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Find an equation of the plane. the plane through the point (β5, 9, 10) and perpendicular to the line...
4 months ago
Q:
Find an equation of the plane. the plane through the point (β5, 9, 10) and perpendicular to the line x = 1 + t, y = 4t, z = 2 β 3t
Accepted Solution
A:
The direction vector of the line
L: x=1+t, y=4t, z=2-3t
isΒ <1,4,-3>
which is also the required normal vector of the plane.
Since the plane passes through point (-5,9,10), the required plane is :
Π 1(x-(-5)+4(y-9)-3(z-10)=0
=>
Π x+4y-3z=1