Question about linear algebra?

Question

here is the link;

http://i12.photobucket.com/albums/a210/hard_assteel/styuff.jpg

how do i do a problem like this?

Thank You

Answer 1

its a vector problem...

u need to convert the equation form into the vector form

Answer 2

It all depends on s and t. Are these values given?

9 Places are to be filled up and accomodate 2 unknowns s and t. There must be something more given.

Answer 3

First, find an arbitrary point:

-2·x - 3·y + 3·z = -17

Choose z = 1:

-2·x - 3·y + 3·(1) = -17

-2·x - 3·y = -17 - 3

-2·x - 3·y = -20

Choose y = 0:

-2·x - 3·(0) = -20

Solve for x:

-2·x = -20

x = 10

This means that the point ( 10, 0, 1 ) exists on this plane.

That gives you values for the first column of this equation.

----

Now you want two vectors (for s and t) that are perpendicular to each other but parallel to the plane.

To make this easier, take the fact that a plane equation of the form:

a·x + b·y + c·z = d

Has a normal vector (perpendicular to its surface) of the form:

< a, b, c >

That means that this plane:

-2·x - 3·y + 3·z = -17

Has the normal vector:

< -2, -3, 3 >

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Now, find a vector that is parellel to the surface:

-2·x - 3·y + 3·z = -17

Choose x = 4 and z = 0:

-2·(4) - 3·y + 3·(0) = -17

-8 - 3·y = -17

-3·y = -9

y = 3

This gives the point (4, 3, 0) on the plane.

Find a vector from here to the previous point ( 10, 0, 1 ) :

< 10 - 4, 0 - 3, 1 - 0 >

= < 6, -3, 1 >

Take the cross product of this vector with the normal vector to find a vector that is parallel to the surface but perpendicular to the above vector that is also parellel to the surface:

< 6, -3, 1 > × < -2, -3, 3 >

= < -6, -20, -24 >

If you want, you can change the size of this with a scalar multiple:

(½) · < -6, -20, -24 >

= < -3, -10, -12 >

----

Using these:

P, = ( 10, 0, 1 )

s = < 6, -3, 1 >

t = < -3, -10, -12 >

Your answer becomes:

(periods for spacing only, and the brackets should obviously look like yours)

[ x ] .. [ 10 ] .. [ 6 .] ..... [ -3 ..]

[ y ] = [ 0 ..] + [ -3 ] s + [ -10 ] t

[ z ] .. [ 1 ..] .. [ 1 .] ..... [ -12 ]

Any value you put in for s and t should give you a point that is on the original plane. Plug them in and then add the rows of the matrices together. (You can try putting in random values for s and t and then putting the results into the original equation to see if they equal -17 in order to test it.)