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 $\cos(\angle GOA)$ (O being the origin) is equal to : Options: (A) $1/\sqrt{30}$ (B) $1/6\sqrt{10}$ (C) $1/\sqrt{15}$ (D) $1/2\sqrt{15}$

### Analyzing the Setup When we look at a triangle in three-dimensional space, it is easy to get lost in the coordinates. We are given three vertices, $A(3, 0, -1)$, $B(2, 10, 6)$, and $C(1, 2, 1)$. Our mission is to find the cosine of the angle between the vectors $\vec{OA}$ and $\vec{OG}$, where $O$ is the origin and $G$ is the centroid of the triangle. The problem states that $M$ is the midpoint of $AC$ and that $G$ divides the median $BM$ in a $2:1$ ratio. In the language of geometry, the point that divides a median in a $2:1$ ratio is the **centroid** of the triangle. ### Finding the Anchor: The Coordinates of $G$ First, we calculate the midpoint $M$ of $AC$ using the midpoint formula: $$M = \left( \frac{3+1}{2}, \frac{0+2}{2}, \frac{-1+1}{2} \right) = (2, 1, 0)$$ Since $G$ is the centroid, we use the centroid formula $G = \frac{A+B+C}{3}$. Plugging in our values: $$G = \left( \frac{3+2+1}{3}, \frac{0+10+2}{3}, \frac{-1+6+1}{3} \right)$$ This simplifies to the coordinates $G(2, 4, 2)$. We now have our anchor point. ### The Vector Bridge We define the position vectors relative to the origin $O(0, 0, 0)$ as $\vec{OA} = 3\hat{i} + 0\hat{j} - 1\hat{k}$ and $\vec{OG} = 2\hat{i} + 4\hat{j} + 2\hat{k}$. To find the angle $\theta$ between them, we use the dot product formula: $$\cos \theta = \frac{\vec{OA} \cdot \vec{OG}}{|\vec{OA}| |\vec{OG}|}$$ ### The Final Calculation First, we compute the dot product: $$\vec{OA} \cdot \vec{OG} = (3)(2) + (0)(4) + (-1)(2) = 6 + 0 - 2 = 4$$ Next, we calculate the magnitudes of the vectors: $$|\vec{OA}| = \sqrt{3^2 + 0^2 + (-1)^2} = \sqrt{10}$$ $$|\vec{OG}| = \sqrt{2^2 + 4^2 + 2^2} = \sqrt{24}$$ Substituting these values into our cosine formula: $$\cos \theta = \frac{4}{\sqrt{10} \sqrt{24}} = \frac{4}{\sqrt{240}}$$ Since $\sqrt{240} = \sqrt{16 \times 15} = 4\sqrt{15}$, the expression simplifies to: $$\cos \theta = \frac{4}{4\sqrt{15}} = \frac{1}{\sqrt{15}}$$ The final result is **$\frac{1}{\sqrt{15}}$**.

JEE Main 2019 (10 April Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Vector Algebra: 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 $A C$ . If $G$ divides $B M$ in the ratio $2 : 1$ , then $cos (∠ GO A)$ (O being the origin) is equal to :

Q: Can I use the centroid formula $\frac{A+B+C}{3}$ instead of the section formula?

Yes, you can! Since $G$ is the centroid, its coordinates are simply the average of the coordinates of the vertices $A$, $B$, and $C$. Using $\frac{A+B+C}{3}$ is often faster and less prone to errors than the section formula. Both methods are mathematically sound, so choose the one you are most comfortable with.

Q: Why does the order of the ratio matter in the section formula?

The section formula $\frac{m\vec{r_2} + n\vec{r_1}}{m+n}$ depends on the ratio $m:n$. If $G$ divides $BM$ in $2:1$, it means $G$ is closer to $M$ than to $B$. If you accidentally swap the ratio to $1:2$, you would be finding a point closer to $B$, which is incorrect. Always double-check which segment is being divided and in what order the ratio is given.

