Show that $a,b,c$ forms the sides of a triangle. Please help on my attempt.

Question

Show that $a=2i+2j+3k,b=3i+j-k,c=i-j-4k$ forms the sides of a triangle.

My attempt: $|a|=\sqrt{17},|b|=\sqrt{11},|c|=\sqrt{18}.$ Since $|c|<|a|+|b|$ using triangle inequality, we can say $a,b,c$ form sides of a triangle.

I am not sure if my attempt is correct. Please help.

Answer 1

To use the triangle inequality, you need to show the sum of any two sides is greater than the third side. Therefore, you also need to verify:

$$|a| ≤ |b| + |c|$$ $$|b| ≤ |c| + |a|$$

Answer 2

In the vector form it is necessary to check the directions. By the triangle law of vector addition , the sum of vectors (in either of direction)should be zero. Then they are guaranteed to be sides of a triangle,then no need to check length or any other conditions. Also your method is not a proof. You can see here $\vec a + \vec c= \vec b$ or $\vec a + \vec c +(-\vec b) = \vec 0$ .