Question

Find a vector that has the opposite direction of u = (1,-2,4), but which has length square root of 3.

Answer 1

Normalize $\vec u$ by dividing $\vec u$ by its magnitude:

$\dfrac{\vec u}{\|\vec u\|}=\dfrac{(1,-2,4)}{\sqrt{1^2+(-2)^2+4^2}}=\dfrac{(1,-2,4)}{\sqrt{21}}$

Multiply by -1 to reverse its direction, and again by $\sqrt3$ to ensure this new vector has the magnitude we want. Notice that $\sqrt{\dfrac3{21}}=\dfrac1{\sqrt7}$.

$-\sqrt3\dfrac{\vec u}{\|\vec u\|}=\dfrac{(-1,2,-4)}{\sqrt7}$