Asked by Jake
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.
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.
Answers
Answered by
Reiny
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)
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)
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.