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I must write an equation for plane which goes through point $B(2; -3; 1)$ perpendicular to line $ \left\{ \begin{array}{c} x=2t-1 \\ y=7-3t \\ z=0 \end{array} \right. $

From that I obtain line direction vector (normal vector of line?) $v=(2; -3; 0)$ And my plane equation is $$2(x-2)-3(y+3)+0=0$$ obtained using formula $A(x-x_0)+B(y-y_0)+C(z-z_0)=0$ (B values are with indexes)

I am pretty sure I have misunderstood something though.

Arnie
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    it looks correct; the point is on the plane, and the normal vector of the plane has the same direction as the line; perhaps you should call the point $P$ instead of $B$, because you used $B$ for something else – J. W. Tanner Jan 01 '20 at 17:20
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    What makes you think your answer is incorrect? – amd Jan 01 '20 at 17:41
  • So what would the final answer be? Let us confirm for you through to the final step of your solution! – Andrew Chin Jan 01 '20 at 21:03
  • Well it's my first time doing analytic geometry problems so I assumed I most likely misunderstood something if I got it this easy... I guess it's correct, thanks for reviewing. – Arnie Jan 01 '20 at 21:14
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    Your answer is correct except (normal vector of line?). The vector is orthogonal to the plane, not the line. – user376343 Jan 07 '20 at 21:49

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