Line A has equation r = <2; 4; 2>t + <1; 0;-2>. Line B passes through (0; 1; 2) and
intersects line A at an angle of pi/2 (i.e., the two lines are perpendicular). Find the
point of intersection.
intersects line A at an angle of pi/2 (i.e., the two lines are perpendicular). Find the
point of intersection.
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Suppose that A and B intersect at (x, y, z). Since (x, y, z) also lies on B (since B intersect A at (x, y, z)), another way to write the direction vector is:
.
You can write the equation for A as a parametric equation:
x = 2t + 1, y = 4t, z = 2t - 2, by expanding the given equation and comparing components.
The dot product of two orthogonal vectors is zero, so:
• <2, 4, 2> = 0 ==> 2x + 4(y - 1) + 2(z - 2) = 0.
Substituting the above parametric equations into this gives:
2(2t + 1) + 4(4t - 1) + 2(2t - 2 - 2) = 0 ==> t = 5/12.
Therefore, the point of intersection occurs at t = 5/12, which yields the point:
(2(5/12) + 1, 4(5/12), 2(5/12) - 2) = (11/6, 5/3, -1/2).
I hope this helps!
You can write the equation for A as a parametric equation:
x = 2t + 1, y = 4t, z = 2t - 2, by expanding the given equation and comparing components.
The dot product of two orthogonal vectors is zero, so:
Substituting the above parametric equations into this gives:
2(2t + 1) + 4(4t - 1) + 2(2t - 2 - 2) = 0 ==> t = 5/12.
Therefore, the point of intersection occurs at t = 5/12, which yields the point:
(2(5/12) + 1, 4(5/12), 2(5/12) - 2) = (11/6, 5/3, -1/2).
I hope this helps!