I have this formula in my formula book:

$$cos(\alpha) = \frac{\lvert \overrightarrow u \cdot \overrightarrow v \rvert}{\lvert \overrightarrow u \rvert \cdot \lvert \overrightarrow v \rvert}$$

Also, this same formula book says:

$$ \overrightarrow u \cdot \overrightarrow b = a_1 \cdot b_1 + a_2 \cdot b_2 + a_3 \cdot b_3 $$

Which confuses me. The result of the $ \overrightarrow u \cdot \overrightarrow b = a_1 \cdot b_1 + a_2 \cdot b_2 + a_3 \cdot b_3 $ operation is in my understanding not a vector, and rather instead a single value, like $5$.

That is confusing, because my formula book says:

$$ \lvert \overrightarrow a \rvert = \sqrt{a_1^2 + a_2^2 + a_3^2 }$$

and therefore, $ \lvert \overrightarrow a \rvert = \sqrt{a_1^2 + a_2^2 + a_3^2 }$ requires a vector and not a single value like $5$.

My question: Where's my missunderstanding?

Note: My mathematical knowledge is limited, i'm on a high school education level

  • $\begingroup$ What requires a vector? $\vec u \cdot \vec b$ is a scalar formed by the inner (aka "dot") product of the vectors $\vec u$ and $\vec b$. $\endgroup$
    – Mark Viola
    Mar 31 '17 at 16:18
  • 4
    $\begingroup$ The vertical bars in the numerator mean the absolute value of a real number, but the vertical bars in the denominator mean the length of a vector. Sometimes double vertical bars are used for length instead. $\endgroup$ Mar 31 '17 at 16:20
  • $\begingroup$ @carmichael561aah, that was the fault, i forgot this principle. If you morph your comment into an answer, i'll mark it as the solution. $\endgroup$
    – toogley
    Mar 31 '17 at 16:35

If $\vec{v} \in \mathbb{R}^n$ is a vector, then

$$\lvert\vec{v}\rvert = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}$$

which is called the Euclidean norm of a vector, often denoted with double bars: $\lVert \vec{v} \rVert$.

If $a \in \mathbb{R}$ is a number, then $\lvert a \rvert$ is just the absolute value function. You can also think of this as a particular instance of the norm of a vector with $n = 1$:

$$\lVert\vec{v}\rVert = \sqrt{v_1^2} = \lvert v_1\rvert.$$

  • 1
    $\begingroup$ @toogley: Just to expand on jacer21's answer: the notation $|a|$ can denote either the vector magnitude $\sqrt{a_1^2 + a_2^2 + \dots}$, or can represent the absolute value of a number. You have to know from external context whether $a$ denotes a vector or a number, and thus which interpretation to use for the vertical bars. $\endgroup$
    – WB-man
    Mar 31 '17 at 16:57

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