Select Answer:

Visualized Solution

Visualize the 3D Setup

Given vertices of $△ A B C$ : $A (3, 0, - 1)$ , $B (2, 10, 6)$ , and $C (1, 2, 1)$ .
Origin is at $O (0, 0, 0)$ .
Objective: Find the cosine of the angle between vectors $O A$ and $O G$ .

Midpoint of $A C$

$M$ is given as the midpoint of side $A C$ .
Midpoint formula: $M = (\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2}, \frac{z _{1} + z _{2}}{2})$

Calculate Coordinates of $M$

For $A (3, 0, - 1)$ and $C (1, 2, 1)$ :
$x = \frac{3 + 1}{2} = 2$
$y = \frac{0 + 2}{2} = 1$
$z = \frac{- 1 + 1}{2} = 0$
$M = (2, 1, 0)$

Locate Point $G$

$G$ divides the median $B M$ in the ratio $2 : 1$ .
Key Concept: The point dividing a median in a $2 : 1$ ratio is the Centroid of the triangle.
We will use the section formula on $B$ and $M$ .

Calculate Coordinates of $G$

Using Section Formula on $B (2, 10, 6)$ and $M (2, 1, 0)$ with ratio $2 : 1$ :
$x_{G} = \frac{2 ( 2 ) + 1 ( 2 )}{2 + 1} = 2$
$y_{G} = \frac{2 ( 1 ) + 1 ( 10 )}{2 + 1} = 4$
$z_{G} = \frac{2 ( 0 ) + 1 ( 6 )}{2 + 1} = 2$
$G = (2, 4, 2)$

Define Vectors $O A$ and $O G$

With origin $O (0, 0, 0)$ , the position vectors are simply the coordinates.
$O A = 3 \hat{i} + 0 \hat{j} - 1 \hat{k}$
$O G = 2 \hat{i} + 4 \hat{j} + 2 \hat{k}$

The Dot Product Formula

To find the angle $θ = ∠ GO A$ , we use the dot product.
$O G \cdot O A = ∣ O G ∣∣ O A ∣ cos θ$
$cos θ = \frac{O G \cdot O A}{∣ O G ∣∣ O A ∣}$

Compute the Dot Product

$O G \cdot O A = (2) (3) + (4) (0) + (2) (- 1)$
$O G \cdot O A = 6 + 0 - 2$
$O G \cdot O A = 4$

Calculate Magnitudes

$∣ O G ∣ = 2^{2} + 4^{2} + 2^{2} = 4 + 16 + 4 = 24$
$∣ O A ∣ = 3^{2} + 0^{2} + (- 1)^{2} = 9 + 0 + 1 = 10$

Substitute into Cosine Formula

$cos θ = \frac{4}{24 10}$
$cos θ = \frac{4}{240}$

Final Simplification

$240 = 16 \times 15 = 415$
$cos θ = \frac{4}{4 15}$
$cos θ = \frac{1}{15}$
This matches option (3).

00:00 / 00:00

The Sigma Insight: Scalar (Dot) Product

Solution Diagram

Analyzing the Setup

When we look at a triangle in three-dimensional space, it is easy to get lost in the coordinates. We are given three vertices,

A (3, 0, - 1)

B (2, 10, 6)

, and

C (1, 2, 1)

Our mission is to find the cosine of the angle between the vectors

O A

and

O G

, where

O

is the origin and

G

is the centroid of the triangle.

The problem states that

M

is the midpoint of

A C

and that

G

divides the median

B M

in a

2 : 1

ratio. In the language of geometry, the point that divides a median in a

2 : 1

ratio is the centroid of the triangle.

Finding the Anchor

The Coordinates of

G

First, we calculate the midpoint

M

A C

using the midpoint formula:

M = (\frac{3 + 1}{2}, \frac{0 + 2}{2}, \frac{- 1 + 1}{2}) = (2, 1, 0)

Since

G

is the centroid, we use the centroid formula

G = \frac{A + B + C}{3}

. Plugging in our values:

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

This simplifies to the coordinates

G (2, 4, 2)

. We now have our anchor point.

