Answer:
The vector of magnitude 3 in the direction of [tex]\vec v = 12\,\hat{i}-5\,\hat{k}[/tex].
Step-by-step explanation:
Let be [tex]\|\vec u\| = 3[/tex] and [tex]\vec v = 12\,\hat{i}-5\,\hat{k}[/tex], the resultant in the direction of [tex]\vec v[/tex] can be obtained by using the following expression:
[tex]\vec u = \frac{\|\vec u\|}{\|\vec v\|} \cdot \vec v[/tex]
Where:
[tex]\|\vec v\|[/tex], [tex]\|\vec u\|[/tex] - Norms of [tex]\vec v[/tex] and [tex]\vec u[/tex], respectively.
Now, we get the norm of [tex]\vec v[/tex] by Pythagorean Theorem:
[tex]\|\vec v\| = \sqrt{12^{2}+(-5)^{2}}[/tex]
[tex]\|\vec v\| = 13[/tex]
The vector is now determined:
[tex]\vec u = \frac{3}{13} \cdot (12\,\hat{i}-5\,\hat{k})[/tex]
[tex]\vec u = \frac{36}{13} \,\hat{i}-\frac{15}{13}\, \hat{k}[/tex]
The vector of magnitude 3 in the direction of [tex]\vec v = 12\,\hat{i}-5\,\hat{k}[/tex].
