1.
The line has direction numbers (2,3,-4)
The direction numbers of the normal of plane b)
are (2,3, -4)
Since the normal is perpendicular to the plane
the given line is perpendicular to the plane
2.
Does (0,0,0 satisfy the equation of plane b) ?
LS = 2(0) + 3(0) - 4(0) = 0
RS = 0 , so YES, it does
your correct answer is b)
Which plane goes through the origin and is perpendicular to the line r=(2,-2,1) + s(2,3,-4), seR?
a) 2x-2y+z=0
b) 2x+3y-4z=0
c) 2x+3y+z-4=0
d) none of the above.
I got D, none of the above. I have substituted and couldn't find the answer equal to zero.
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