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Math Help - Vectors

  1. #1
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    Unhappy Vectors

    (a) Find parametric equations for the line of intersection between the planes x + y = 0 and 3y + 8z = 6.

    (b) Find the point of intersection of the line found in part (a) and the plane 2x + 5y + 4z = 5.
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  2. #2
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    a: n_{1}=(1,1,0), \;\ n_{2}=(3,0,8)

    These are the normals to the planes.

    Find the cross product: n_{1}\times n_{2}=(8,-8,-3)

    Now, let z=0 in the two equations and solve for x and y.

    Finish?.
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  3. #3
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    Another way: the two planes are x + y = 0 and 3y + 8z = 6. Solve for, say, x and z in terms of y: x= -y and 8z= 6- 3y so z= (3/4)- (3/8)y.

    Now take t= y/8 as parameter. x= -8t, y= 8t, z= (3/4)- 3t.

    To find the point where that line intersects the plane 2x + 5y + 4z = 5, replace x, y, and z in that equation by their parametric expressions: 2(-8t)+ 5(8t)+ 4((3/4)- 3t)= -16t+ 40t- 12t+ 3= 12t+ 3= 5. 12t= 2, t= 1/6. Now put that value of t back into the parametric equations: x= -8(1/6)= -4/3, y= 8(1/6)= 4/3, z= (3/4)- 3(1/6)= 3/4- 2/4= 1/4. The point of intersection is (-4/3, 4/3, 1/4).

    Note that those values of x, y, and z satisfy the equations of all three planes. -4/3+ 4/3= 0, 3(4/3)+ 8(1/4)= 4- 2= 6, and 2(-4/3)+ 5(4/3)+ 4(1/4)= -8/3+ 20/3+ 1= 12/3+ 1= 4+ 1= 5.
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  4. #4
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    I'm confused.
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