Let the position vectors of three vertices of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let the vertices be $A, B, C$. The centroid $G$ of the triangle is given by the average of the position vectors of its vertices:$G = \frac{(4\over

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) Let the vertices be $A, B, C$. The centroid $G$ of the triangle is given by the average of the position vectors of its vertices:$G = \frac{(4\overrightarrow{p} + \overrightarrow{q} - 3\overrightarrow{r}) + (-5\overrightarrow{p} + \overrightarrow{q} + 2\overrightarrow{r}) + (2\overrightarrow{p} - \overrightarrow{q} + 2\overrightarrow{r})}{3}$$G = \frac{(4-5+2)\overrightarrow{p} + (1+1-1)\overrightarrow{q} + (-3+2+2)\overrightarrow{r}}{3} = \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{3}$We know that the orthocenter $(O)$,centroid $(G)$,and circumcenter $(C)$ are collinear,and the centroid divides the line segment joining the orthocenter and circumcenter in the ratio $2:1$. Thus,$G = \frac{1 \cdot O + 2 \cdot C}{3}$,which implies $3G = O + 2C$.Given $O = \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{4}$ and $C = \alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r}$,we have:$3 \left( \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{3} \right) = \frac{\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}}{4} + 2(\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r})$$\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r} - \frac{1}{4}(\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}) = 2(\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r})$$\frac{3}{4}(\overrightarrow{p} + \overrightarrow{q} + \overrightarrow{r}) = 2(\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r})$$\alpha \overrightarrow{p} + \beta \overrightarrow{q} + \gamma \overrightarrow{r} = \frac{3}{8}\overrightarrow{p} + \frac{3}{8}\overrightarrow{q} + \frac{3}{8}\overrightarrow{r}$Comparing coefficients,we get $\alpha = \frac{3}{8}, \beta = \frac{3}{8}, \gamma = \frac{3}{8}$.Therefore,$\alpha + 2\beta + 5\gamma = \frac{3}{8} + 2(\frac{3}{8}) + 5(\frac{3}{8}) = \frac{3 + 6 + 15}{8} = \frac{24}{8} = 3$.

List-$I$	List-$II$
$(P)$ $\|\vec{v}\|^2$ is equal to	$(1)$ $0$
$(Q)$ If $\alpha=\sqrt{3}$,then $\gamma^2$ is equal to	$(2)$ $1$
$(R)$ If $\alpha=\sqrt{3}$,then $(\beta+\gamma)^2$ is equal to	$(3)$ $2$
$(S)$ If $\alpha=\sqrt{2}$,then $t+3$ is equal to	$(4)$ $3$
	$(5)$ $5$

Similar Questions

If $ABCDEF$ is a regular hexagon,then $\overrightarrow{AD} + \overrightarrow{EB} + \overrightarrow{FC} = .....$

Let $a, b, c$ be three non-coplanar vectors such that $r_1 = a - b + c$,$r_2 = b + c - a$,$r_3 = c + a + b$,and $r = 2a - 3b + 4c$. If $r = \lambda_1 r_1 + \lambda_2 r_2 + \lambda_3 r_3$,then:

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be any three non-coplanar vectors. If $m$ and $n$ are scalars such that $\vec{a}+\vec{b}=m \vec{d}-\vec{c}$ and $\vec{b}+\vec{c}=n \vec{a}-\vec{d}$,then $3 \vec{a}+2 \vec{b}+2 \vec{c}+\vec{d}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module