Question

The centroid of a triangle of the triangle OAB is denoted by G. If o is the origin and line(OA) = 4i + 3j, line(OB) = 6i - j,find line(OG) in terms of the unit vectors I and j

a. 10i - 3j

b. 1/2(10i−2j)
c. 10i + 2j

d. 1/3(10i+2j)

Answer 1

Answer on Question #51026 – Math – Vector Calculus

Problem

The centroid of the triangle OAB is denoted by $G$. If $o$ is the origin and $\text{line(OA)} = 4i + 3j$, $\text{line(OB)} = 6i - j$, find $\text{line(OG)}$ in terms of the unit vectors $i$ and $j$.

a. $10i - 3j$

b. $1/2(10i - 2j)$

c. $10i + 2j$

d. $1/3(10i + 2j)$

Solution

Vector $\overline{OB} = \overline{OA} + \overline{AB}$, hence $\overline{AB} = \overline{OB} - \overline{OA}, \overline{AM} = \frac{1}{2}\overline{AB}$.

Vector $\overline{OM} = \overline{OA} + \overline{AM} = \overline{OA} + \frac{1}{2}\overline{AB} = \overline{OA} + \frac{1}{2}(\overline{OB} - \overline{OA}) = \frac{1}{2}(\overline{OA} + \overline{OB})$.

Let OM be the median of the triangle OAB. By properties of centroid, $OG = \frac{2}{3} OM$ then.

Thus,

$$\overline{OG} = \frac{2}{3} \overline{OM} = \frac{2}{3} \cdot \frac{1}{2} (\overline{OA} + \overline{OB}) = \frac{1}{3} (\overline{OA} + \overline{OB}) = \frac{1}{3} (4i + 3j + 6i - j) = \frac{1}{3} (10i + 2j) = \frac{10}{3}i + \frac{2}{3}j$$

Answer: d. $1/3(10i + 2j)$

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