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    How can I prove that the vector form of a line completely within a 3D plane.

    A plane has vector equation ‘r = 2i + 3j + lambda(i - 2j + k) + mu(2i - j + 3k)

    And a line has vector equation ‘r = (2i + 6j + k) + t(5i -7j + 6k)

    Show that the line lies entirely within the plane.
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    If you show that the line and plane have a common point (they intersect), and that they are parallel, then you have shown that the plane contains the line
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    (Original post by physcamy)
    If you show that the line and plane have a common point (they intersect), and that they are parallel, then you have shown that the plane contains the line
    How could I determine if both the line and plane is parallel if the normal vector (vector perp to plane) is not given?

    And how would I get a common point for both if it actually has an infinite number of common point that exits?

    I’m a bit confused, sorry.

    Could you tell me bit more if that’s okay.
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    (Original post by Yatayyat)
    How could I determine if both the line and plane is parallel if the normal vector (vector perp to plane) is not given?

    And how would I get a common point for both if it actually has an infinite number of common point that exits?

    I’m a bit confused, sorry.

    Could you tell me bit more if that’s okay.
    From the equation of the plane, you have 2 vectors which are parallel to the plane (the ones with the λ and μ scalar factors). You can then find a vector normal to the plane by finding the cross product of these two vectors. In terms of the common point you could simply pick a random value of t and get the position vector of any point on the line as all of the points on the line will also lie in the plane. Show that this point also lies in the plane (it would probably be simpler to show this using the dot product format of the equation of the plane which you could work out using the normal vector you have already found). I hope that’s a bit clearer!
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    it is sufficient to show that two points on the line lie in the plane
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    (Original post by physcamy)
    From the equation of the plane, you have 2 vectors which are parallel to the plane (the ones with the λ and μ scalar factors). You can then find a vector normal to the plane by finding the cross product of these two vectors. In terms of the common point you could simply pick a random value of t and get the position vector of any point on the line as all of the points on the line will also lie in the plane. Show that this point also lies in the plane (it would probably be simpler to show this using the dot product format of the equation of the plane which you could work out using the normal vector you have already found). I hope that’s a bit clearer!
    Okay I got the normal vector to be (-5,-1, 3) when using the cross product and I have used z=1

    I then picked a random point as you said of the vector line, I did t=1 and got the point (7,-1,7)

    But I’m still quite uncertain with the last step. When you multiply the normal vector with the point on the line?
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    (Original post by Yatayyat)
    Okay I got the normal vector to be (-5,-1, 3) when using the cross product and I have used z=1

    I then picked a random point as you said of the vector line, I did t=1 and got the point (7,-1,7)

    But I’m still quite uncertain with the last step. When you multiply the normal vector with the point on the line?
    You can get the equation of the plane in dot product format using r•n=a•n where a is the position vector of a point on the plane ie. (2i + 3j) from the parametric equation of the plane. Then dot the position vector of the point on the line with the normal vector and show that it is consistent with the equation of the plane ie. you get the same result as a•n. This shows that the point lies in the plane.
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    (Original post by physcamy)
    You can get the equation of the plane in dot product format using r•n=a•n where a is the position vector of a point on the plane ie. (2i + 3j) from the parametric equation of the plane. Then dot the position vector of the point on the line with the normal vector and show that it is consistent with the equation of the plane ie. you get the same result as a•n. This shows that the point lies in the plane.
    So following on from that, using a•n = r•n, I did get two consistent answers.

    It was (2,3,0) • (-5,-1,3) = (7,-1,7) • (-5,-1,3)

    It came out as -13 for each.


    Would that be okay?
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    (Original post by Yatayyat)
    So following on from that, using a•n = r•n, I did get two consistent answers.

    It was (2,3,0) • (-5,-1,3) = (7,-1,7) • (-5,-1,3)

    It came out as -13 for each.


    Would that be okay?
    Yep that’s it
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    (Original post by physcamy)
    Yep that’s it
    Thanks so much for your help!

    I appreciate it! 🙂
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    (Original post by Yatayyat)
    Thanks so much for your help!

    I appreciate it! 🙂
    No problem - glad I could help I can assure you that vector questions become easier with practice!
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