linear algebra - Parallel vectors

Question

The vector [c^2, c^3, c^4] is parallel to the vector [1,-2,4] with the same direction.

Not sure how to solve this since it asks to find C.

Answer 1

Thus, $$c^2=k,$$ $$c^3=-2k$$ and $$c^4=4k,$$ which gives $$c^3=-2c^2$$ or $$c^2(c+2)=0$$ and $c=-2$, $c=0$ they are valid.

Answer 2

As an alternative by cross product

$$\begin{vmatrix}e_1&e_2&e_3\\1&-2&4\\c^2&c^3&c^4\end{vmatrix}=(-2c^4-4c^3)e_1-(c^4-4c^2)e_2+(c^3+2c^2)e_3=\vec 0$$

that is for $c\neq0$

$$\begin{cases} 2c^4+4c^3=0\\c^4-4c^2=0\\c^3+2c^2=0\end{cases}\implies\begin{cases} c=-2\\c=\pm2\\c=-2\end{cases}\implies c=-2$$

Answer 3

Correct me if wrong :

Assume $c \not =0$, real.

$(c^2,c^3,c^4) = c^2(1,c,c^2).$

Vector $(1,c,c^2)$ is parallel to $(c^2,c^3,c^4)$ (In the same direction).

Comparison of $(1,c,c^2)$ with $(1,-2,4)$ implies: $c=-2.$