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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.

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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 $\vec a + \vec c +(-\vec b) = \vec 0$ .

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  • $\begingroup$ $\vec a + \vec b + (-\vec c) \ne 0$, but $\vec a + \vec c + (-\vec b) = 0$. $\endgroup$ – Toby Mak Mar 24 at 5:58
  • $\begingroup$ Just corrected:) $\endgroup$ – Tojrah Mar 24 at 6:02
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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:

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

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