Let \(\overrightarrow{v} = <v_1, v_2, \dots, v_n>\) be a vector of \(n\) numbers. The length of \(\overrightarrow{v}\) is \[||\overrightarrow{v}|| = \sqrt{v_1^2 + v_2^2 + \dots + v_n^2}\]
This measure of length is from the Euclidean distance or \(l^2\) norm. Recall the distance formula for two points on the plane: Given points \((x_1, y_1)\) and \((x_2, y_2),\) the distance between them is \[\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\] A vector \(\overrightarrow{v} = <x, y>\) has a tail at \((0,0),\) so plugging in \((0,0)\) for one of the points we get \[||\overline{v}|| = \sqrt{x^2 + y^2}\]
The formula for a \(3\)-dimensional vector \(\overline{v} = (x,y,z)\) is \[||\overline{v}|| = \sqrt{x^2 + y+2 + z^2}\]
The messiest part of the formula for the length of a vector is the square root. Often times, such as when you are comparing the length of two vectors, you don't actually need to compute the square root.
Which is longer, \(\overrightarrow{v} = < 2, 3, 2 >\) or \(\overrightarrow{w} = < 0, 4, 2 >?\)
If \(||\overrightarrow{v}||^2 > ||\overrightarrow{w}||^2\) then \(||\overrightarrow{v}|| > ||\overrightarrow{w}||,\) so it is enough to find \(||\overrightarrow{v}||^2\) and \(||\overrightarrow{w}||.\) \[||\overrightarrow{v}||^2 = 2^2 + 3^2 + 2^2 = 17\] \[||\overrightarrow{w}||^2 = 0^2 + 4^2 + 2^2 = 20\] Therefore, \(||\overrightarrow{w}|| > ||\overrightarrow{v}||.\)
Enter a vector to see its length.
If \(\overrightarrow{v}\) is \(2\)-dimensional, the length can be found using the pythagorean theorem.
For example, consider the vector with \(x\)-value \(2\) and \(y\)-value \(1.\) \[\overrightarrow{v} = (2,1)\]
Notice that the vector is the hypotenuse of a triangle with base length \(2\) and height \(1.\) Using the Pythagorean Theorem \(a^2 + b^2 = c^2\) with \(a = 2,\) \(b = 1,\) and with \(c = ||\overrightarrow{v}||,\) \[2^2 + 1^2 = ||\overrightarrow{v}||^2\] Taking the square root gives \[||\overline{v}|| = \sqrt{1^2 + 2^2} = \sqrt{5} \approx 2.24\]
Compute the square of the length:
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