Find the condition that the given vectors are parallel. Find the condition that the vectors $\vec a=k\hat{i}+l\hat{j}$ and $\vec b=l\hat{i}+k\hat{j}$ are parallel.
Answer is $l^2=k^2$. 
But I dont know how to find it using formula.
I searched everywhere but cant find something similar to this.

Edit - 

This question is not from the scalar product questions. So please help me.
 A: Two vectors  $\vec v $ and $\vec u $ are parallel only if their cross product is zero.

Using this concept, we have $$\vec a \times \vec b =\vec 0$$ $$k^2 - l^2 =0 \Rightarrow l^2=k^2$$ and we are done!!

Handicap of cross product is not a problem. If two vectors are parallel, then one of them will be a multiple of the other. So divide each one by its magnitude to get a unit vector. If they're parallel, the two unit vectors will be the same. 
We can see this is satisfied by $\vec a $ and $\vec b $, so they are parallel. Hope it helps. 
A: If $\overrightarrow{a}=ki+lj$ is parallel to $\overrightarrow{b}=li+kj$ then, by definition, there is $\alpha$ such that 
$$\overrightarrow{a}=\alpha\cdot \overrightarrow{b} \rightarrow ki+lj=(\alpha l)i+(\alpha k)j \rightarrow (k-\alpha l)i+(l-\alpha k)j=0$$
once we have a null vector then each coordinate is zero. 
That give us: 
$k=\alpha l \quad (1)$ 
$l=\alpha k\quad (2)$. 
Putting $(2)$ in $(1)$ we get:
$$k=\alpha l=\alpha^2 k \rightarrow k(1-\alpha^2)=0$$
so $k=0$ or $\alpha = \pm 1$. 
1) If $k=0$ then, from $(2)$, we get $l=0$ and then $l^2=k^2$
2) If $\alpha = \pm 1$ then, from $(2)$, we get $k=\pm l$ and square both sides and get $l^2=k^2$.
