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Chapter 16: Problem 1480

If \(\mathrm{G}(0)\) is centroid of \(\triangle \mathrm{ABC}\), then \(\underline{\mathrm{GA}}+\underline{\mathrm{GB}}+\underline{\mathrm{GC}}=\) (A) \(\underline{0}\) (B) 0 (C) \(\underline{x}+y+\underline{z}\) (D) \([(\underline{x}+y+\underline{z}) / 3]\)

Short Answer

Expert verified
\(\underline{GA} + \underline{GB} + \underline{GC} = 0\)

Step by step solution

01

Recall the Centroid formula

The Centroid of a triangle with vertices A, B, and C is the point where the three medians of the triangle intersect. It can be found using the formula: \( G(x, y, z) = \frac{1}{3} (A + B + C) \) where A, B, and C are the coordinates of the vertices of the triangle.
02

Given Information

In this exercise, we are given that G(0) is the centroid of triangle ABC. In other words, the position vector of G, denoted as \(\underline{G}\), is 0. Therefore, the coordinates of G can be represented as (0, 0, 0) or: \( \underline{G} = 0 \)
03

Rearrange Centroid formula

First, we need to find the position vectors \(\underline{GA}\), \(\underline{GB}\), and \(\underline{GC}\). From the Centroid formula, we can rearrange the formula: \( 3 \cdot \underline{G} = \underline{A} + \underline{B} + \underline{C} \) Since we know that the position vector \(\underline{G}\) = 0, we insert this into our formula: \( 3 \cdot 0 = \underline{A} + \underline{B} + \underline{C} \)
04

Find position vectors GA, GB, and GC

The sum of position vectors GA, GB, and GC can be represented by: \( \underline{GA} + \underline{GB} + \underline{GC} = 3 \cdot 0 \) According to the given information, G(0) is the centroid of the triangle, meaning that the sum of the position vectors GA, GB, and GC is equal to 0.
05

Conclusion

Our final answer is: \( \underline{GA} + \underline{GB} + \underline{GC} = 0 \) The correct answer is (B) 0.

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Most popular questions from this chapter

If angle between two units vectors \(\underline{a}\) and \(\underline{b}\) is \(\theta\), then \(\cos (\theta / 2)=0<\theta<\pi\) (A) \(|\underline{\mathrm{a}}+\underline{\mathrm{b}}|\) (B) \((1 / 2)|\underline{a}+\underline{b}|\) (C) \((1 / 2)|\underline{\mathrm{a}}+\underline{\mathrm{b}}|^{2}\) (D) \(|\underline{a}+\underline{b}|^{2}\)

The direction cosine of \(\mathrm{x}=\mathrm{ay}+\mathrm{b}, \mathrm{z}=\mathrm{cy}+\mathrm{d}\) (A) \(\pm\left[a / \sqrt{\left(a^{2}+c^{2}+1\right)}\right], \pm\left[1 / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left.\pm\left[\mathrm{c} / \sqrt{(} \mathrm{a}^{2}+\mathrm{c}^{2}+1\right)\right]\) (C) \(\left[(-a) / \sqrt{\left(a^{2}+c^{2}+1\right)}\right],\left[(-1) / \sqrt{ \left.\left(a^{2}+c^{2}+1\right)\right]}\right.\) \(\left[(-c) / \sqrt{\left(a^{2}+c^{2}+1\right)}\right]\) (D) None of these

The equation of plane passing through point \((1,-1,2)\) and \(\underline{r}=(1,1,1)+\mathrm{k}(2,1,2), \mathrm{k} \in \mathrm{R}\) is (A) \(5 \mathrm{x}-2 \mathrm{y}-4 \mathrm{z}+1=0\) (B) \(5 \mathrm{x}+2 \mathrm{y}+4 \mathrm{z}+1=0\) (C) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}+1=0\) (D) \(5 \mathrm{x}-2 \mathrm{y}+4 \mathrm{z}=1\)

If vector \(\underline{r}\) forms an angle \(\alpha, \beta, \gamma\) with \(\mathrm{x}, \mathrm{y}, \mathrm{z}\) -axis then \(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=\) (A) 1 (B) 2 (C) \(-1\) (D) \(-2\)

Equation of line passing through \((0,0,0)\) and parallel to Y-axis is (A) \((\mathrm{x} / 0) \overline{=(\mathrm{y}} / 1)=(\mathrm{z} / 1)\) (B) \((\mathrm{x} / 0)=(\mathrm{y} / 1)=(\mathrm{z} / 0)\) (C) \((\mathrm{x} / 0)=(\mathrm{y} / 0)=(\mathrm{z} / 1)\) (D) \((\mathrm{x} / 1)=(\mathrm{y} / 0)=(\mathrm{z} / 1)\)
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