The magnitude of the vector is [tex]4 \sqrt{2}[/tex]
Explanation:
It is given that the vector is [tex]4 i-4 j[/tex]
We need to find the magnitude of the vector.
The magnitude of the vector can be determined by
[tex]|\vec{v}|=\sqrt{(4)^{2}+(-4)^{2}}[/tex]
Simplifying the terms, we get,
[tex]|\vec{v}|=\sqrt{16+16}[/tex]
Adding the terms within the square root, we get,
[tex]|\vec{v}|=\sqrt{32}[/tex]
The value of 32 can be written as prime factorization of 2,
[tex]|\vec{v}|=\sqrt{2^{5}}[/tex]
[tex]|\vec{v}|=\sqrt{2^{4} \cdot 2}[/tex]
Simplifying the square root, we have,
[tex]|\vec{v}|=4\sqrt{2}[/tex]
Thus, the magnitude of the vector is [tex]4 \sqrt{2}[/tex]
