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.Find the plane equation that contains the X axis and the point A(4,-3,-1) The correct answer is y-3z=0 Anyone could explain me why is that the answer and is not -3y-1z=0? Thanks you in advanced! Sorry for my English, I'm Spanish

Ceci
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1 Answers1

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It will suffice to find two linearly independent points in the plane. One such point is the unit vector in the $x$-direction. That would be $(1,0,0)$. Another is the point $(4,-3,-1)$. Go ahead and take their cross-product to produce a vector normal to the plane. You know that each vector in this plane is then orthogonal to the normal vector. Can you get the equation for the plane from that?

Alex Ortiz
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  • Sorry,What is cross product? – Ceci Sep 07 '16 at 01:07
  • @Ceco The cross-product of $(a,b,c)$ with $(d,e,f)$, written $(a,b,c)\times(d,e,f) = (bf - ce, cd - af, ae - bd).$ – Alex Ortiz Sep 07 '16 at 01:10
  • I could do it but the answer is y-3z=0 and i optein y+3z=0 :( thanks for answer – Ceci Sep 07 '16 at 01:23
  • @Ceco, I computed the cross product and determined the vector $(0,1,-3)$ as being normal to the plane. Check your calculation. With this in hand, do you see how to get the equation of the plane? – Alex Ortiz Sep 07 '16 at 01:32