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

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.
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tun1t2j
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 $\stackrel{\to }{a}+\stackrel{\to }{c}=\stackrel{\to }{b}$ or $\stackrel{\to }{a}+\stackrel{\to }{c}+\left(-\stackrel{\to }{b}\right)=\stackrel{\to }{0}$