Question

1.consider the point a=(-1,0,1), b=(0,-2,3),and c=(-4,4,1) to be vertices of a triangle∆.evaluate all side lengths of ∆.

2.let ∆ be the triangle with vertices the points p=(3,1,-1),q=(2,0,3)and r=(1,1,1).determine whether ∆ is a right angle triangle.if it is not ,explain with reason,why?

Answer 1

Solution.

$1. a=(-1;0;1); b=(0;-2;3); c=(-4;4;1);$

$ab=\sqrt{(0+1)^2+(-2-0)^2+(3-1)^2}=3;$

$bc=\sqrt{(-4-0)^2+(4+2)^2+(1-3)^2}=2\sqrt{14};$

$ac=\sqrt{(-4+1)^2+(4-0)^2+(1-1)^2}=5;$

$2. p=(3,1,-1),q=(2,0,3), r=(1,1,1);$

$pq=\sqrt{(2-3)^2+(0-1)^2+(3+1)^2}=3\sqrt{2};$

$pr=\sqrt{(1-3)^2+(1-1)^2+(1+1)^2}=2\sqrt{2};$

$qr=\sqrt{(1-2)^2+(1-0)^2+(1-3)^2}=\sqrt{6};$

If the triangle is right-angled, then the square of the hypotenuse is equal to the sum of the squares of the legs:

$pq^2=pr^2+qr^2;$

$18\not=8+6=14,$ the triangle is not right-angled.

Answer: $1) 3;2\sqrt{14};5;$

$2)$ A triangle is not right-angled because Pythagoras' theorem does not hold.