Question

Prove that for any two vectors $\overrightarrow a $and $\overrightarrow b $we always have, $|\overrightarrow a + \overrightarrow b | \leqslant |\overrightarrow a | + |\overrightarrow b |$ (triangle inequality)

Answer 1

Hint:The question can be started by taking the term $|\overrightarrow a + \overrightarrow b |$ and squaring it. Since, we know that the identity that the square of modulus of a vector is equal to the dot product of it with itself we can use this identity for solving the question.

Complete step-by-step answer:
We are given that $\overrightarrow a $and $\overrightarrow b $are two vectors of triangle and also we need to prove that $|\overrightarrow a + \overrightarrow b | \leqslant |\overrightarrow a | + |\overrightarrow b |$ is true for any value of these two vectors.
We will start of by understanding what this triangle inequality means.
Triangle inequality says that the sum of two sides of a triangle is always greater than the third side. Here the term $|\overrightarrow a + \overrightarrow b |$ is referred to as the third side and $|\overrightarrow a | + |\overrightarrow b |$are the sum of the rest of the two sides.
The question can be started by taking the term $|\overrightarrow a + \overrightarrow b |$ and squaring it. We get,
\[|\overrightarrow a + \overrightarrow b {|^2} = (\overrightarrow a + \overrightarrow b ).(\overrightarrow a + \overrightarrow b )\]……….. Here, the property used is $|\overrightarrow c {|^2} = \overrightarrow {c.} \overrightarrow c $
\[|\overrightarrow a + \overrightarrow b {|^2} = (\overrightarrow a + \overrightarrow b ).(\overrightarrow a + \overrightarrow b )\]
\[|\overrightarrow a + \overrightarrow b {|^2} = \overrightarrow a .\overrightarrow a + \overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow a + \overrightarrow b .\overrightarrow b .............(1)\]
Since, dot product of vectors are commutative which means\[\overrightarrow a .\overrightarrow b = \overrightarrow b .\overrightarrow a \]. We will use it in equation (1),
\[|\overrightarrow a + \overrightarrow b {|^2} = \overrightarrow a .\overrightarrow a + 2\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow b .\]
Now, we have a dot product of two vectors as a multiplication of their magnitudes and cosine of angle between the vectors. We get,
\[|\overrightarrow a + \overrightarrow b {|^2} = |\overrightarrow a {|^2} + 2\overrightarrow a .\overrightarrow b + |\overrightarrow b {|^2}\]……….here, the property used is \[\overrightarrow a .\overrightarrow a = |\overrightarrow a {|^2}\]
\[|\overrightarrow a + \overrightarrow b {|^2} = |\overrightarrow a {|^2} + 2|\overrightarrow a ||\overrightarrow b |\operatorname{Cos} \theta + |\overrightarrow b {|^2}\]
Where, \[\theta\] is the angle between the vectors $\overrightarrow a $and$\overrightarrow b $.
Now, we know that
 \[\operatorname{Cos} \theta \leqslant 1\]
Multiplying \[2|\overrightarrow a ||\overrightarrow b |\] both sides we get,
\[2|\overrightarrow a ||\overrightarrow b |\operatorname{Cos} \theta \leqslant 2|\overrightarrow a ||\overrightarrow b |\]
Adding $|\overrightarrow a {|^2} + |\overrightarrow b {|^2}$on both the sides we get,
\[2|\overrightarrow a ||\overrightarrow b |\operatorname{Cos} \theta + |\overrightarrow a {|^2} + |\overrightarrow b {|^2} \leqslant 2|\overrightarrow a ||\overrightarrow b | + |\overrightarrow a {|^2} + |\overrightarrow b {|^2}\]
Forming the perfect squares we have,
\[|\overrightarrow a + \overrightarrow b {|^2} \leqslant {(|\overrightarrow a | + |\overrightarrow b |)^2}\]
Taking square root both the sides we get,
\[|\overrightarrow a + \overrightarrow b | \leqslant (|\overrightarrow a | + |\overrightarrow b |)\]
Hence proved.

Note:The dot products in vector are commutative but this property shouldn’t be applied when we take cross products. Since, the cross product of vectors are not commutative.