The points (5, -4, 2), (4, -3, 1), (7, -6, 4) | vedclass

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

Question

(D) Let the given points be $A(5, -4, 2)$,$B(4, -3, 1)$,$C(7, -6, 4)$,and $D(8, -7, 5)$.Calculate the vectors:$\vec{AB} = (4-5, -3-(-4), 1-2) = (-1, 1

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Let the given points be $A(5, -4, 2)$,$B(4, -3, 1)$,$C(7, -6, 4)$,and $D(8, -7, 5)$.Calculate the vectors:$\vec{AB} = (4-5, -3-(-4), 1-2) = (-1, 1, -1)$$\vec{BC} = (7-4, -6-(-3), 4-1) = (3, -3, 3)$$\vec{CD} = (8-7, -7-(-6), 5-4) = (1, -1, 1)$$\vec{DA} = (5-8, -4-(-7), 2-5) = (-3, 3, -3)$Note that $\vec{BC} = -3\vec{AB}$ and $\vec{DA} = 3\vec{AB}$.Since the vectors are collinear,the points $A, B, C, D$ lie on a single straight line.Therefore,they do not form a polygon (rectangle,square,or parallelogram).Hence,option $(d)$ is correct.

The points $(5, -4, 2), (4, -3, 1), (7, -6, 4)$ and $(8, -7, 5)$ are the vertices of

Similar Questions

The extremities of a diagonal of a parallelogram are the points $(3, -4)$ and $(-6, 5)$. If the third vertex is $(-2, 1)$,then the fourth vertex is:

Three vertices of a parallelogram $ABCD$ are $A(3, -1, 2)$,$B(1, 2, -4)$,and $C(-1, 1, 2)$. Find the coordinates of the fourth vertex $D$.

Points $(1, 1, 1), (-2, 4, 1), (-1, 5, 5)$ and $(2, 2, 5)$ are the vertices of a

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module