Audrey Hall

2023-03-25

How to find the angle between the vector and $x-$axis?

Kasztelanwhd7

Beginner2023-03-26Added 5 answers

Measurement of the angle between $x-$axis and vector:

Angle between vectors $\alpha ={\mathrm{cos}}^{-1}\left(\frac{a.\to b\to}{|a\to ||b\to |}\right)$.

x-axis is represented by $i\to $.

Let vector be $xi\to +yj\to +zk\to $

$a\to .b\to =i\to .(xi\to +yj\to +zk\to )a\to .b\to =x\left[\because (i\to .i\to =1),(i\to .j\to =0),(i\to .k\to =0)\right]\left(a,\to \right)={1}^{2}[\because |v\to |={x}^{2}+{y}^{2}+{z}^{2}]\left(a,\to \right)=1|b\to |={x}^{2}+{y}^{2}+{z}^{2}\alpha ={\mathrm{cos}}^{-1}\left(\frac{x}{1.{x}^{2}+{y}^{2}+{z}^{2}}\right)$

Consequently, the vector's angle with the x-axis is ${\mathrm{cos}}^{-1}\left(\frac{x}{1.{x}^{2}+{y}^{2}+{z}^{2}}\right)$ .

Angle between vectors $\alpha ={\mathrm{cos}}^{-1}\left(\frac{a.\to b\to}{|a\to ||b\to |}\right)$.

x-axis is represented by $i\to $.

Let vector be $xi\to +yj\to +zk\to $

$a\to .b\to =i\to .(xi\to +yj\to +zk\to )a\to .b\to =x\left[\because (i\to .i\to =1),(i\to .j\to =0),(i\to .k\to =0)\right]\left(a,\to \right)={1}^{2}[\because |v\to |={x}^{2}+{y}^{2}+{z}^{2}]\left(a,\to \right)=1|b\to |={x}^{2}+{y}^{2}+{z}^{2}\alpha ={\mathrm{cos}}^{-1}\left(\frac{x}{1.{x}^{2}+{y}^{2}+{z}^{2}}\right)$

Consequently, the vector's angle with the x-axis is ${\mathrm{cos}}^{-1}\left(\frac{x}{1.{x}^{2}+{y}^{2}+{z}^{2}}\right)$ .

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