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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.

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  • $\begingroup$ Hint: the direction vector of the intersection line is given by the cross product (or vector product) of the two normal vectors. $\endgroup$ – Gribouillis Aug 24 '17 at 17:47
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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 $(0,0,0)$.

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  • $\begingroup$ I like and appreciate this answer because it does not involve any vector algebra involving cross products and such, because those things are not mentioned anywhere before this section and the only preliminary in the preface is noted as Calculus. So I think one should be able to solve this question without involving the cross product as you have shown. I'd like to ask why it is acceptable to subtract twice the plane equation from the first, because if $P_1=0$ and $P_2=0$ then $P_1-P_2=0$ and not $P_1-2P_2=0$? $\endgroup$ – Hushus46 Aug 24 '17 at 18:30
  • $\begingroup$ Nevermind, I understand why it works. Its solving a system of equations. $\endgroup$ – Hushus46 Aug 24 '17 at 18:42
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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.

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