Thanks! Please explain!
2007-09-03 05:06:59 · 0 answers · asked by Arch-RF d 1 in Science & Mathematics ➔ Mathematics
The normal vector n, to the given plane
x + 3y - 6z - 8 = 0
is
n = <1, 3, -6>
The vector n is also the directional vector of the line thru point P(4, -5, 20) and perpendicular to the given plane.
The equation of the line L is:
L = P + tn = <4, -5, 20> + t<1, 3, -6>
L = <4 + t, -5 + 3t, 20 - 6t>
Now convert the equation of the line to symmetric form by solving for t.
L:
x = 4 + t
y = -5 + 3t
z = 20 - 6t
Symmetric form of equation.
t = (x - 4)/1 = (y + 5)/3 = (z - 20)/-6
2007-09-03 18:48:26 · answer #1 · answered by Northstar 7 · 1⤊ 0⤋
Symmetric Equation Of A Line
2017-01-04 08:48:42 · answer #2 · answered by hazelton 4 · 0⤊ 0⤋
RE:
What is the symmetric equation of the line through point (4,-5,20) & perpendicular to the plane x+3y-6z-8=0.?
Thanks! Please explain!
2015-08-02 04:12:02 · answer #3 · answered by Anonymous · 0⤊ 0⤋