AN} }}{{\overrightarrow {AC} }} = \dfrac{1}{5}$
So, the value of $K$ is $\dfrac{1}{5}$
Therefore from the above explanation the correct option is [A] $\dfrac{1}{5}$.

Note: Two vectors are said to be orthogonal if they are perpendicular to each other.
That means the angle between them is ${90^0}$.
We know that $\cos {90^0} = 0$
Therefore, $\overrightarrow a \times \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos {90^0} = 0$
That means, two vectors $\overrightarrow a $ and $\overrightarrow b $are perpendicular if $\overrightarrow a \times \overrightarrow b = 0$