# Thread: vectors part 2 ( new at all this)

1. ## vectors part 2 ( new at all this)

1) Points P and Q have position vectors OP = 5i - 3j + 2k andOQ = 2i+ 3j, respectively. Find the vectorQP, and determine its direction cosines. Hence find the unit
vector in the direction of
QP.

2) Find the direction cosines of the vector joining the point
(-2,2,8) (i.e.,
x = ¡2, y = 2, z = 8) to the point (1,0,2).

3) Find the parametric vector and Cartesian equations of

the line with direction cosines
(2/7,-3/7, 6/7) which
passes through the point
A with position vector i + 2j -k.
Calculate the coordinates of the point where the line
meets the plane
x = -1. A second line passes through
the points
B and C with respective coordinates (3,-2, 1)
and (2, 1, 4). Find the Cartesian parametric equations of
this line and show that the two lines intersect. Calculate
the coordinates of the point of intersection.

4) In a Cartestian coordinate system, point A is given by
(-3, 0,-5) and point B by (3, 4, 3). Find the Direction
Cosines of the line
AB and write down the vector
equation for this line.
A second line passes through point C with coordinates
(2, 5, 5), and is in the direction (2i+ 3j+ 6k). Show
that these two lines intersect at a point D and write down
the coordinates of D.
Find a vector that is perpendicular to both
AB and CD.

2. Hello, raza88sab!

1) Points $\displaystyle P$ and $\displaystyle Q$ have position vectors $\displaystyle \overrightarrow{OP} \:= \:5i - 3j + 2k$ and $\displaystyle \overrightarrow{OQ}\: =\: 2i + 3j$, respectively.
(a) Find the vector $\displaystyle \overrightarrow{QP}.$
(b) Determine its direction cosines.
(c) Find the unit vector in the direction of $\displaystyle \overrightarrow{QP}.$
We have: .$\displaystyle \begin{Bmatrix}\overrightarrow{OP} & = & \langle 5,\,\text{-}3,\,2\rangle \\ \overrightarrow{OQ} & = & \langle 2,\,3,\,0\rangle\end{Bmatrix}$

(a) Then: .$\displaystyle \overrightarrow{QP} \:=\:\langle2,\,3,\,0\rangle - \langle5,\text{-}3,\,2\rangle \;=\;\langle5,\,\text{-}3,\,2\rangle$

(b) The direction cosines of $\displaystyle \langle a,\,b,\,c\rangle$ are: .$\displaystyle \begin{array}{ccc}\cos\alpha & = & \frac{a}{\sqrt{a^2+b^2+c^2}} \\ \cos\beta &=& \frac{b}{\sqrt{a^2+b^2+c^2}} \\ \cos\gamma &=& \frac{c}{\sqrt{a^2+b^2+c^2}} \end{array}$
We have: .$\displaystyle |\overrightarrow{QP}| \:=\:\sqrt{(\text{-}3)^2 + 6^2 + (\text{-}2)^2} \:=\:7$

. . The direction cosines are: .$\displaystyle \left[\text{-}\frac{3}{7},\:\frac{6}{7},\:\text{-}\frac{2}{7}\right]$

(c) The unit vector in the direction of $\displaystyle \overrightarrow{QP}$ is: .$\displaystyle \frac{\overrightarrow{QP}}{|\overrightarrow{QP}|} \;=\;\frac{\langle5,\,\text{-}3,\,2\rangle}{7}$
. . Therefore: .$\displaystyle \left\langle\frac{5}{7},\:-\frac{3}{7},\:\frac{2}{7}\right\rangle$

3. Originally Posted by raza88sab
...

3) Find the parametric vector and Cartesian equations of
the line with direction cosines (2/7,-3/7, 6/7) which
passes through the point A with position vector i + 2j -k.
Calculate the coordinates of the point where the line
meets the plane x = -1.
A second line passes through
the points B and C with respective coordinates (3,-2, 1) and (2,1, 4). Find the Cartesian parametric equations of
this line and show that the two lines intersect. Calculate
the coordinates of the point of intersection.

...
Hi,

the parametric equation of the line:

$\displaystyle <x,y,z>=<1, 2, -1> + r \cdot \left \langle \frac27 , -\frac37, \frac67 \right \rangle$

$\displaystyle \boxed{\begin{array}{l}x=1+\frac27 \cdot r \\y=2-\frac37 \cdot r\\z=-1+\frac67 \cdot r\end{array}}$

You know that x = -1. Take the 1rst equation to calculate r: $\displaystyle r = -7$. Plug in this value to calculate the coordinates of the point of intersection P: $\displaystyle P\left( -1, 5, -7\right)$

Equations of the 2nd line:

$\displaystyle \langle x,y,z\rangle=\langle 2,1,4 \rangle + s \cdot \left \langle \langle 2,1,4 \rangle - \langle 3, -2,1 \rangle \right \rangle = \langle 2,1,4 \rangle + s \cdot \langle -1, 3, 3 \rangle$

and:

$\displaystyle \boxed{\begin{array}{l}x=2-s\\y=1 + 3s\\z=4+3s\end{array}}$

The direction vectors of the 2 lines are linearly independent that means the lines are not parallel.
To calculate the point of intersection I use the first two equatios, calculate r and s and check if these 2 values satisfy the third equation too. If so, the two lines intersect:

$\displaystyle \left|\begin{array}{l}1+\frac27 \cdot r=2-s\\2-\frac37 \cdot r=1+3s\end{array}\right.$ . I've got $\displaystyle s = -\frac13$ and $\displaystyle r = \frac{14}{3}$. These two values satisfy the 3rd equation:

$\displaystyle -1+\frac67 \cdot r = 4+3s ~\implies~-1+\frac67 \cdot \frac{14}{3} = 4+3 \cdot \left(-\frac13\right)$

Therefore the point of intersection is:

$\displaystyle <x,y,z>=<1, 2, -1> + \left(\frac{14}{3}\right) \cdot \left \langle \frac27 , -\frac37, \frac67 \right \rangle = \left \langle \frac73, 0, 3 \right \rangle$