Intersecting vector lines

1. Feb 21, 2010

lemon

1.Consider the two straight lines:
2λi+(1+4λ)j+(3-3λ)k
(4+μ)i+(3+5μ)j+(-6+μ)k

a) Show lines are skew when a=3.
b) Find value of a for which lines intersect and state coordinates of point of intersection

2.
2λi+(1+4λ)j+(3-3λ)k=(4+μ)i+(3+5μ)j+(-6+μ)k

comparing i components:
(1)
2λ=4+μ
λ=(4+μ)/2

comparing j components:
(2)
1+4λ=3+5μ
λ=(2+5μ)/4

(1)=(2)
(4+μ)/2=(2+5μ)/4
16+4μ=4+10μ
μ=2

sub into (1)
λ=(4+μ)/2
λ=(4+2)/2
λ=3

Different values so the lines are skew

d) Find a value of a for which the lines intersect and state the coordinates of the point of intersection.
r1=2λi+(4λ+1)j+(a-3λ)k

r2=(μ+4)i+(5μ+3)j+(μ-6)k

I need to find the value of a for which the lines intersect and state the coordinates of point of intersection.

λ=3
μ=2

LHS a-3λ=μ-6
a-3(3)=2-6
a-9=-4
a=5

r2=4i+3j-6k+μ[i+5j+k]
r2=4i+3j-6k+2[i+5j+k]
r2=4i+3j-6k+2i+10j+2k
6i+13j-4k

coordinates of point of intersection are:
(6, 13, -4)

could somebody tell me if i'm on the right track here please?

2. Feb 22, 2010

songoku

Hi lemon

It should be : 2λi+(1+4λ)j+(a-3λ)k

What do you mean by "different values" ? You only find one value of λ and μ so how can you say "different values" ? To what did you compare or check it?

The rest of your work is nice.

3. Feb 22, 2010

lemon

ahh yes. I should say:
With the values of λ and μ the coefficients of K become:
LHS=3-3(3)=-6
RHS=-6+2=-4
LHS≠RHS, and so the lines are skew.

I'm not sure how I would compare of check this?

Last edited: Feb 22, 2010
4. Feb 22, 2010

songoku

If I don't misinterpret your post, the value of λ and μ should not be the same for the line to be skew. Check it using the third equation, comparing the k components. You'll end up LHS is not the same as RHS.

5. Feb 22, 2010

lemon

yes. Thank you songoku. The values are different and so the lines are skew.