The incenter of the triangle ABC,whose vertic | vedclass

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

Question

(C) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.Calculate the side lengths:$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sq

Vedclass · Accepted Answer

$\left(-\frac{3}{2}, \frac{1}{2}, 1\right)$

Vedclass · Answer

(C) Let the vertices be $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$.Calculate the side lengths:$a = BC = \sqrt{(-2 - (-2))^2 + (0 - 0)^2 + (2 - 0)^2} = \sqrt{0 + 0 + 4} = 2$$b = AC = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (2 - 1)^2} = \sqrt{4 + 4 + 1} = 3$$c = AB = \sqrt{(-2 - 0)^2 + (0 - 2)^2 + (0 - 1)^2} = \sqrt{4 + 4 + 1} = 3$The incenter $I$ is given by $\frac{aA + bB + cC}{a + b + c}$.$I = \frac{2(0,2,1) + 3(-2,0,0) + 3(-2,0,2)}{2 + 3 + 3}$$I = \frac{(0,4,2) + (-6,0,0) + (-6,0,6)}{8}$$I = \frac{(-12, 4, 8)}{8} = \left(-\frac{12}{8}, \frac{4}{8}, \frac{8}{8}\right) = \left(-\frac{3}{2}, \frac{1}{2}, 1\right)$.

The incenter of the triangle $ABC$,whose vertices are $A(0,2,1)$,$B(-2,0,0)$,and $C(-2,0,2)$ is

Similar Questions

Three identical balls of radius $2 \, cm$ each are placed on a table such that they touch each other as well as the table. Now a fourth ball of the same radius is placed above these three balls. The height of the highest point on the fourth ball above the table is -

$A(2,3,-4), B(-3,3,-2), C(-1,4,2)$ and $D(3,5,1)$ are the vertices of a tetrahedron. If $E, F, G$ are the centroids of its faces containing the point $A$, then the centroid of the triangle $EFG$ is

If a line makes angles $\alpha, \beta, \gamma, \delta$ with the four diagonals of a cube,then find the value of $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + \cos^2 \delta$.

Find the distance of the point $P(3, 8, 2)$ from the line $\frac{x-1}{2} = \frac{y-3}{4} = \frac{z-2}{3}$ measured parallel to the plane $3x + 2y - 2z + 15 = 0$.

$A$ square $ABCD$ with diagonal length $2a$ is folded along the diagonal $AC$ such that the planes $DAC$ and $BAC$ are perpendicular to each other. What is the shortest distance between $DC$ and $AB$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module