Answer to Question #209777 in Analytic Geometry for Jaguar

Question

(4.1) 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 ∆.

(4.2) 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.

$(4.1)A=(-1;0;1), B=(0;-2;3), C=(-4;4;1);$

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

$BC=\sqrt{(-4-0)^2+(4+2)^2+(1-3)^2}=\sqrt{56}=2\sqrt{14};$

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

$(4.2) P=(3;1;-1), Q=(2;0;3), R=(1;1;1)$

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

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

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

$PQ^2=QR^2+PR^2;$

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

Answer: $(4.1) AB=3; BC=2\sqrt{14}; AC=5;$

$(4.2)$ A triangle is not right-angled because Pythagoras' theorem does not hold.