# Algebra - involving equations of lines and planes

## 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.

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danago
Gold Member
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.

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
Gold Member
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.

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.