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Question: By vector method, prove that the medians of a triangle are concurrent....

By vector method, prove that the medians of a triangle are concurrent.

Explanation

Solution

Median is a line joining a vertex to the midpoint of the opposite side.
The centroid (point of concurrence) divides the
Section Formula: The position vector c\vec{c} of the point C dividing two points A and B with position vectors a\vec{a} and b\vec{b} respectively, in the ratio AC:CB=m:nAC:CB=m:n internally, is given by c=na+mbn+m\vec{c}=\dfrac{n\vec{a}+m\vec{b}}{n+m} .
The vector AB\vec{AB} from a point A with position vector a\vec{a} to a point B with position vector b\vec{b} , is given by AB=ba\vec{AB}=\vec{b}-\vec{a} .
Vectors are added using the Triangle Law of vector addition:
PQ+QR=PR\vec{PQ}+\vec{QR}=\vec{PR}

Complete step-by-step answer:
Let's say that we have a triangle ABC with AP, BQ and CR as its medians.

Let's say that the position vectors of the points are:
A=2aA=2\vec{a}
B=2bB=2\vec{b}
C=2cC=2\vec{c}
Using the section formula, the position vectors of the points P, Q and R will be:
P=2b+2c2=b+cP=\dfrac{2\vec{b}+2\vec{c}}{2}=\vec{b}+\vec{c}
Q=2a+2c2=a+cQ=\dfrac{2\vec{a}+2\vec{c}}{2}=\vec{a}+\vec{c}
R=2a+2b2=a+bR=\dfrac{2\vec{a}+2\vec{b}}{2}=\vec{a}+\vec{b}
Let's say that the point M divides the vectors AP\vec{AP} , BQ\vec{BQ} and CR\vec{CR} in the ratio AM:MP=k:1AM:MP=k:1 , BM:MQ=l:1BM:MQ=l:1 and CM:MR=m:1CM:MR=m:1 respectively.
Using the section formula, the position vector of M will be:
k(b+c)+2ak+1\dfrac{k\left( \vec{b}+\vec{c} \right)+2\vec{a}}{k+1} OR l(a+c)+2bl+1\dfrac{l\left( \vec{a}+\vec{c} \right)+2\vec{b}}{l+1} OR m(a+b)+2cm+1\dfrac{m\left( \vec{a}+\vec{b} \right)+2\vec{c}}{m+1}
It can be observed that k=l=m=2k=l=m=2 , satisfy all these three values for the position vector of M.
Therefore, it can be concluded that the three median vectors AP\vec{AP} , BQ\vec{BQ} and CR\vec{CR} are concurrent at a point M, which also divides them in the ratio 2:12:1 .

Note: The direction of a vector is important: AB=BA\vec{AB}=-\vec{BA} .
The position vector c\vec{c} of the mid-point C of two points A and B with position vectors a\vec{a} and b\vec{b} respectively, is given by c=a+b2\vec{c}=\dfrac{\vec{a}+\vec{b}}{2} .
The vector sum of two or more vectors is also called their resultant. It is the result of adding the vectors together using the triangle law of addition.
The vector component along the x-axis, of a vector r\vec{r} which makes an angle θ\theta with the positive direction of the x-axis, is rsinθ|r|\sin \theta and the component along the y-axis is rcosθ|r|\cos \theta , where r|r| denotes the magnitude of r\vec{r} .