Question 1207403: The lines l and m have vector equations
r = 2i-j+4k+s(i+j- k) and r= -2i + 2j+ k+ t(-2i+j+ k)
respectively・
i) Show that l and m do not intersect.
The point P lies on l and the point Q has position vector 2i - k.
(ii) Given that the line PQ is perpendicular to l, find the position vector of P.
(iii) Verify that Q lies on m and that PQ is perpendicular to m.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
I'll focus on problem 1 only.
i,j,k are the unit vectors that point in the positive x, positive y, and positive z directions respectively.
The component form of each unit vector would be
i = (1,0,0)
j = (0,1,0)
k = (0,0,1)
Each vector is of magnitude 1.
magnitude = length
The unit vectors help set up a 3D coordinate system to locate any point in 3-space.
Let's find the component form of line L.
2i-j+4k+s(i+j-k)
2(1,0,0)-(0,1,0)+4(0,0,1)+s( (1,0,0)+(0,1,0)-(0,0,1) )
(2,0,0)+(0,-1,0)+(0,0,4)+s( (1,0,0)+(0,1,0)+(0,0,-1) )
(2,0,0)+(0,-1,0)+(0,0,4)+s(1,1,-1)
(2,0,0)+(0,-1,0)+(0,0,4)+(s,s,-s)
(2+s,-1+s,4-s)
Where 's' represents any real number. Think of it like the timestamp.
At time s = 0 we will be at location (2,-1,4) because of this scratch work
(2+s,-1+s,4-s)
(2+0,-1+0,4-0)
(2,-1,4)
At time s = 1, we are then going to be at the point (3,0,3) since
(2+s,-1+s,4-s)
(2+1,-1+1,4-1)
(3,0,3)
And so on.
Each point mentioned in this paragraph resides on line L.
Let's find the component form of line M.
-2i+2j+k+t(-2i+j+k)
-2(1,0,0)+2(0,1,0)+(0,0,1)+t(-2(1,0,0)+(0,1,0)+(0,0,1))
(-2,0,0)+(0,2,0)+(0,0,1)+t((-2,0,0)+(0,1,0)+(0,0,1))
(-2,0,0)+(0,2,0)+(0,0,1)+t(-2,1,1)
(-2,0,0)+(0,2,0)+(0,0,1)+(-2t,t,t)
(-2-2t,2+t,1+t)
Where t represents the timestamp marker.
t = 0 leads to (-2,2,1)
t = 1 leads to (-4,3,2)
etc
I skipped showing the scratch work, but it's going to be very similar to what is found in the previous paragraph.
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Here are the component forms of lines L and M
L = (2+s,-1+s,4-s)
M = (-2-2t,2+t,1+t)
If the two lines intersected at some point P, then the x components of L and M must match up. So must the y components, and z components as well.
For example, let's say they intersect at the point (7,8,9)
This would mean the x components of L and M must be 7, the y components 8, and z components 9.
e.g. 2+s = 7 and -2-2t = 7 which leads to 2+s = -2-2t
Equate each pair of components to form a system of equations
2+s = -2-2t
-1+s = 2+t
4-s = 1+t
Let's isolate t in the third equation
4-s = 1+t
t = 4-s-1
t = 3-s
Then plug this into the first equation
2+s = -2-2t
2+s = -2-2(3-s)
2+s = -2-6+2s
2+s = -8+2s
s-2s = -8-2
-s = -10
s = 10
Repeat for the second equation also
-1+s = 2+t
-1+s = 2+3-s
-1+s = 5-s
s+s = 5+1
2s = 6
s = 6/2
s = 3
But this contradicts the previous result of s = 10.
Therefore this system
2+s = -2-2t
-1+s = 2+t
4-s = 1+t
does not have any solutions. The system is inconsistent.
This proves that lines L and M do not intersect.
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