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Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of these two vectors?

Can the sum of the magnitudes of two vectors ever be equal to the magnitude of the sum of these two vectors?

Grade:upto college level

1 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago
Yes, the sum of magnitude of two vectors can be equal to the magnitude of sum of these two vectors.
Consider two vectors\overrightarrow{a} = a_{x}\widehat{i} +a _{y}\widehat{y} and such that the sum of both the vectors is given by another vector as:
\overrightarrow{c} = \overrightarrow{a} + \overrightarrow{b}
Substitute\overrightarrow{a} = a_{x}\widehat{i} + a_{y}\widehat{j} and , \overrightarrow{b} = b_{x}\widehat{i} + b_{y}\widehat{j}
\overrightarrow{c} = a_{x}\widehat{i} + a_{y}\widehat{j} + b_{x}\widehat{i} + b_{y}\widehat{j}
= (a_{x} + b_{x})\widehat{i} + (a_{y} + b_{y})\widehat{j}

The magnitude of vector \overrightarrow{c}is:
|\overrightarrow{c}| = \sqrt{(a_{x}+b_{x})^{2} + (a_{y}+b_{y})^{2}}
The magnitude of vectors is: \overrightarrow{a} = a_{x}\widehat{i} + a_{y}\widehat{j}
|\overrightarrow{a}| = \sqrt{(a_{x})^{2}+ (a_{y})^{2}}
The magnitude of vectors\overrightarrow{b} = b_{x}\widehat{i} + b_{y}\widehat{j} is:
|\overrightarrow{b}| = \sqrt{(b_{x})^{2}+ (b_{y})^{2}}
If the sum of magnitude of vector\overrightarrow{a} and magnitude of vector\overrightarrow{b} is equal to magnitude of vector\overrightarrow{c} , then
|\overrightarrow{c}| = |\overrightarrow{b}| + |\overrightarrow{a}|
Square both sides and expand,

|\overrightarrow{c}|^{2} = \left ( |\overrightarrow{b}| \right + |\overrightarrow{a}|)^{2}
= |\overrightarrow{a}|^{2} + |\overrightarrow{b}|^{2} + 2 |\overrightarrow{a}| |\overrightarrow{b}|
Substitute|\overrightarrow{c}| = \sqrt{(a_{x}+ b_{x})^{2} + (a_{y}+ b_{y})^{2}} , |\overrightarrow{a}| = \sqrt{(a_{x})^{2} + (a_{y})^{2}}and |\overrightarrow{b}| = \sqrt{(b_{x})^{2} + (b_{x})^{2}},


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Square both sides,
Therefore, if two vectors are such that the condition2a_{x}a_{y}b_{x}v_{y} = a_{x}^{2}b_{y}^{2}+ a_{y}^{2}b_{x}^{2} is satisfied, then the magnitude of the sum of two vectors is equal to the sum of magnitude of the vectors.

For example, assume \overrightarrow{b} = -\widehat{i}-\widehat{j}and\overrightarrow{b} = -\widehat{i}-\widehat{j} , then the condition 2a_{x}a_{y}b_{x}b_{y} = a_{x}^{2} b_{y}^{2}+ a_{y}^{2} b_{x}^{2} gives 2 = 2 , therefore for these vectors that sum of magnitude of each vector is equal to the magnitude of the sum of the vectors

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