The Vector Bridge

We define the position vectors relative to the origin

O (0, 0, 0)

O A = 3 \hat{i} + 0 \hat{j} - 1 \hat{k}

and

O G = 2 \hat{i} + 4 \hat{j} + 2 \hat{k}

To find the angle

θ

between them, we use the dot product formula:

cos θ = \frac{O A \cdot O G}{∣ O A ∣∣ O G ∣}

The Final Calculation

First, we compute the dot product:

O A \cdot O G = (3) (2) + (0) (4) + (- 1) (2) = 6 + 0 - 2 = 4

Next, we calculate the magnitudes of the vectors:

∣ O A ∣ = 3^{2} + 0^{2} + (- 1)^{2} = 10

∣ O G ∣ = 2^{2} + 4^{2} + 2^{2} = 24

Substituting these values into our cosine formula:

cos θ = \frac{4}{10 24} = \frac{4}{240}

Since

240 = 16 \times 15 = 415

, the expression simplifies to:

cos θ = \frac{4}{4 15} = \frac{1}{15}

The final result is $\frac{1}{15}$ .

Common Traps & FAQs

Wait, why do we call

G

the centroid? Is it always true that the point dividing a median in a

2 : 1

ratio is the centroid?

Yes, absolutely! By definition, the centroid of a triangle is the point of intersection of its medians, and it always divides each median in a

2 : 1

ratio, measured from the vertex to the midpoint of the opposite side. Recognizing this property is a powerful shortcut in JEE problems, saving you time and reducing the chance of calculation errors.

Can I use the centroid formula

\frac{A + B + C}{3}

instead of the section formula?

Yes, you can! Since

G

is the centroid, its coordinates are simply the average of the coordinates of the vertices

A

B

, and

C

. Using

\frac{A + B + C}{3}

is often faster and less prone to errors than the section formula. Both methods are mathematically sound, so choose the one you are most comfortable with.

Why does the order of the ratio matter in the section formula?

The section formula

\frac{m r _{2} + n r _{1}}{m + n}

depends on the ratio

m : n

. If

G

divides

B M

2 : 1

, it means

G

is closer to

M

than to

B

. If you accidentally swap the ratio to

1 : 2

, you would be finding a point closer to

B

, which is incorrect. Always double-check which segment is being divided and in what order the ratio is given.

Animated Solution for Mathematics - Vector Algebra: 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 $A C$ . If $G$ divides $B M$ in the ratio $2 : 1$ , then $cos (∠ GO A)$ (O being the origin) is equal to :

Select Answer:

Visualized Solution

Visualize the 3D Setup

Midpoint of $A C$

Calculate Coordinates of $M$

Locate Point $G$

Calculate Coordinates of $G$

Define Vectors $O A$ and $O G$

The Dot Product Formula

Compute the Dot Product

Calculate Magnitudes

Substitute into Cosine Formula

Final Simplification

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Finding the Anchor

The Vector Bridge

The Final Calculation

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Let $p$ and $q$ be the position vectors of $P$ and $Q$ respectively, with respect to $O$ and $∣ p ∣ = p, ∣ q ∣ = q$ . The points $R$ and $S$ divide $P Q$ internally and externally in the ratio $2 : 3$ respectively. If $O R$ and $O S$ are perpendicular then

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Select Answer:

Visualized Solution

Visualize the 3D Setup

Midpoint of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Calculate Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } M

Locate Point .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Calculate Coordinates of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } G

Define Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OA and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } OG

The Dot Product Formula

Compute the Dot Product

Calculate Magnitudes

Substitute into Cosine Formula

Final Simplification

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

Finding the Anchor

The Vector Bridge

The Final Calculation

Similar Questions

Comprehension Passage

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$∣ O X \times O Y ∣ =$

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