Question

Let A (3, 0, – 1), B (2, 10, 6) and C (1, 2, 1) be the vertices of a triangle and M be the midpoint of AC. If G divides BM in the ratio, 2:1, then $$\cos \left(\mathrm{\angle }GOA\right)$$ (O being the origin) is equal to:
$$\left(a\right){\displaystyle \frac{1}{\sqrt{30}}}$$
$$\left(b\right){\displaystyle \frac{1}{6\sqrt{10}}}$$
$$\left(c\right){\displaystyle \frac{1}{\sqrt{15}}}$$
$$\left(d\right){\displaystyle \frac{1}{2\sqrt{15}}}$$

Answer 1

Hint: We are given M is the midpoint of AC. So, we get the line BM is the median as G is the point on BM on dividing it into 2:1. We will get by definition of the centroid that G is the centroid then the coordinate of G is given by $$x={\displaystyle \frac{x_1+x_2+x_3}{3}},y={\displaystyle \frac{y_1+y_2+y_3}{3}},z={\displaystyle \frac{z_1+z_2+z_3}{3}}.$$ Then for finding cos of the angle ROA, we use $$\cos \left(\mathrm{\angle }ROA\right)={\displaystyle \frac{\left(OR\right).\left(OA\right)}{\left|OR\right|\left|OA\right|}}.$$ To do so we will find the magnitude of OR and OA and then find the dot product OR.OA.

Complete step-by-step answer:
We are given the coordinates of the triangle ABC as A (3, 0, – 1), B (2, 10, 6) and C (1, 2, 1). We have M as the midpoint of AC and G is the point on BM that divides BM in the ratio 2:1. We know that as M is the midpoint, so BM is the median of triangle ABC. Now, the point on the median which divides the median in 2:1 is known as the centroid.
As G lies on the median BM and divides BM in the ratio 2:1, so it means G is the centroid.

Now we know that coordinate of the centroid G is given as
$$x={\displaystyle \frac{x_1+x_2+x_3}{3}},y={\displaystyle \frac{y_1+y_2+y_3}{3}},z={\displaystyle \frac{z_1+z_2+z_3}{3}}$$
where $$(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)$$ are the coordinates of the vertex.
As the vertex, we have A (3, 0, – 1), B (2, 10, 6), C (1, 2, 1). So, we get the coordinate of G as
$$x={\displaystyle \frac{3+2+1}{3}}={\displaystyle \frac{6}{3}}=2$$
$$y={\displaystyle \frac{2+10+2}{3}}={\displaystyle \frac{12}{3}}=4$$
$$z={\displaystyle \frac{-1+6+1}{3}}={\displaystyle \frac{6}{3}}=2$$
So, we have the coordinate as G (2, 4, 2).
Now, we have to find the cos of $$\mathrm{\angle }GOA$$ where O is the origin.

We know that the cos angle between the two vectors is given as
$$\cos \left(\mathrm{\angle }XOY\right)={\displaystyle \frac{OX.OY}{\left|OX\right|\left|OY\right|}}$$
So, for $$\mathrm{\angle }GOA$$ we will get,
$$\cos \left(\mathrm{\angle }GOA\right)={\displaystyle \frac{OG.OA}{\left|OG\right|\left|OA\right|}}$$
So, for G (2, 4, 2) and O (0, 0, 0) we have,
$$OG=(2-0)i+(4-0)j+(2-0)k$$
$$\Rightarrow OG=2i+4j+2k$$
For A (3, 0, – 1) and O (0, 0, 0), we have,
$$OA=(3-0)i+(0-0)j+(-1-0)k$$
$$\Rightarrow OA=3i-k$$
Now,
$$\left|OG\right|=\sqrt{2^2+4^2+2^2}$$
$$\Rightarrow \left|OG\right|=\sqrt{4+16+4}$$
$$\Rightarrow \left|OG\right|=\sqrt{24}$$
And
$$\left|OA\right|=\sqrt{3^2+{(-1)}^2}$$
$$\Rightarrow \left|OA\right|=\sqrt{10}$$
Also,
$$OA.OG=(3i-k)(2i+4j+2k)$$
$$\Rightarrow OA.OG=6-2$$
$$\Rightarrow OA.OG=4$$
Putting this value in $$\cos \left(\mathrm{\angle }GOA\right).$$
$$\cos \left(\mathrm{\angle }GOA\right)={\displaystyle \frac{OA.OG}{\left|OA\right|\left|OG\right|}}$$
$$\Rightarrow \cos \left(\mathrm{\angle }GOA\right)={\displaystyle \frac{4}{\left|\sqrt{10}\right|\left|\sqrt{24}\right|}}$$
After simplification we get,
$$\Rightarrow \cos \left(\mathrm{\angle }GOA\right)={\displaystyle \frac{4}{2\times 2\times \sqrt{15}}}$$
$$\Rightarrow \cos \left(\mathrm{\angle }GOA\right)={\displaystyle \frac{1}{\sqrt{15}}}$$

So, the correct answer is “Option C”.

Note: To simplify the square root we need to factorize it. $$\sqrt{24}$$ can be written as $$\sqrt{24}=\sqrt{2\times 2\times 2\times 3}$$ and $$\sqrt{10}=\sqrt{2\times 5}.$$ So,
$$\sqrt{24}\times \sqrt{10}=\sqrt{2\times 2\times 2\times 2\times 3\times 5}$$
2 comes out two times as it makes a pain that gives us $$2\times 2\times \sqrt{15}.$$ This we use while simplifying. Also, remember that |OG| means the magnitude of OG which is given as $$\sqrt{x^2+y^2+z^2}$$ if OG is given as $$x\hat{i}+y\hat{j}+2\widehat{k.}$$