Pia R.
asked • 08/26/16find 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 :((
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1 Expert Answer
Kendra F. answered • 08/26/16
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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
Two direction vectors which are perpendicular to the required line are the normal of the given plane, <1, 1, -1> and a direction vector for the given line. A direction vector for the line 3x=2y=z is found by finding two points. I went with (0,0,0) and (2,3,6). So the direction vector is <2,3,6>.
to find a direction vector for the line we need, we take the cross product.
v = <1, 1, -1> x <2, 3, 6> = <9, -8, 1>
v = <1, 1, -1> x <2, 3, 6> = <9, -8, 1>
Write the required line out in parametric form using the given point as (x0, y0, z0).
parametric form:
x = x0 + at
y = y0 + bt
z = z0 + ct
x = 1 + 9t
y = -1 - 8t
z = 1 + t
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
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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Nicolas M.
08/26/16