vector in space

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September 20th 2010, 06:40 AM
bobey

vector in space

Question :

describe the set of points in space whose coordinate satisfy the given equation or pair of equations:

(i) z=2y (ii) 3x=4y, z=1

if (i) or (ii) represents a line in space, give a unit vector that is parallel to the line. If (i) or (ii) represents a plane, give a unit vector that is normal to the plane.

My attempt :

z=2y is a line parallel to x-axis consisting of all points of the form (0,z,z)

3x=4y, z=1 is a plane perpendicular to the z=1 axis consisting of all points of the form (3x,3x,1)

(i) is a line, thus the unit vector that is parallel to the line is v/|v|= (0i-2j+k)/5

(ii) is a plane, thus a unit vector that is normal to the plane is v/|v|= (3i-4j+k)/26

is my answer is correct??? pls help me... tq
September 20th 2010, 09:50 AM
Ackbeet

I'm afraid you're a bit off. A location in 3-dimensional space must be described by three variables. But a position on a line requires only one variable. A position on a plane requires two variables. So, if you're talking about a line, how many equations must you have in order to knock down the number of variables from 3 to 1? If you're talking about a plane, how many equations must you have in order to knock down the number of variables from 3 to 2?
September 20th 2010, 12:35 PM
yungman

I am no expert as Ackbeet.

I believe (1) is a plane because $-\infty < x < \infty$

(2) is a line because z=1 and $y=\frac{3}{4}x$
September 20th 2010, 01:46 PM
Soroban

Hello, bobey!

Quote:

Describe the set of points in space whose coordinates
satisfy the given equation or pair of equations:

. . $(a)\; z\:=\:2y \qquad\qquad (b)\;\begin{array}{ccc}3x&=&4y \\ z&=&1\end{array}$

if (a) or (b) represents a line in space, give a unit vector that is parallel to the line.
If (a) or (b) represents a plane, give a unit vector that is normal to the plane.

In (a), we have: . $2y - z \:=\:0$

This is the equation of a plane.
It passes through the origin and has normal vector: . $\vec n \:=\:\langle 0,2,-1\rangle$

The unit normal vector is:

. . $\displaystyle \frac{\vec n}{|\vec n|} \;=\;\frac{\langle 0,2,-1\rangle}{\sqrt{0^2+2^2+(\text{-}1)^2}} \;=\;\frac{\langle 0,2,-1\rangle}{\sqrt{5}} \;=\;\left\langle 0,\,\frac{2}{\sqrt{5}},\,-\frac{1}{\sqrt{5}}\right\rangle$

In (b), we have: . $\begin{Bmatrix}3x &=& 4y \\ z &=& 1 \end{Bmatrix}$
We can write the equations like this: . $\begin{Bmatrix} x &=& x \\ y &=& \frac{3}{4}x \\ z &=& 1 \end{Bmatrix}$

$\text{On the right, replace }\,x\text{ with }\,t\!:\;\;\begin{Bmatrix}x &=& t \\ y &=& \frac{3}{4}t \\ z &=& 1\end{Bmatrix}$

These are the parametric equations of a line.

It passes through $(0,0,1)$ and has direction vector $\vec v \:=\:\langle 1,\frac{3}{4},0\rangle$

The unit vector is:

. . $\displaystyle \frac{\vec v}{|\vec v|} \;=\;\frac{\langle 1,\frac{3}{4},0\rangle}{\sqrt{1^2+(\frac{3}{4})^2 + 0^2}} \;=\;\frac{\langle1,\frac{3}{4},0\rangle}{\frac{5} {4}} \;=\;\left\langle \frac{4}{5},\,\frac{3}{5},\,0\right\rangle$