equation of a plane with 3 points on axis

Question

Well I'am stuck with one question about finding equation of a plane.. Few months ago I got at my exam question that sounded something like this:

Find the equation of a plane if it cuts segment on x-axis,y-axis and z-axis in the ratio of 4:3:2 retrospective and if... rest of the text isn't important

Sorry for bad explanation of the task, I had to translate it from my native language, if you don't understand something feel free to ask..

So finally, my question goes like this:

By this segments in question, am I allowed to take points with cordinates $(4,0,0), (0,3,0), (0,0,2)$

I know everything what I need to find equation of a plane,but I always get confused with segments like this, that's the thing on which I need help

Thanks

Answer 1

Yes, you are allowed to assume that those points lie on a desired plane satisfying the conditions of the problem. Notice that what the problem wants you to understand is that the given ratio $m:n:l$ is sort of like a generalization of the concept of slope in plane geometry when dealing with lines. Just like you can find a family of parallel lines with varying intercepts when you have been given the slope, you can find a family of parallel planes with the given ratio $m:n:l$, passing through different points in space.

To solve the problem, you can follow Deepesh Meena's solution. If you're looking for another way to solve the problem, you can first find two vectors created by the three points $(4,0,0)$, $(0,3,0)$ and $(0,0,2)$ and then calculate their cross product to find the normal vector of the plane spanned by them. Note that this plane will contain all the three points because it contains both vectors.

More precisely, in your problem, we can write

$$\vec{u}=(4,0,0)-(0,3,0) =(4,-3,0)$$ $$\vec{v}=(4,0,0)-(0,0,2) =(4,0,-2)$$

We know that $\vec{u} \times \vec{v} = (6,8,12)$ is orthogonal to both $\vec{u}$ and $\vec{v}$. Hence, it is parallel to the normal vector of the plane containing $\vec{u}$ and $\vec{v}$. You can normalize $\vec{u}\times\vec{v}$ if you want, but it isn't necessary in this case.

So, the equation of a plane satisfying the given ratio can now be found as follows:

$$(6,8,12)\cdot\left((x,y,z)-(4,0,0)\right)=0$$

$$6(x-4)+8y+12z=0$$ $$6x + 8y + 12 z = 24$$ $$\frac{x}{4} + \frac{y}{3} + \frac{z}{2} = 1$$

Notice that the numbers in the denominator match your initial $4:3:2$ ratio.

Answer 2

Yes, you are correct why don't you take general points $(4k,0,0),(0,3k,0),(0,0,2k)$ and calculate the equation of the plane passing through this points.

Suppose the equation of the plane is $ax+by+cz=d$

Now put these points in the plane you get

$$a(4k)=d$$ $$b(3k)=d$$ $$c(2k)=d$$

Pur back all the values in the first equation

Thus equation of your plane is $$\frac{x}{4}+\frac{y}{3}+\frac{z}{2}=k$$