The centroid of the triangle OAB is denoted by G. If O is the origin and OA=4i + 3j ,OB=6i-j. Find OG in terms of the unit vectors, I and j.
The centroid of a triangle is located at 2/3 of the length of any median.
Median=(OA+OB)/2=<5,1>
2/3 from vertex (origin)=<10/3,2/3>.
Another way:
It turns out that the coordinates of the centroid of a triangle is the mean of the x and y coordinates of the vertices.
In this case, the coordinates are:
<0,0>
<4,3>
<6,-1>
The mean of x=10/3, mean of y=2/3
So G is at <10/3,2/3>.
To find the centroid G of the triangle OAB, we need to take the average of the coordinates of points O, A, and B.
Let's break down the coordinates of each point:
Point O: (0, 0)
Point A: 4i + 3j
Point B: 6i - j
To find the coordinates of G, we add up the corresponding components of O, A, and B, and then divide by 3.
Coordinates of G =
(0 + (4i) + (6i))/3 + (0 + (3j) + (-j))/3
Simplifying this expression, we get:
Coordinates of G =
(10i)/3 + (2j)/3
Therefore, OG = 10i/3 + 2j/3.
To find the centroid G of triangle OAB, we need to calculate the average of the coordinates of points O, A, and B. The centroid is given by the formula:
G = (O + A + B) / 3
First, let's calculate the position vector of point O:
O = OA + OB
O = (4i + 3j) + (6i - j)
Simplifying, we get:
O = (4 + 6)i + (3 - 1)j
O = 10i + 2j
Next, let's calculate the position vector of point A:
A = (4i + 3j)
Finally, let's calculate the position vector of point B:
B = (6i - j)
Now, we can find the position vector OG by substituting the values of O, A, and B into the formula for the centroid:
OG = (O + A + B) / 3
Substituting the values we calculated earlier:
OG = [(10i + 2j) + (4i + 3j) + (6i - j)] / 3
Simplifying, we get:
OG = [(10 + 4 + 6)i + (2 + 3 - 1)j] / 3
OG = (20i + 4j) / 3
Therefore, OG in terms of the unit vectors i and j is:
OG = (20/3)i + (4/3)j