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Calculus and Beyond Homework Help
Thx for help tonight yall 3 more questions X_x
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[QUOTE="Mark44, post: 4512273, member: 147785"] Yes. And what would that value be? This problem is much simpler than you're making it. I don't think the concept of a normal to a plane has gelled in your mind just yet. If the plane is perpendicular to the x-axis, give me one normal to this plane. Do you understand what the variables in this equation represent? A(x - x[SUB]0[/SUB]) + B(y - y[SUB]0[/SUB]) + C(z - z[SUB]0[/SUB]) = 0 In case you don't, [B]n[/B] = <A, B, C> is a normal to the plane, and P[SUB]0[/SUB](x[SUB]0[/SUB], y[SUB]0[/SUB], z[SUB]0[/SUB]) is a particular point on the plane. The equation comes from the fact that if P(x, y, z) is an arbitrary point on the plane, then the vector PP[SUB]0[/SUB] = <x - x[SUB]0[/SUB], y - y[SUB]0[/SUB], z - z[SUB]0[/SUB]> is perpendicular to [B]n[/B]. Therefore, [B]n[/B] ##\cdot## PP[SUB]0[/SUB] = 0. The equation above comes directly from this dot product. [/QUOTE]
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Calculus and Beyond Homework Help
Thx for help tonight yall 3 more questions X_x
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