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    http://www.ocr.org.uk/Images/57750-q...hematics-3.pdf

    I'm attempting question 6.

    So I've done part 1.

    For part 2, I thought that it would be the normal vector in the opposite direction so, (-1,0,2), but it still uses the same normal vector of (1,0,-2). Why is that?
    And also for part 3, I'm not to sure how to go about finding the distance, I thought it would be the distance of the second plane - the distance of the first over the magnitude of the normal vector, but when I do it I get the wrong answer .
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    (Original post by Music99)
    http://www.ocr.org.uk/Images/57750-q...hematics-3.pdf

    I'm attempting question 6.

    So I've done part 1.

    For part 2, I thought that it would be the normal vector in the opposite direction so, (-1,0,2), but it still uses the same normal vector of (1,0,-2). Why is that?
    And also for part 3, I'm not to sure how to go about finding the distance, I thought it would be the distance of the second plane - the distance of the first over the magnitude of the normal vector, but when I do it I get the wrong answer .
    You can multiply the normal by any non-zero scalar and you still have a normal.
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    (Original post by BabyMaths)
    You can multiply the normal by any non-zero scalar and you still have a normal.
    Yeah, but I mean why is it for part 2, that you use the normal (-1,0,2) which gives a value for d of -3 and not the normal (1,0,-2) which gives a d value of 3, how do you know what one to use?
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    (Original post by Music99)
    Yeah, but I mean why is it for part 2, that you use the normal (-1,0,2) which gives a value for d of -3 and not the normal (1,0,-2) which gives a d value of 3, how do you know what one to use?
    First plane

    r.<1,0,-2> = d

    Substitute a point in the first plane..

    <3,4,-1>.<1,0,-2>=d to get d=5. The equation is r.<1,0,-2> = 5.

    Second plane

    r.<1,0,-2> = d

    Substitute a point in the second plane..

    <5,1,1>.<1,0,-2>=d to get d=3. The equation is r.<1,0,-2> = 3.
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    It makes no difference, you get x-2z = 3 or -x+2z = -3.
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    (Original post by BabyMaths)
    It makes no difference, you get x-2z = 3 or -x+2z = -3.
    Ah okay thank you, would you mind helping me with another question?

    http://www.ocr.org.uk/Images/59548-q...hematics-3.pdf

    Question number 6, the last bit.

    The first part is r.(1,2,0)=4

    The next bit is 74.4 degrees.

    So for the last bit, I worked out vector GF which is 4i so then M is 2i?

    And then I'm not to sure where to go from there, I know i need to find the point of intersection of the line and plane, but not too sure how.
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    If I didn't read the previous questions, I would find the vector equation of a line AM with respect to O, then find the cartesian equation of a plane with respect to O using normal OCxOE, then find the common parameters.

    But, given the answers to the previous questions, I suspect that you have to draw some similar triangles.
 
 
 
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