Vectors 1 Question 3
3. Let $A(3,0,-1), B(2,10,6)$ and $C(1,2,1)$ be the vertices of a triangle and $M$ be the mid-point of $A C$. If $G$ divides $B M$ in the ratio $2: 1$, then $\cos (\angle G O A)$ ( $O$ being the origin) is equal to
(2019 Main, 10 Jan I)
(a) $\frac{1}{\sqrt{15}}$
(b) $\frac{1}{2 \sqrt{15}}$
(c) $\frac{1}{\sqrt{30}}$
(d) $\frac{1}{6 \sqrt{10}}$
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Answer:
Correct Answer: 3. (c)
Solution:
$$ \begin{aligned} & \text { Key Idea Use the angle between two non-zero vectors } \mathbf{a} \text { and } \mathbf{b} \\ & \text { is given by } \cos \theta=\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \text { and coordinates of the centroid i.e. } \\ & \left(\frac{x _1+x _2+x _3}{3}, \frac{y _1+y _2+y _3}{3}, \frac{z _1+z _2+z _3}{3}\right) \text { of a triangle formed } \\ & \text { with vertices; }\left(x _1, y _1, z _1\right),\left(x _2, y _2, z _2\right) \text { and }\left(x _3, y _3, z _3\right) \text {. } \end{aligned} $$
Given vertices of a $\triangle A B C$ are $A(3,0,-1), B(2,10,6)$ and $C(1,2,1)$ and a point $M$ is mid-point of $A C$. An another point $G$ divides $B M$ in ratio $2: 1$, so $G$ is the centroid of $\triangle A B C$.
$\therefore G\left(\frac{3+2+1}{3}, \frac{0+10+2}{3}, \frac{-1+6+1}{3}\right)=(2,4,2)$.
Now, $\cos (\angle G O A)=\frac{\mathbf{O G} \cdot \mathbf{O A}}{|\mathbf{O G}||\mathbf{O A}|}$, where $O$ is the origin.
$\because \quad \mathbf{O G}=2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} \Rightarrow|\mathbf{O G}|=\sqrt{4+16+4}=\sqrt{24}$
and $\quad|\mathbf{O A}|=3 \hat{\mathbf{i}}-\hat{\mathbf{k}}$
$$ \begin{aligned} & \Rightarrow|\mathbf{O A}|=\sqrt{9+1}=\sqrt{10} \text { and } \mathbf{O G} \cdot \mathbf{O A}=6-2=4 \\ & \therefore \quad \cos (\angle G O A)=\frac{4}{\sqrt{24} \sqrt{10}}=\frac{1}{\sqrt{15}} \end{aligned} $$
