The points $(3, -1, -5)$, $(1, 4, 19)$, and $(5, 3, 7)$ lie on a unique plane, where does it cross the $z$-axis.
I have tried to use the cross product after forming two vectors to solve this problem but am still not able to reach the correct answer.
Finding where three points cross $z$-axis
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linear-algebra
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0Drawing a diagram can really help here. Wikipedia also has an article on [Plane-Line Intersection](https://en.wikipedia.org/wiki/Line%E2%80%93plane_intersection). – 2017-01-13
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0I tried to follow that method and I got z = 15. Which is the wrong answer. – 2017-01-13
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0@user2175095 Since you said $z=15$ is a wrong answer, what was the correct answer? – 2017-01-13
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0@user2175095 I got the answer $z=5$. What do you think? – 2017-01-13
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0I submitted 5, and it was the correct answer, thanks! – 2017-01-13
3 Answers
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You can just parameterize the lines through the points and solve the simultaneous equations in the first two coordinates for $s,t$, then compute $z$ $$(3,-1,-5)+s(1-3,4-(-1),19-(-5))+t(5-3,3-(-1),7-(-5))=(0,0,z)$$
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The equation of the plane which passes through the points $(3,-1,-5)$, $(1,4,19)$, and $(5,3,7)$ is given by $2x-4y+z-5=0$. This plane passes the $z$-axis at the point $(0,0,5)$.
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Let $\mathbf n$ be a normal to the plane, which you can find by taking cross products. The normal form of the equation of a plane in $\mathbb R^3$ says that for all points $\mathbf x$ on the plane, $\mathbf n\cdot\mathbf x$ is constant. You’re looking for a point of the form $(0,0,z)$, so solve $\mathbf n\cdot(0,0,z)=\mathbf n\cdot\mathbf p$ for $z$, where $\mathbf p$ is any of the three given points (or any other known point on the plane, for that matter), yielding $z=\mathbf n\cdot\mathbf p/n_z$.