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# 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)

Last updated date: 13th Jun 2024
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Answer
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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.