Finding unknowns in basic vectors Given that $u = 3i + 2j$  and $v = 2i + \lambda j $ determine $\lambda$ such that:
(a) $u$ and $v$ are at right angles (this means perpendicular I presume?)
(b) $u$ and $v$ are parallel
Whilst working out the first one, I obtained the value of -3 for $\lambda$. However I am not sure whether this is correct and am not quite sure how to tackle question (b) from this exercise.
What I can't seem to grasp is, how on earth does the value of lambda determine whether these points on the XY plane are parallel or not?
Would the multiple of the two vectors result in a 0 for perpendicular vectors? If it indeed so then the shift-cosine of 0 would give us 90 degrees (indeed perpendicular). But what value would the multiple of both of those vectors require to be in order to be parallel? 
 A: $\lambda$ is simply the scalar value of the $\textbf{j}$-component of the vector $\textbf{v}$. The only thing that the question asks of you, is to determine the values of $\lambda$ which will satisfy the conditions in (a) and (b) respectively.
For a):
We know that two vectors are perpendicular if the angle between them are $90\unicode{xb0}$. Where can we use this fact? Clearly the DOT PRODUCT must spring to mind, since we know the dot product of two vectors, $u$ and $v$ are given by:

$$u \cdot v = ||u|| \ ||v|| \ \cos{\theta}$$

Now we know $\theta=$ $90\unicode{xb0}$ and $\cos$($90\unicode{xb0}$)$=0$, so we must have $$u \cdot v = 0$$
We thus need to find $\lambda$ such that
\begin{align}u\cdot v &= 0 \\ \therefore (3)(2)+ (2)(\lambda) &=0 \\ \therefore 2\lambda &= -6 \\ \therefore \lambda &= -3 \end{align}
For b):
Two vectors are parallel to one another if their CROSS PRODUCT is zero.
Thus we need to find $\lambda$ such that
\begin{align} u \times v &= 0 \\ \therefore \begin{vmatrix}i & j \\ 3 & 2\\ 2 & \lambda\end{vmatrix} &= 0 \\ \therefore 3\lambda - 4 &= 0 \\ \therefore 3\lambda &= 4 \\
\therefore \lambda &= \frac{4}{3}\end{align}
A: The following identity of the dot product is useful

$$ u.v = ||u||\,||v||\cos(\theta). $$

