Vectors #3

• Apr 28th 2014, 11:36 PM
Fratricide
Vectors #3
The parallelepiped has OA, OB and OC as three edges and remaining vertices X, Y, Z and D as shown in the diagram.
OA = 5i
OB = i + 3k
and OC = i + 4j

a)
Write down the position vectors of X, Y, Z and D in terms of i, j and k and calculate the length of OD and OY.
I've done this part, but I included it so you know I have these answers if they're needed.

b) Calculate the size of angle OZY.
I know I need to use the formula u.v = |u||v|cosθ, but what vectors do I use for u and v? I tried using OZ and OY, and OZ and ZY, but neither gave me the correct answer. However, if I subtracted my "OZ and ZY" answer from 180, I got the answer in the book, but I would never have done this without already knowing the answer.

c) The point P divides CZ in the ration $\lambda : 1$, i.e. CP:PZ =
$\lambda : 1$.
i. Give the position vector of P.
The problem here lies with my confusion surrounding the ratio statement. I'm new to vectors and have rarely used such ratios in the past.

ii. Find $\lambda$ if OP is perpendicular to CZ.
I should be able to do this after doing part i.

Thanks heaps.
• Apr 29th 2014, 02:10 AM
emakarov
Re: Vectors #3
Quote:

Originally Posted by Fratricide
b) Calculate the size of angle OZY.
I know I need to use the formula u.v = |u||v|cos[COLOR=#252525][FONT=sans-serif]θ, but what vectors do I use for u and v? I tried using OZ and OY, and OZ and ZY, but neither gave me the correct answer.

Why don't you look at a text that descrives the formula u.v = |u||v|cos θ to understand what the relationship is between vectors u, v and angle θ? You can consult your textbook or Wikipedia here and here. Again, what does the text say concerning how θ is defined by the two vectors? Can you say that angle OZY, which you need to find, is defined by vectors OZ and OY, or by OZ and ZY, in this way?

Quote:

Originally Posted by Fratricide
c) The point P divides CZ in the ration $\lambda : 1$, i.e. CP:PZ = $\lambda : 1$.
i. Give the position vector of P.

If $CP:PZ = \lambda$, then
$OP=\frac{OC+\lambda OZ}{\lambda+1}$
It may be easier to remember this formula in a more general case, where $CP:PZ = \lambda:\mu$; then
$OP=\frac{\mu OC+\lambda OZ}{\lambda+\mu}$
Note that $OC$ and $OZ$ are multiplied by $\lambda$ and $mu$ in a criss-cross manner. Indeed, $\lambda$ corresponds to the initial segment $CP$ and $\mu$ corresponds to the final segment $PZ$, yet the radius-vector $OC$ of the initial point $C$ is multiplied by $\mu$, while the radius-vector $OZ$ of the final point $Z$ is multiplied by $\lambda$. This is justified by the limit cases: when $\lambda=0$, all contribution is done by vector $OC$ and $P=C$; when $\mu=0$, all contribution is done by $OZ$ and $P=Z$. Of course, $OC$, $OZ$ and so on above are vectors, not just segment lengths.

Quote:

Originally Posted by Fratricide
ii. Find $\lambda$ if OP is perpendicular to CZ.

Use the fact that $OP\cdot CZ=0$.
• Apr 29th 2014, 02:18 AM
SlipEternal
Re: Vectors #3
Throughout this response, when I use a line segment as a reference to a vector, I am referencing a vector with direction from the left endpoint to the right endpoint of the line segment with magnitude equal to the length of the line segment.

For b) The vectors you want are the two vectors that start at angle's vertex and go out. So, the "starting point" for both vectors should be Z, which is the vertex of the angle. The two vectors you want are ZO and ZY. When you tried OZ, you were essentially using negative u (which is why when you subtracted that answer from 180 degrees, you got the correct answer).

For c) i. Hopefully you understand how vector addition works. The position vector of P corresponds to line segment OP. Using only line segments, OP = OC + CP = OZ + ZP.

You can write OP = p1i + p2j + p3k and OZ = z1i + z2j + z3k

So, CP = OP - OC = (p1-1)i + (p2-4)j + p3k and
PZ = OZ - OP = (z1 - p1)i + (z2 - p2)j + (z3 - p3)k

You know that $\dfrac{|CP|}{|PZ|} = \lambda = \dfrac{ \sqrt{(p_1-1)^2 + (p_2-4)^2 + p_3^2} }{ \sqrt{ (z_1-p_1)^2 + (z_2 - p_2)^2 + (z_3 - p_3)^2 } }$

You also know that CZ = OZ - OC = (z1 - 1)i + (z2 - 4)j + z3k is in the same direction as both CP and PZ, so the dot product of each should be the product of their magnitudes:

(z1-1)(p1-1) + (z2-4)(p2-4) + z3p3 = |CP||CZ|
and
(z1-1)(z1-p1) + (z2-4)(z2-p2) + z3(z3-p3) = |PZ||CZ|

You said you solved for OZ in part a), so just plug in and solve for p1, p2, and p3

A bit easier: if you study affine geometry, you will find you can use $OP = \dfrac{1}{\lambda+1}OC + \dfrac{\lambda}{\lambda+1}OZ$ where OC and OZ represent the position vectors for C and Z respectively.
This gives $OP = \dfrac{1+\lambda z_1}{\lambda+1}\vec{i} + \dfrac{4+\lambda z_2}{\lambda+1}\vec{j} + \dfrac{\lambda z_3}{\lambda+1}\vec{k}$.
• Apr 29th 2014, 04:03 AM
Fratricide
Re: Vectors #3
Thank you both.