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)

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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)

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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.

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)

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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)

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