Find a plane that goes through three given points

Question

The Question

Data is collected on a person’s income (thousands of dollars), their age, and the value of their home (thousands of dollars). We would like to predict home value, H, as a function of income and age. The following three ordered pairs are found: (100,40,275), (60, 30, 180), and (110, 50, 305). The relationship is assumed to be linear. Find a plane to model the data.

My Attempt

My first thought is to turn these three points into two vectors and then get the cross-product of said two vectors. This results in a vector normal to the plane which we need for the equation: $n_1(x-x_0) + n_2(y-y_0) + n_3(z-z0) = 0$

Where n1, n2, and n3 are the components of the normal vector. Now what trips me up is the wording in the question that asks that the "Home value be a function of income and age" but the plane that this gets me is a function of income, age, AND home value which doesn't make much sense considering we are calculating home value.

Answer 1

You can always write $z$ in terms of the other values. Maybe it's easier to explain in 2D. Assume that you have the equation $ax+by+c=0$. This is the equation of a line. You can still write $y$ in terms of $x$ as $$y=-\frac ab x-\frac cb$$ The only requirement is that $b\ne 0$. Your 3D problem is equivalent, you just add one more parameter. The equation of your plane will become $$z=\alpha x+\beta y+\gamma$$ If you plug in the three triplets in the above equation, you get a system of three linear equations with three unknowns ($\alpha,\ \beta,\ \gamma$).

Answer 2

The graph of a function $f:X \rightarrow Y$ is the set $(x,f(x))$ with $x \in X$ and $f(x) \in Y$. Your mistake is assuming that these points are in $X$ entirely which is not the case as they are point on the graph which is a subset of $X \times Y$.

Answer 3

This is a multivariable regression problem. For example, let $y = [100, 60, 110]$, and $X = \begin{bmatrix} 1 &40 & 275 \\ 1 & 30 & 180 \\ 1 & 50 & 305 \end{bmatrix}$ (the ones column is added so we can also fit an additive constant). Then the regression coefficients $\theta = [\theta_1, \theta_2, \theta_2]$ in $X\theta = y$ can be found with $\theta = (X^TX)^{-1}X^Ty$, and to predict new data $x = [1, x_1, x_2]$, you simply take the dot product of $x$ with $\theta$: $x\cdot \theta = \theta_1 + \theta_2\cdot x_1 + \theta_3 \cdot x_2$.

There are ways to calculate $\theta$ in the case that $X^TX$ is not invertible. Most standard software will use these methods to calculate the coefficients, so that you will always obtain a solution for the coefficients.

Answer 4

Though I'm not sure, this may be correct. Think of the plane as representing all combinations of home value, income and age that is attainable given the points you used to construct the plane. If you think of home value as being measured along one of the axis, say the z-axis, then all the values on the plane with a certain z-coordinate represents combinations of income and age predicting that fixed home value. That is, for a given $z=z_0$ all combinations of $x$ and $y$ coordinates yielding that $z$-value. Since all values of $z$ can be attained this way, $z$ is a function of $x$ and $y$.