SATHEE: Vectors 1 Question 3

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 $\cos (∠ G O A)$ ( $O$ being the origin) is equal to

(2019 Main, 10 Jan I)

(a) $\frac{1}{\sqrt{15}}$

(b) $\frac{1}{2 \sqrt{15}}$

(d) $\frac{1}{6 \sqrt{10}}$

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Answer:

Correct Answer: 3. (a)

Solution:

$\begin{aligned} Key Idea Use the angle between two non-zero vectors a and b \\ is given by \cos θ = \frac{a \cdot b}{| a | | b |} and coordinates of the centroid i.e. \\ (\frac{x_{1} + x_{2} + x_{3}}{3}, \frac{y_{1} + y_{2} + y_{3}}{3}, \frac{z_{1} + z_{2} + z_{3}}{3}) of a triangle formed \\ with vertices; (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2}, z_{2}) and (x_{3}, y_{3}, z_{3}) . \end{aligned}$

Given vertices of a $△ A B C$ 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 $△ A B C$ .

$∴ G (\frac{3 + 2 + 1}{3}, \frac{0 + 10 + 2}{3}, \frac{- 1 + 6 + 1}{3}) = (2, 4, 2)$ .

Now, $\cos (∠ G O A) = \frac{O G \cdot O A}{| O G | | O A |}$ , where $O$ is the origin.

$∵ O G = 2 \hat{i} + 4 \hat{j} + 2 \hat{k} \Rightarrow | O G | = \sqrt{4 + 16 + 4} = \sqrt{24}$

and $| O A | = 3 \hat{i} - \hat{k}$

$\begin{aligned} \Rightarrow | O A | = \sqrt{9 + 1} = \sqrt{10} and O G \cdot O A = 6 - 2 = 4 \\ ∴ \cos (∠ G O A) = \frac{4}{\sqrt{24} \sqrt{10}} = \frac{1}{\sqrt{15}} \end{aligned}$

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 $\cos (∠ G O A)$ ( $O$ being the origin) is equal to

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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 AC. If G divides BM in the ratio 2:1, then cos⁡(∠GOA) ( O being the origin) is equal to

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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 $\cos (∠ G O A)$ ( $O$ being the origin) is equal to