Pia R.

# find equation of line through the point (1,-1,1) perpendicular to the line 3x=2y=z and parallel to plane x+y-z=0

i have no idea solving the problem :(( Nicolas M.

Please, check if correct:  3x = 2y = z   Should it have equal symbols or arithmetic symbols (i.e., + , -)?
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08/26/16

Pia R.

just equal signs :)
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08/26/16

By: Patient & Knowledgeable Math & Science Tutor

Pia R.

how did you get (2,3,6) ?? what do you mean by finding two points? how do i find them?
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08/27/16 Kendra F.

Hi Pia,

There is only one independent variable on a line. Once you pick one of the coordinate, like z = 6, then x and y must be 2 and 3 respectively. 3(2) = 2(3) = 6
pick x = 0 then y and z  must be zero also. 3(0) = 2(0) = 0

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08/27/16

Pia R.

So I could also choose z=1 and end up having x=1/3 y=1/2?
and z=2 and end up having z=2/3 x=1

and also what you meant for independent variable like z is the variable without coefficient? and then from there i pick 1 2 0 or 6 to substitute for it?

oh. and you're so awesome, thank you. how did you know the steps for every question??? lol
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08/27/16 Kendra F.

What I meant is that two of the coordinates are dependent on the other one. So yes, you can pick any value for one and the other two will have values. let y = 1/2 and 3(1/3) = 2(1/2) = 1, P(1/3, 1/2, 1)

It's the same as if you had a line in 2-dimensional space. 3x = 2y is y = 3x/2 and you can find points on the line by choosing a value for one coordinate and solving for the other.
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08/27/16

Pia R.

wouldnt be v = <-9,8,-1>?
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08/31/16

Pia R.

wouldnt it be
x=1-9t
y=1+8t
z=-1-t
??
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08/31/16 Kendra F.

<1,1,-1> x <2,3,6> = <9,-8,1>

Changing the order;

<2,3,6> x <1,1,-1> = <-9,8,-1>

This makes a vector going in the opposite direction. It shouldn't matter if you use this for your line although you should use the point, (1,-1,1) as (x0, y0, z0).
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08/31/16 Kendra F.

You have to use the given point, (1,-1,1) as (x0, y0, z0) so that it is included in the line as well.. at t = 0, (1, -1, 1).
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08/31/16

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