Question

Let $A(2,-3,0),B(2,-1,1),C(0,1,4)$ be the vertices of $\Delta ABC$. Then angle between the median $BD$ and the base $AC$ is

• A. ${30}^o$
• B. ${45}^o$
• C. ${60}^o$
• D. ${90}^o$

Answer 1

Answer: (B)

${45}^o$

$D$ is mid point of $AC$
So, $D=(1,-1,2)$
$\overrightarrow{BD}=(-1,0,1)$
$\overrightarrow{AC}=(-2,4,4)$

$\overrightarrow{BD}\cdot \overrightarrow{AC}=|\overrightarrow{BD}||\overrightarrow{AC}|\cos \theta$

$\Rightarrow 6=\sqrt{2}\sqrt{6}\cos \theta$
$\Rightarrow \cos \theta =\frac{1}{\sqrt{2}}$
$\Rightarrow \theta ={45}^{\circ }$