1. ## Vector Math Help

The question is,

Let K be a scalar and let U = (Ux,Uy,Uz). Prove that ||kU|| = |k| ||U||

I'm not entirely sure what the question is asking here, I'm getting confused with the meaning of the bars ||kU|| and |k| and ||U||.

I'd think ||kU|| would be U = (k * Ux, k* Uy, k * Uz).

Would it be better to prove this scenario by assigning the variables values, like let k = 6, U = (-5, 4, 8). So ||6U|| = (6 * -5, 6 * 4, 6 * 8).

I think my problem might be stemming from not understanding the question or I might just be losing the plot.

2. Hi RedKMan

Usually in proofing, you're not allowed to substitute numerical value. Maybe this will do :

If $U=\left(\begin{array}{cc}U_x\\U_y\\U_z\end{array}\ right)$ , then :

$|U| = \sqrt{(U_x)^2+(U_y)^2+(U_z)^2}$

For : $kU = \left(\begin{array}{cc}kU_x\\kU_y\\kU_z\end{array} \right)$

$|kU| = \sqrt{(kU_x)^2+(kU_y)^2+(kU_z)^2}$

$=\sqrt{k^2((U_x)^2+(U_y)^2+(U_z)^2)}$

$=|k| \sqrt{(U_x)^2+(U_y)^2+(U_z)^2}$

$=|k||U|$

3. Hi Songoku,

Thanks for reply, I am still struggling to understand this.

I'm not sure how your answer proves that ||kU|| = |k| ||U||.

4. Originally Posted by RedKMan
I'm not sure how your answer proves that ||kU|| = |k| ||U||.
It does

songoku has proven $||kU|| = |k|\:||U||$ or in other words : the modulus of the product of a scalar k and a vector U is equal to the product of the absolute value of k and the modulus of U