The points A(2, -1, 4),B(1, 0, -1),C(1, 2, 3) | vedclass

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

Question

(D) Given points are $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$.First,we calculate the lengths of the sides using the distance formula

Vedclass · Accepted Answer

parallelogram

Vedclass · Answer

(D) Given points are $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$.First,we calculate the lengths of the sides using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.$AB = \sqrt{(1-2)^2 + (0-(-1))^2 + (-1-4)^2} = \sqrt{(-1)^2 + 1^2 + (-5)^2} = \sqrt{1+1+25} = \sqrt{27}$.$BC = \sqrt{(1-1)^2 + (2-0)^2 + (3-(-1))^2} = \sqrt{0^2 + 2^2 + 4^2} = \sqrt{0+4+16} = \sqrt{20}$.$CD = \sqrt{(2-1)^2 + (1-2)^2 + (8-3)^2} = \sqrt{1^2 + (-1)^2 + 5^2} = \sqrt{1+1+25} = \sqrt{27}$.$DA = \sqrt{(2-2)^2 + (-1-1)^2 + (4-8)^2} = \sqrt{0^2 + (-2)^2 + (-4)^2} = \sqrt{0+4+16} = \sqrt{20}$.Since $AB = CD = \sqrt{27}$ and $BC = DA = \sqrt{20}$,the opposite sides are equal.Now,we check the diagonals $AC$ and $BD$.$AC = \sqrt{(1-2)^2 + (2-(-1))^2 + (3-4)^2} = \sqrt{(-1)^2 + 3^2 + (-1)^2} = \sqrt{1+9+1} = \sqrt{11}$.$BD = \sqrt{(2-1)^2 + (1-0)^2 + (8-(-1))^2} = \sqrt{1^2 + 1^2 + 9^2} = \sqrt{1+1+81} = \sqrt{83}$.Since $AC 
eq BD$,the diagonals are not equal.Therefore,the points $A, B, C$,and $D$ form a parallelogram.

The points $A(2, -1, 4)$,$B(1, 0, -1)$,$C(1, 2, 3)$,and $D(2, 1, 8)$ form a

Similar Questions

If the mid-points of the sides $AB, BC, CA$ of a triangle are $(1, 5, -1), (0, 4, -2), (2, 3, 4)$ respectively,then the length of the median drawn from $C$ to $AB$ is

If $A(2, 4, -1)$,$B(3, 6, -1)$,and $C(4, 5, 1)$ are three consecutive vertices of a parallelogram,then its fourth vertex is

The centroid of a triangle with vertices $A(3,4,5)$, $B(6,7,2)$, and $C(x, y, z)$ is $(3,2,3)$. Then $x+y+z=$

Find the centroid of a triangle,the mid-points of whose sides are $D(1, 2, -3)$,$E(3, 0, 1)$,and $F(-1, 1, -4)$.

If $G(3, -5, r)$ is the centroid of triangle $ABC$ where $A(7, -8, 1)$,$B(p, q, 5)$,and $C(q+1, 5p, 0)$ are the vertices of the triangle,then the values of $p, q, r$ are respectively . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module