Verify that the points (0, 7, 10),(-1, 6, 6), | vedclass

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

Question

(N/A) Let the points be $A(0, 7, 10)$,$B(-1, 6, 6)$,and $C(-4, 9, 6)$.Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:

Vedclass · Accepted Answer

Vedclass · Answer

(N/A) Let the points be $A(0, 7, 10)$,$B(-1, 6, 6)$,and $C(-4, 9, 6)$.
Using the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$:
$AB = \sqrt{(-1-0)^2 + (6-7)^2 + (6-10)^2} = \sqrt{(-1)^2 + (-1)^2 + (-4)^2} = \sqrt{1 + 1 + 16} = \sqrt{18} = 3\sqrt{2}$
$BC = \sqrt{(-4 - (-1))^2 + (9-6)^2 + (6-6)^2} = \sqrt{(-3)^2 + (3)^2 + (0)^2} = \sqrt{9 + 9 + 0} = \sqrt{18} = 3\sqrt{2}$
$CA = \sqrt{(0 - (-4))^2 + (7-9)^2 + (10-6)^2} = \sqrt{(4)^2 + (-2)^2 + (4)^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$
Now,check for the Pythagoras theorem condition $AB^2 + BC^2 = AC^2$:
$AB^2 + BC^2 = (3\sqrt{2})^2 + (3\sqrt{2})^2 = 18 + 18 = 36$
$AC^2 = (6)^2 = 36$
Since $AB^2 + BC^2 = AC^2$,the points $A, B,$ and $C$ form a right-angled triangle.

Verify that the points $(0, 7, 10)$,$(-1, 6, 6)$,and $(-4, 9, 6)$ are the vertices of a right-angled triangle.

Similar Questions

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

If the origin is the centroid of the triangle whose vertices are $A(2, p, -3)$,$B(q, -2, 5)$,and $C(-5, 1, r)$,then

The circumcentre of the triangle formed by the points $A(3,4,5)$,$B(2,3,1)$,and $C(-1,6,1)$ is:

If $(1,0,3), (2,1,5), (-2,3,6)$ are the mid-points of the sides of a triangle, then the centroid of the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module