The incenter and centroid of the triangle,who | vedclass

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

Question

(D) Given vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$.First,calculate the lengths of the sides:$a = BC = \sqrt{(0-0)^2 + (3-

Vedclass · Accepted Answer

$(0,2,3),\left(0,2, \frac{8}{3}\right)$

Vedclass · Answer

(D) Given vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$.First,calculate the lengths of the sides:$a = BC = \sqrt{(0-0)^2 + (3-0)^2 + (4-4)^2} = \sqrt{0+9+0} = 3$$b = CA = \sqrt{(0-0)^2 + (3-3)^2 + (4-0)^2} = \sqrt{0+0+16} = 4$$c = AB = \sqrt{(0-0)^2 + (0-3)^2 + (4-0)^2} = \sqrt{0+9+16} = \sqrt{25} = 5$Now,the Incenter $I$ is given by $\left(\frac{ax_1+bx_2+cx_3}{a+b+c}, \frac{ay_1+by_2+cy_3}{a+b+c}, \frac{az_1+bz_2+cz_3}{a+b+c}\right)$:$I = \left(\frac{3(0)+4(0)+5(0)}{3+4+5}, \frac{3(3)+4(0)+5(3)}{3+4+5}, \frac{3(0)+4(4)+5(4)}{3+4+5}\right)$$I = \left(\frac{0}{12}, \frac{9+0+15}{12}, \frac{0+16+20}{12}\right) = \left(0, \frac{24}{12}, \frac{36}{12}\right) = (0, 2, 3)$The Centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$:$G = \left(\frac{0+0+0}{3}, \frac{3+0+3}{3}, \frac{0+4+4}{3}\right) = \left(0, \frac{6}{3}, \frac{8}{3}\right) = \left(0, 2, \frac{8}{3}\right)$Thus,the incenter and centroid are $(0, 2, 3)$ and $\left(0, 2, \frac{8}{3}\right)$ respectively.

The incenter and centroid of the triangle,whose vertices are $A \equiv(0,3,0), B \equiv(0,0,4)$,and $C \equiv(0,3,4)$,are respectively given by

Similar Questions

Given $\triangle ABC$ such that $A = 2\hat{i} - \hat{j} + \hat{k}$,$B = \hat{i} - 3\hat{j} - 5\hat{k}$,and $C = 3\hat{i} - 4\hat{j} - 4\hat{k}$,then $\triangle ABC$ is:

In $\triangle ABC$,the midpoints of the sides $AB, BC$ and $CA$ are respectively $(l, 0, 0), (0, m, 0)$ and $(0, 0, n)$. Then,$\frac{AB^2+BC^2+CA^2}{l^2+m^2+n^2}$ is equal to

If $A(1, 2, -3)$,$B(2, 3, -1)$,and $C(3, 1, 1)$ are the vertices of $\triangle ABC$,then $\left|\frac{\cos A}{\cos B}\right| = $

In a triangle $ABC$,if the midpoints of sides $AB, BC, CA$ are $(3,0,0), (0,4,0), (0,0,5)$ respectively,then $AB^2+BC^2+CA^2=$

The circumradius of the triangle formed by the points $A(2, -1, 1)$,$B(1, -3, -5)$,and $C(3, -4, -4)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module