# Math Help - Finding a parallel vector

1. ## Finding a parallel vector

i'm looking at past year papers for sections which i can't do and i dont know how to do this question. i need to know under what topic is this and if possible how to get to the answer as well so i have an example to figure out how it goes. The answer is at the bottom. Thanks.
(a) We are given a point A(1, 1, 0) and the line
l : (x − 1)/2= y =(2 − z)/3
(i) Find a vector, v, parallel to l.
(ii) Find the cartesian equation of the plane that contains A and l.
(b) Two vectors are given by a = (0, 1,−1) and b = (1,−1, 0). Determine the area of the parallelogram with edges a and b.

(a)(i) v = (2, 1,−3)
(ii) A point on the line is P(1, 0, 2)
->
AP × v =
i j k
0 −1 2
2 1 −3
= i + 4j + 2k
The equation is therefore of the form
x + 4y + 2z = d
As A is in the plane, the equation is
x + 4y + 2z = 5
(b)
a × b =
i j k
0 1 −1
1 −1 0
= −i − j − k
Area = ||a × b|| =sqrt(3)

2. "Past year papers" for what courses? It seems strange that you would be looking at "past year papers" without knowing the area of math that they involve.

These look to me like basic problems in "multi-variable Calculus", though introductory- they don't involve, yet, any Calculus.

(a) We are given a point A(1, 1, 0) and the line
l : (x − 1)/2= y =(2 − z)/3
(i) Find a vector, v, parallel to l.[/quote]
Notice that you could find parametric equations for this line by setting each of these values equal to parameter t: (x- 1)/2= t so x= 1+ 2t. y= t. (2-z)/3= t so 2- z= 3t and z= 2- 3t. That could also be written as a vector equation: (x, y, z)= (1+2t, t, 2- 3t)= (1, 0, 2)+ t(2, 1, -3).

The point of that is to show that (2, 1, -3), formed from the denominators of the original equation, is a vector pointing in the same direction as the line. Notice that you could find parametric equations for this line by setting each of these values equal to parameter t: (x- 1)/2= t so x= 1+ 2t. y= t. (2-z)/3= t so 2- z= 3t and z= 2- 3t. That could also be written as a vector equation: (x, y, z)= (1+2t, t, 2- 3t)= (1, 0, 2)+ t(2, 1, -3).
The point of that is to show that the vector (2, 1, -3) points in the direction of the line.

(ii) Find the cartesian equation of the plane that contains A and l.
The equation of a plane containing point $(x_0, y_0, z_0)$ and having normal vector (U, V, W) is $U(x-x_0)+ V(y- y_0)+ W(z- z_0)= 0$. The vector you already have, (2, 1, -3) is in this plane, not normal (perpendicular) to it. You need to construct that normal vector. But you also know that (1, 0, 2) (taking t=0 in the equation of the line) is a point on the line and so in the plane and that A(1, 1, 0) is another point in the plane. The vector from (1, 0, 2) to (1, 1, 0) is (0, 1, -2).

Take the cross product of those two vectors to get the normal vector and use either of those points as $(x_0, y_0, z_0)$.

(b) Two vectors are given by a = (0, 1,−1) and b = (1,−1, 0). Determine the area of the parallelogram with edges a and b.
This is easy: the area of a parallelogram is the length of the cross product of the two vectors.

Two see that, recall that the area of a parallelogram is "hw" where w is the "width", the length of one side, and h is the "height", measured perpendicular to that side. Dropping a line from the end of u perpendicular to v gives a right triangle having the length of u, |u|, as hypotenuse. If the angle between u and v is $\theta$ then $sin(\theta)= \frac{h}{|u|}$ so $h= |u|sin(\theta)$. With $w= |v|$, the area of the parallelogram is $|u||v|sin(\theta)$, precisely the length of the cross product of u and v.

3. im doing a linear algebra course. im trying to figure out what topic this is under so i might get to revise on that topic.