Answer:
![\^v=-(3)/(5)i+(4)/(5)j](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9l4mbocydb673r0in5wvk7dkkktpdw0hh9.png)
Explanation:
We have the following vector
![v=3√(2)i-4√(2)j](https://img.qammunity.org/2020/formulas/mathematics/middle-school/39t8pi2gg5bia6zwa8e98bs1qr8d89j0k6.png)
First we calculate its magnitude
The magnitude of the vector v will be
![|v|=\sqrt{(3√(2))^2 + (4√(2))^2}\\\\|v|=√(9*2+16*2)\\\\|v|=√(18+32)\\\\|v|=5√(2)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/gse9l5qpaf8gvza4bei1frumdqmetf2ezz.png)
Now to create a unitary vector in the opposite direction to v, we divide the vector v between the negative of its magnitude
we call this new vector "
"
![\^v=(3√(2))/(-5√(2))i-(4√(2))/(-5√(2))j](https://img.qammunity.org/2020/formulas/mathematics/middle-school/5gt1vmgijsiq5ozmbnxpl8kt3rvyrg9tuo.png)
![\^v=-(3)/(5)i+(4)/(5)j](https://img.qammunity.org/2020/formulas/mathematics/middle-school/9l4mbocydb673r0in5wvk7dkkktpdw0hh9.png)