Algebra - involving equations of lines and planes

[DanielT29]

Homework Statement

Hi guys,

I was helping a friend in grade 12 with his Linear Algebra, when I got stumped on some questions I wasn't quite to sure of. I forgot some of my old linear algebra so I need you guys to help me out:) Here's the question:

1)The two line with equations L1: r = (1,2,-4) + t(k+1,3k+1,k-3) and L2: x=2-3s, y=1-10s, z=3-5s are given.

a) Determine the value of k if these two lines are parallel and perpendicular.

2)

a)does the point P(2,4,2) lie on the line x=2t,y=3+t, z=1+t?

b)if the point (a,b,-3) lies on the line find values of a and b

3) Determine the equation of the plane through A(2,1-5) perpendicular to both 3x-2y+2=8 and 4x+6y-5z=10.

4) What is the scalar equation of the plane parallel to the plane 3x-9y+z-12=0 and including the point (-3,7,1).
2. The attempt at a solution

1. Parallel : I took the t(k+1,3k+1,k-3) as tv(vector) and since it had to be parallel with the line, I took the vector from the other equation x=2-3s, y=1-10s, z=3-5s, which is -3,-10,-5 and i equated to the respective x y z in the first line. So:

k+1 = -3 -> -4
3k+1 = -10 -> -11/3
k-3 = -5 -> -2

Perpendicular: I need your help I didn't know how to do it, do I have to make it somehow equal 0? Something about dot product being 0 not sure need advice guys!

2.

a)So i subed in 2, 4 ,2 into the respective x y and z, if they all had the same t I said the point lied on the same line, and it did.

b) I subed in -3 for z = 1+t , solved for t, then used that t to solve for the x and y in the other provided equations.

3. I took the vectors from the plane equations so v1 = (3,-2,2) and v2= (4,-6,-5) then I did cross product and I got v3 = ( -2,23,-10)( someone confirm this). After that I made it into the equation -2+23y-10+D = 0, I put in the points it went through into their respective x y and z coordinated and solved for z.

4) A bit confused on this one can someone give me a tip I think it's just basically the same equation except you'd have to solve for a new D, so 3x-9y+z-12=0 would be 3x-9y+z-D=0 then you sub in the new points solve for a new D. I think that's how you do it. Can you guys help me:)

Thanks a lot guys I appreciate every response in advance.

 

[danago]

For (a), just note that, as you said, if two vectors are perpendicular, their dot product are zero. So basically make the "slope" components of the lines have a zero dot product.

For 4, you are pretty much right. In general, a plane which is normal to the vector (a,b,c) and passes through the point (p,q,r) has the equation:

a(x-p) + b (y-q) + c(z-r)=0
or
ax + by + cz = ap + bq + cr

If the two planes are parallel, they will clearly have the same normal vector; (3,-9,1) in this case, and you now have a new (p,q,r) to find the constant term.

 

[DanielT29]

O, thanks. Um just a small question what did you mean by their "slope" components? Did you mean something like this

k+1/-3 =0
3k+1/-10 = 0
k-3/-5 = 0

If you did can you tell me how you got that, cause i thought dot product was multiplication of components of a vector, but I guess doing it both ways you'd still have to make the k equal a value to make the equation 0 ex. (k+1)(3) = 0 k=-1. So is this what you mean?

 

[danago]

Sorry, what i meant was the vector that is parallel to each line. Basically, lines have the form

r = a + tv

Where v is the "slope component" i.e. it is parallel to the line. So in this case, the first line is parallel to (k+1,3k+1,k-3), and the second line is parallel to (-3,-10,-5). So those two vectors should have a zero dot product.

 

[DanielT29]

Ahh haa, thanks. So it would be:

(k+1)(-3)+(3k+1)(-10)+(k-3)(-5) = 0

Then solve for k, so in this case there's is only one value for k..unless I did the parallel one wrong. Can you check this out for me. If you apply the dot product for being parallel then the it should equal 1 or -1. But I suppose you take 1.

Thanks for your response.