# basic linear algebra questions

• Apr 17th 2009, 07:32 PM
Hikari Clover
basic linear algebra questions
let ABC be a triangle and let p be the midpoint of side AB
let http://i52.photobucket.com/albums/g37/mmmmms/1-7.jpg and http://i52.photobucket.com/albums/g37/mmmmms/2-6.jpg

express http://i52.photobucket.com/albums/g37/mmmmms/5-3.jpg and http://i52.photobucket.com/albums/g37/mmmmms/6-2.jpg in terms of http://i52.photobucket.com/albums/g37/mmmmms/7-1.jpg and http://i52.photobucket.com/albums/g37/mmmmms/8-1.jpg and http://i52.photobucket.com/albums/g37/mmmmms/9-1.jpg

hence prove that the midpoint of the hypotenuse of a right-angle triangle is equidistant from the three vertices

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let p be the plane with cartesian equation 2x+y-3z=9 and let A,B be points with coordinates (1,-2,3) and (-2,1,-1)
l is the line passing through the point A and B

find the coordinates of the point C where the line l intersects the plane p

i have no ideas how to find the intersection between a line and a plane

thx
• Apr 17th 2009, 11:43 PM
running-gag
Quote:

Originally Posted by Hikari Clover

Hi

$\overrightarrow{AP}^2 = \frac12\:\overrightarrow{AB}\cdot\frac12\:\overrig htarrow{AB}\cdot = \frac14\:\overrightarrow{AB}\cdot\overrightarrow{A B}$

$\overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB} = -\vec{a} + \vec{b}$

$\overrightarrow{AP}^2 = \frac14\:\left(-\vec{a} + \vec{b})^2\right) = \frac14\:\left(\vec{a}^2 + \vec{b}^2 -2 \vec{a} \cdot \vec{b}\right)$

$\overrightarrow{CP} = \frac12\:\left(\overrightarrow{CA} + \overrightarrow{CB}\right) = \frac12\:\left(\vec{a} + \vec{b}\right)$

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Quote:

Originally Posted by Hikari Clover
let p be the plane with cartesian equation 2x+y-3z=9 and let A,B be points with coordinates (1,-2,3) and (-2,1,-1)
l is the line passing through the point A and B

find the coordinates of the point C where the line l intersects the plane p

i have no ideas how to find the intersection between a line and a plane

thx

AB coordinates are (-3,3,-4)
Therefore one parametric equation of l is :
x = -3t + 1
y = 3t -2
z = -4t +3

C(x,y,z) is on the plane iff 2x+y-3z=9
C is on l iff there exists t such that
x = -3t + 1
y = 3t -2
z = -4t +3

Substitute x,y,z in the Cartesian equation of the plane to get one linear equation. Solve for t and substitute in x,y,z expressions.

Spoiler:

2(-3t + 1)+(3t -2)-3(-4t +3)=9 gives t=2
x = -5
y = 4
z = -5
C(-5,4,-5)
• Apr 18th 2009, 04:20 AM
Hikari Clover
Quote:

Originally Posted by running-gag
Hi

$\overrightarrow{AP}^2 = \frac12\:\overrightarrow{AB}\cdot\frac12\:\overrig htarrow{AB}\cdot = \frac14\:\overrightarrow{AB}\cdot\overrightarrow{A B}$

$\overrightarrow{AB} = \overrightarrow{AC} + \overrightarrow{CB} = -\vec{a} + \vec{b}$

$\overrightarrow{AP}^2 = \frac14\:\left(-\vec{a} + \vec{b})^2\right) = \frac14\:\left(\vec{a}^2 + \vec{b}^2 -2 \vec{a} \cdot \vec{b}\right)$

$\overrightarrow{CP} = \frac12\:\left(\overrightarrow{CA} + \overrightarrow{CB}\right) = \frac12\:\left(\vec{a} + \vec{b}\right)$

==================================================
AB coordinates are (-3,3,-4)
Therefore one parametric equation of l is :
x = -3t + 1
y = 3t -2
z = -4t +3

C(x,y,z) is on the plane iff 2x+y-3z=9
C is on l iff there exists t such that
x = -3t + 1
y = 3t -2
z = -4t +3

Substitute x,y,z in the Cartesian equation of the plane to get one linear equation. Solve for t and substitute in x,y,z expressions.

Spoiler:

2(-3t + 1)+(3t -2)-3(-4t +3)=9 gives t=2
x = -5
y = 4
z = -5
C(-5,4,-5)

hey , thx for ur replying(Clapping)

but for the first question,did u forget to put that absolute value sign? or it doesnot matter?
• Apr 18th 2009, 05:58 AM
running-gag
Quote:

Originally Posted by Hikari Clover
hey , thx for ur replying(Clapping)

but for the first question,did u forget to put that absolute value sign? or it doesnot matter?

Are you talking about the modulus ?

$||\overrightarrow{AB}||^2 = \overrightarrow{AB}\cdot\overrightarrow{AB} = \overrightarrow{AB}^2 = AB^2$
• Apr 18th 2009, 06:17 AM
Hikari Clover
oh i got it
thanks so much ^_^