The magnitude of the vector is [tex]4 \sqrt{2}[/tex]
Explanation:
The given vector is [tex]4 i-4 j[/tex]
We shall find the value of the magnitude of the vector.
The magnitude of the vector [tex]4 i-4 j[/tex] can be determined by
[tex]|\vec{v}|=\sqrt{(4)^{2}+(-4)^{2}}[/tex]
Squaring the terms, we get,
[tex]|\vec{v}|=\sqrt{16+16}[/tex]
Adding the terms, we have,
[tex]|\vec{v}|=\sqrt{32}[/tex]
Let us simplify [tex]\sqrt{32}[/tex] by prime factorization of 2, we get,
[tex]|\vec{v}|=\sqrt{2^{5}}[/tex]
Simplifying, we get,
[tex]|\vec{v}|=\sqrt{2^{2} \cdot 2^2\cdot 2}[/tex]
[tex]|\vec{v}|=2\times2(\sqrt{2})[/tex]
[tex]|\vec{v}|=4 \sqrt{2}[/tex]
Thus, the magnitude of the vector is [tex]4 \sqrt{2}[/tex]
