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QUESTION: I have no clue what to do for this problem. the equation y= -3 is confusing me.

Find an equation for the plane containing the line L: (x-4)/(-4)= (z+3)/2 ; y= -3 and the point Q(-3,2,-2).

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Subject:Linear algebra

Question:

I have no clue what to do for this problem. the equation y= -3 is confusing me.

Find an equation for the plane containing the line L: (x-4)/(-4)= (z+3)/2 ; y= -3 and the point Q(-3,2,-2).

Write  x = -4t + 4  and   z = 2t - 3   and  y = 3 + 0t  <<<< 0t is the clue.

Your normal vector is thus   <4,0,2>  or  <2,0,1>

That should do it.

---------- FOLLOW-UP ----------

QUESTION: Hi, thanks for the help. It is super clear.

I got 4x+0y+2z=10 as the equation for the plane but it is telling me that it is wrong. I got this answer by plugging the point(4,3,-3) into the equation 4x+0y+2z=0.

I have no idea why it is wrong.

QUESTION: Hi, thanks for the help. It is super clear.

I got 4x+0y+2z=10 as the equation for the plane but it is telling me that it is wrong. I got this answer by plugging the point(4,3,-3) into the equation 4x+0y+2z=0.

I have no idea why it is wrong.
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#### Paul Klarreich

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