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Finding length or magnitude using vector addition and the Pythagorean theorem.

I am trying to understand why vector addition and the Pythagorean theorem are giving different results?

vector_addition

Vector Addition :

According to diagram (A) : $\vec{a} + \vec{b} = \vec{c}$

now suppose : magnitude of $\vec{a} = 3$, magnitude of $\vec{b} = 4$ then

$\|\vec{c}\| = \|\vec{a}\| + \|\vec{b}\|$

$\|\vec{c}\| = 3 + 4$

magnitude of c = 7

Pythagorean Theorem

Now when we consider this as a triangle shown in diagram (B)

Similarly, length of $a = 3$, length of $b = 4$

so according to Pythagorean theorem

$\text{hypotenuse}^2 = \text{opposite}^2 + \text{adjacent}^2$

$c^2 = a^2 + b^2$

i.e.

$c^2 = 3^2 + 4^2$

$c = \sqrt{9 + 16} = \sqrt{25}$

length of c = 5

Why there is inconsistency in the results? I am doing something wrong?

Thanks!

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    $\begingroup$ In general $\|a\|+\|b\|\neq \|a+b\|$. What is true however is the "triangle inequality" for norms, saying $\|a+b\|\leq \|a\|+\|b\|$. In terms of euclidean distance, equality only holds if $a$ and $b$ are in the same "direction." $\endgroup$ – JMoravitz Apr 19 '16 at 1:43
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    $\begingroup$ Letting $a=(3,0)$ and $b=(0,4)$, we have $c=a+b=(3,0)+(0,4)=(3,4)$. The magnitude of $c$ (with respect to euclidean distance) is $\|(3,4)\|=\sqrt{3^2+4^2}=\sqrt{25}=5$, as expected. $\endgroup$ – JMoravitz Apr 19 '16 at 1:46
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    $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Apr 19 '16 at 2:01
  • $\begingroup$ thank you for updating the question. I will go through the tutorial! $\endgroup$ – user2259784 Apr 19 '16 at 17:36
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Vector Magnitude

The magnitude of a vector $(\vec{a} + \vec{b})$ is obtained in a similar way to that for Pythagorean triangles. It is incorrect to simply add up magnitudes of $\vec{a}$ and $\vec{b}$.

Rather, $\| \vec{c} \| = \sqrt {\| \vec{a} \| + \| \vec{b} \|} $ (See link).

Khan Academy

As @JMoravitz has already demonstrated.

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    $\begingroup$ As a side note, for magnitude, it is recommended to use \|a\| as opposed to ||a||. Compare the following: $\|a\|~~~~~||a||$ $\endgroup$ – JMoravitz Apr 19 '16 at 1:54
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    $\begingroup$ Thank you for the note. I shall adjust. $\endgroup$ – Sharky Apr 19 '16 at 1:56

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