Question #209777

(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?


1
Expert's answer
2021-06-23T14:28:00-0400

Solution.

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

AB=(0+1)2+(20)2+(31)2=3;AB=\sqrt{(0+1)^2+(-2-0)^2+(3-1)^2}=3;

BC=(40)2+(4+2)2+(13)2=56=214;BC=\sqrt{(-4-0)^2+(4+2)^2+(1-3)^2}=\sqrt{56}=2\sqrt{14};

AC=(4+1)2+(40)2+(11)2=5;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)(4.2) P=(3;1;-1), Q=(2;0;3), R=(1;1;1)

PQ=(23)2+(01)2+(3+1)2=32;PQ=\sqrt{(2-3)^2+(0-1)^2+(3+1)^2}=3\sqrt{2};

QR=(12)2+(10)2+(13)2=6;QR=\sqrt{(1-2)^2+(1-0)^2+(1-3)^2}=\sqrt{6};

PR=(13)2+(11)2+(1+1)2=22;PR=\sqrt{(1-3)^2+(1-1)^2+(1+1)^2}=2\sqrt{2};

PQ2=QR2+PR2;PQ^2=QR^2+PR^2;

186+8;1814,18\not =6+8; 18\not=14, therefore, the triangle is not right-angled.

Answer: (4.1)AB=3;BC=214;AC=5;(4.1) AB=3; BC=2\sqrt{14}; AC=5;

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


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