Question

Show that vectors a = (4, 0, - 1 ) , b = (1,3, 4), and c = (-5, - 3 , - 3 ) form the sides of a triangle. Is this a right angle triangle? Calculate the area of the triangle.

Answer 1

use herons formula or 1/2| formula

Answer 2

Given :

3 vector points :

$a(4,0,-1) , b(1,3,4) , c(-5,-3,-3)$

To Check :

(1). If the points making a triangle or not.

(2). If the given points make right angle triangle.

(3). Area of triangle.

Step-by-step explanation:

(1). First we will calculate the distances between given points :

$C$ = distance $ab$ = $\sqrt{(4-1)^2 + (0-3)^2 + (-1-4)^2} = \sqrt{9 + 9 + 25} = \sqrt{43} = 6.56$

$A$ = distance $bc$$= \sqrt{(1-(-5))^2 + (3-(-3))^2 + (4-(-3))^2} = \sqrt{36 + 36 + 49} = \sqrt{121} = 11$

$B$ = distance ac = $\sqrt{(4-(-5))^2 + (0-(-3))^2 + (-1 - (-3))^2} = \sqrt{81 + 9 + 4} = \sqrt{94} = 9.7$

Here,

$A + B > C\\B + C > A\\C + A > B$

Hence , it is a triangle

(2). here,

$A^2 + B^2 \neq C^2\\B^2 + C^2 \neq A^2\\C^2 + A^2 \neq B^2$

Hence, it is not a right angle triangle

(3).

Area of a triangle = $\sqrt{S(S-A)(S-B)(S-B)}$

where, S = $\frac{6.56 + 9.7 + 11}{3} = 9.08$

Area =

$\sqrt{(9.08 - 9.7)(9.08-11)(9.08-6.56)} = \sqrt{9.08*0.62*4.92*2.52} = \sqrt{69.798} = 8.35$

Learn More About Triangle's Properties :

https://brainly.in/question/14856390

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