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# Maths: Vectors and planes Watch

1. Hi everyone

I am struggling to do this problem in math:

Show that the plane r.(3i-2j+k)=1 contains the line r=3i+3j-2k+t(i+j-k).

Does anyone has any suggestions? Any help will be greatly appreciated

Thank you
2. I have two ways: a "cheaty" way, and a more "this is what we're examining" way.
The cheaty way: two points on the line are (3,3,-2) and (4,4,-3). You can check by substituting in the values that both these points lie on the plane. Hence the entire line lies on the plane, since that's the only way a line can intersect a plane twice.

The alternative way:
You need to do two things: show that t(i+j-k) is parallel to the plane, and that (3,3,-2) lies on the plane. That will tell you that the line lies in the plane.
To show that the point is on the plane should be straightforward by substituting in the values. The parallel line: if you substitute the form of the line, r=3i+3j-2k+t(i+j-k) into the equation r.(3i-2j+k)=1, you get that 1=1, as required. (I'm not certain whether you actually have to show that the point is in the plane; this should actually work without that step.)
3. I did something weird and I am not sure if it is correct: I have made the equation of the plane equal to the equation of the line so I could solve for t. As a result I found out that t was 0. Does this suggest that the plane and the line do not intersect? Can you use this method in order to prove that the line is in the plane?

(Original post by Smaug123)
I have two ways: a "cheaty" way, and a more "this is what we're examining" way.
The cheaty way: two points on the line are (3,3,-2) and (4,4,-3). You can check by substituting in the values that both these points lie on the plane. Hence the entire line lies on the plane, since that's the only way a line can intersect a plane twice.

The alternative way:
You need to do two things: show that t(i+j-k) is parallel to the plane, and that (3,3,-2) lies on the plane. That will tell you that the line lies in the plane.
To show that the point is on the plane should be straightforward by substituting in the values. The parallel line: if you substitute the form of the line, r=3i+3j-2k+t(i+j-k) into the equation r.(3i-2j+k)=1, you get that 1=1, as required. (I'm not certain whether you actually have to show that the point is in the plane; this should actually work without that step.)
4. (Original post by beatrice15)
I did something weird and I am not sure if it is correct: I have made the equation of the plane equal to the equation of the line so I could solve for t. As a result I found out that t was 0. Does this suggest that the plane and the line do not intersect? Can you use this method in order to prove that the line is in the plane?

You can't just set the equation of the plane equal to that of the line: the reason it's invalid to compare the two is that r.(3i-2j+k) = 1 is a scalar, whereas r = 3i+3j-2k + t(i+j+k) is a vector, so you can't compare them. It's just like saying (3,4,2) = 5; it just doesn't make sense mathematically. What you can do is substitute the line equation wherever r appears in the plane equation, because then you're just swapping in the "same expression" - r is a vector and 3i+3j-2k + t(i+j+k) is a vector, so it's fine to interchange between the two.
5. Thank you this helped me a lot

(Original post by Smaug123)
You can't just set the equation of the plane equal to that of the line: the reason it's invalid to compare the two is that r.(3i-2j+k) = 1 is a scalar, whereas r = 3i+3j-2k + t(i+j+k) is a vector, so you can't compare them. It's just like saying (3,4,2) = 5; it just doesn't make sense mathematically. What you can do is substitute the line equation wherever r appears in the plane equation, because then you're just swapping in the "same expression" - r is a vector and 3i+3j-2k + t(i+j+k) is a vector, so it's fine to interchange between the two.
6. (Original post by Smaug123)
I have two ways: a "cheaty" way, and a more "this is what we're examining" way.
The cheaty way: two points on the line are (3,3,-2) and (4,4,-3). You can check by substituting in the values that both these points lie on the plane. Hence the entire line lies on the plane, since that's the only way a line can intersect a plane twice.

The alternative way:
You need to do two things: show that t(i+j-k) is parallel to the plane, and that (3,3,-2) lies on the plane. That will tell you that the line lies in the plane.
To show that the point is on the plane should be straightforward by substituting in the values. The parallel line: if you substitute the form of the line, r=3i+3j-2k+t(i+j-k) into the equation r.(3i-2j+k)=1, you get that 1=1, as required. (I'm not certain whether you actually have to show that the point is in the plane; this should actually work without that step.)
Any Website or tutorial that can help revise all the concepts?

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