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Math Help - A Vector Problem

  1. #1
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    A Vector Problem

    Consider the planes  5x+4y+1z = 1 and  5x+1z = 0

    (A) Find the unique point P on the y-axis which is on both planes. (__, __, __)

    (B) Find a unit vector U with positive first coordinate that is parallel to both planes. ___I + ___J + ___K

    (C) Use the vectors found in parts (A) and (B) to find a vector equation for the line of intersection of the two planes, r(t) = ___I + ___J + ___K
    I was able to find part A by setting x and z to 0 and solve for y:
     5(0) + 4y + 1(0) = 1
     y = \frac{1}{4}  => (0, \frac{1}{4}, 0)

    Need help solving part B and C
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  2. #2
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    Quote Originally Posted by viet View Post
    I was able to find part A by setting x and z to 0 and solve for y:
     5(0) + 4y + 1(0) = 1
     y = \frac{1}{4}  => (0, \frac{1}{4}, 0)

    Need help solving part B and C
    The equations of the two planes are given in normal form with

    \overrightarrow{n_1} = (5,4,1) and

    \overrightarrow{n_2} = (5,0,1)

    The vector which is perpendicular to both normal vectors must be parallel to both planes indicating the direction of the line of intersection:

    \vec v = \overrightarrow{n_1} \times \overrightarrow{n_2} = (4,0,-20)

    |\vec v| = \sqrt{16+400} = \sqrt{416}=4 \cdot \sqrt{26}

    The unit vector \vec U = \dfrac{(4,0,-20)}{4 \cdot \sqrt{26}}=\left(\dfrac1{\sqrt{26}}\ ,\ 0\ , -\dfrac5{\sqrt{26}}\right)

    You already know that the line of intersection passes through the point \left( 0, \dfrac14 ,0\right)

    Therefore

    \vec r = \left( 0, \dfrac14 ,0\right) + t \cdot \left(\dfrac1{\sqrt{26}}\ ,\ 0\ , -\dfrac5{\sqrt{26}}\right)
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