Intersection of two subspaces (Friedberg Chapter 1.3)

Question

Let $W_1$ and $W_2$ be subspaces of $\mathbb{R}^3$ under addition and scalar multiplication such that:

$$W_1=\{(a_1,a_2,a_3)\in \mathbb{R}^3:2a_1-7a_2+a_3=0\}$$

$$W_2=\{(a_1,a_2,a_3)\in \mathbb{R}^3:a_1-4a_2-a_3=0\}$$

Describe $W_1\cap W_2$

My thoughts:

Clearly $W_1$ and $W_2$ are planes with normal vectors $(2,-7,1)$ and $(1,-4,-1)$, respectively.

It must be that their intersection is a line, but I don't know how to show this or express the equation of this line.

I tried setting their planar equations equal:

$$2a_1-7a_2+a_3=a_1-4a_2-a_3$$ $$ \Rightarrow a_1-3a_2+2a_3=0$$

But is this not just another plane, which obviously can't be correct? I think there is something simple that I am overlooking.

For reference, this question is from "Linear Algebra Second Edition" by Friedberg, Insel, Spence, Chapter 1 Section 3 Exercise 9

Any hints or guidance would be greatly appreciated.

Answer 1

You can parametrize the variables as follows:

$2a_1-7a_2+a_3-2(a_1-4a_2-a_3)=0 \rightarrow a_2=-3a_3$

Substituting this into any equation leads $a_1=-11a_3$, which shows that $(-11t, -3t, t)$ for all $t\in\mathbb{R}$ provides a complete solution for the system of equations for $W_1$ and $W_2$.

Actually, if you learn more you will know that for finite-dimensional vector space $W_1, W_2$, it holds that $\dim(W_1+W_2)=\dim(W_1)+\dim(W_2)-\dim(W_1\cap W_2)$ (It is in chapter 1 somewhere of the book). Here the sum of sets is defined as $S+T=\{s+t:s\in S, t\in T\}$. Since any element in $\mathbb{R}^3$ can be expressed as $w_1+w_2$ where $w_1 \in W_1$ and $w_2 \in W_2$, we have $\dim(W_1+W_2)=3$. Since $\dim(W_1)=\dim(W_2)=2$, we have $\dim(W_1 \cap W_2)=1$, which can be an evidence that $W_1 \cap W_2$ is a line that passes $(0,0,0)$.

Answer 2

You have two equations of plane and both passes through the origin which are $2a_1-7a_2+a_3=0$ and $a_1-4a_2-a_3=0$.Clearly they are not parallel and must intersect at a straight line that passes through the origin.Let $(a,b,c) $ be a point on that line.It has direction ratio$ (a-0,b-0,c-0)=(a,b,c)$.That line lies on the 1st plane and hence perpendicular to the normal of the plane.Note that the direction ratio of the normal is$ (2,-7,1)$.So $2a-7b+c=0$,similarly from second plane you get $a-4b-c=0$. Solving these two equations we get $(11,3,-1)×c $,c is scaler.This is the line that you desired.