
Determining the Length of a Vector
Welcome to today's lesson where we'll dive into a nifty process to determine the length also known as the magnitude, of a vector. We'll explore two methods, one employing Pythagoras' theorem and the other utilizing the scalar product of a vector with itself.
Let's kick things off with a hands-on example.
Imagine a vector in a plane, say
$$ \vec{v} = \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \end{pmatrix} $$
These x and y components provide the vector's coordinates on the Cartesian plane.
Now visualize this vector in the plane.
Here's where it gets interesting. The vector forms a triangle relative to the axes.
The legs of this triangle align with the x=4 and y=3 axes, and our vector |v| neatly serves as the hypotenuse.
According to Pythagoras' theorem, which I'm sure you're familiar with, the hypotenuse (that's our vector length) is the square root of the sum of the squares of the legs.
$$ hypotenuse = \sqrt{leg_A^2+leg_B^2} $$
So, with Pythagoras in our corner, we can calculate the vector length.
In this case, we have leg A as x=4 and leg B as y=3, while our hypotenuse is the magnitude of the vector |v|.
$$ | \vec{v} | = \sqrt{x^2 + y^2}$$
Substituting x=3 and y=4
$$ | \vec{v} | = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$
And there we have it! We've successfully computed the length of the vector.
In the world of linear algebra, the length of a vector is also referred to as its magnitude |v| or its norm ||v||.
Alternative method
Ready for a twist? Let's consider an alternative method.
You can also determine the vector length through the dot product of the vector with itself. Here's how it works:
$$ \vec{v} \cdot \vec{v} = x \cdot x + y \cdot y = x^2 +y^2 $$
You'll notice the scalar product of the vector with itself equals the sum of the squares x2+y2
Referencing Pythagoras' theorem again, the sum of these squares equals the squared magnitude of the vector |v|2=x2+y2
To put it into practice, let's substitute x=3 and y=4
$$ \vec{v} \cdot \vec{v} = 3^2 +4^2 = 9 + 16 = 25 $$
The last step involves computing the square root of the scalar product of the vector with itself.
$$ \sqrt{ \vec{v} \cdot \vec{v} } = \sqrt{25} = 5 $$
And voila! The final result is the magnitude or, in simpler terms, the length (module) of the vector.
I hope you found this lesson from Nigiara enlightening. Stick around as we unravel more fascinating aspects of vectors in the lessons to come.