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

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.

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 $$\vec a + \vec c +(-\vec b) = \vec 0$$ .
• $\vec a + \vec b + (-\vec c) \ne 0$, but $\vec a + \vec c + (-\vec b) = 0$. – Toby Mak Mar 24 at 5:58
$$|a| ≤ |b| + |c|$$ $$|b| ≤ |c| + |a|$$