Name: Parallelogram Law for Addition of Two Vectors
Uploaded: 2019-07-19
Duration: 5 min 1 s
Description: Now, we know one of the methods to add the two vectors. It is time to learn another law to add the two vectors. Explore this video to learn about it.

Question

Prove by vector method that: the medians of a triangle are concurrent.

Accepted Answer

Let $A, B and C$ be vertices of a triangle.

Let $D, E and F$ be the midpoints of the sides $B C, A C and A B$

Let $\vec{O A} = \vec{a}, \vec{O B} = \vec{b}, \vec{O C} = \vec{c}, \vec{O D} = \vec{d}, \vec{O E} = \vec{e} and O F = \vec{f}$ be position vectors of points $A, B, C, D, E and F$ respectively.

Therefore, by Midpoint formula, we have: $\vec{d} = \frac{\vec{b} + \vec{c}}{2}, \vec{e} = \frac{\vec{a} + \vec{c}}{2} and \vec{f} = \frac{\vec{a} + \vec{b}}{2}$

$∴ 2 \vec{d} = \vec{b} + \vec{c}, 2 \vec{e} = \vec{a} + \vec{c} and 2 \vec{f} = \vec{a} + \vec{b}$

$∴ 2 \vec{d} + \vec{a} = \vec{a} + \vec{b} + \vec{c}$

$\Rightarrow 2 \vec{e} + \vec{b} = \vec{a} + \vec{b} + \vec{c}$

$\Rightarrow 2 \vec{f} + \vec{c} = \vec{a} + \vec{b} + \vec{c}$

Now, $\frac{2 \vec{d} + \vec{a}}{3} = \frac{2 \vec{e} + \vec{b}}{3} = \frac{2 \vec{f} + \vec{c}}{3} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$

Let $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$ , then we have; $\vec{g} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} = \frac{(2) \vec{d} + (1) \vec{a}}{2 + 1}$ $= \frac{(2) \vec{e} + (1) \vec{b}}{2 + 1} = \frac{(2) \vec{f} + (1) \vec{c}}{2 + 1}$

So, if $G$ is the point whose position vector is $\vec{g},$ then from the above equation it is clear that the point $G$ lies on the medians $A D, B E, C F$ and it divides them internally in the ratio $2 : 1$ .

Hence, the medians of a triangle are concurrent.

Prove by vector method that: the medians of a triangle are concurrent.

Important Questions on Vectors

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