# 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?

• HINT: If two vectors are perpendicular, then what will their dot product equal? Similarly, if two vectors are parallel, then what will their dot product equal? – Mufasa Jan 27 '15 at 23:05
• @Mufasa Well, if they're perpendicular than I presume their dot product will be equal to 0, however when parallel, -1 or 1? – Juxhin Jan 27 '15 at 23:06
• Excuse me, I skipped a few steps when I gave you that answer. 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? – Juxhin Jan 27 '15 at 23:15
• @Mufasa - That is indeed what is confusing me haha, I'm having a mental tug of war between 180 degrees or 0 degrees – Juxhin Jan 27 '15 at 23:20
• I do not think so, wouldn't the former be -1 and latter 1? – Juxhin Jan 27 '15 at 23:21

$\lambda$ 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}

• Thank you very much Dillon, I'm simply trying to understand every corner of the question and answer prior to accepting. – Juxhin Jan 27 '15 at 23:32
• Which parts do you still feel uncertain about? I will try and edit the answer accordingly to try and explain :) – user860374 Jan 27 '15 at 23:35
• Alright thanks again for your edit. I wanted to also ask if the cross-product is still equal to zero in 3-dimensional space? Otherwise I am able to both understand and visualize the scenario in 2-dimensional space fairly well. – Juxhin Jan 27 '15 at 23:35
• Yes it is still equal to zero in 3-dimensional space :) . – user860374 Jan 27 '15 at 23:36
• Also, the reason for this also being valid in 3-dimensional space, is because we are using a geometrical interpretation of both the dot product and cross product. This geometrical interpretation can be used for any dimension up until $\mathbb{R}^3$ after that we rely on a more "theoretical" approach, which we do not need to cover right now. – user860374 Jan 27 '15 at 23:42

The following identity of the dot product is useful

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

• I've amended a small part going into a bit more detail as to why my results are slightly confusing me, however I shall add my comment in here to grab your attention directly. "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?" – Juxhin Jan 27 '15 at 23:17