If A(2, 4, -1),B(3, 6, -1),and C(4, 5, 1) are | vedclass

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

Question

(D) Let the vertices of the parallelogram be $A(2, 4, -1)$,$B(3, 6, -1)$,$C(4, 5, 1)$,and $D(x, y, z)$.Since the diagonals of a parallelogram bisect e

Vedclass · Accepted Answer

$(3, 3, 1)$

Vedclass · Answer

(D) Let the vertices of the parallelogram be $A(2, 4, -1)$,$B(3, 6, -1)$,$C(4, 5, 1)$,and $D(x, y, z)$.Since the diagonals of a parallelogram bisect each other,the midpoint of diagonal $AC$ is the same as the midpoint of diagonal $BD$.Midpoint of $AC = \left( \frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2} \right) = \left( 3, \frac{9}{2}, 0 \right)$.Midpoint of $BD = \left( \frac{3+x}{2}, \frac{6+y}{2}, \frac{-1+z}{2} \right)$.Equating the midpoints:$\frac{3+x}{2} = 3 \implies 3+x = 6 \implies x = 3$.$\frac{6+y}{2} = \frac{9}{2} \implies 6+y = 9 \implies y = 3$.$\frac{-1+z}{2} = 0 \implies -1+z = 0 \implies z = 1$.Thus,the fourth vertex $D$ is $(3, 3, 1)$.Therefore,option $(D)$ is correct.

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

Similar Questions

$\triangle ABC$ has vertices at $A \equiv (2, 3, 5)$,$B \equiv (-1, 3, 2)$ and $C \equiv (\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the axes,then the values of $\lambda$ and $\mu$ respectively are

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| = $

The perimeter of the triangle with vertices at $(1,0,0), (0,1,0)$ and $(0,0,1)$ is

If $D(2, 1, 0)$,$E(2, 0, 0)$,and $F(0, 1, 0)$ are mid-points of the sides $BC$,$CA$,and $AB$ of $\triangle ABC$,respectively,then the centroid of $\triangle ABC$ is

If $A(1, 1, 2), B(2, 1, 2), C(2, 2, 1)$,then $A, B, C$ are $........$

Vedclass Test Series

Exam Paper Generator

Online Exam Module