Finding line where two planes intersect

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

However, I could not get that. By getting the system in REF, I got $x = 3 + \frac{3}{7}z$ and $y = \frac{1}{7}z$. Therefore z is a free variable so $x = 3 + \frac{3}{7}t = 0$ and $y = \frac{1}{7}t$.

Thus equation of line is $x\hat i + y\hat j + z\hat k = 3\hat i + (\frac{3}{7}\hat i + \frac{1}{7}\hat j)t$

Does anybody please know what I did wrong here?

Many thanks!

 

[anuttarasammyak]

Homework Statement: Please see below
Relevant Equations: Please see below

By getting the system in REF, I got x=3+37z=0 and y=17z.

You are right to get
$$x=3+\frac{3}{7}z$$
but why do you make it zero ?

 

[ChiralSuperfields]

You are right to get
$$x=3+\frac{3}{7}z$$
but why do you make it zero ?

Thank you for your reply @anuttarasammyak !

Sorry my mistake. That was not meant to be set to zero.

Many thanks!

 

[Mark44]

Homework Statement: Please see below
Relevant Equations: Please see below

For this problem,
View attachment 325740
The solution is,
View attachment 325741
However, I could not get that. By getting the system in REF, I got $x = 3 + \frac{3}{7}z$ and $y = \frac{1}{7}z$. Therefore z is a free variable so $x = 3 + \frac{3}{7}t = 0$ and $y = \frac{1}{7}t$.

I got the same thing but took it one step further, with z = t. So the equation of the line of intersection is $L = (\frac 3 7 t + 3)\hat i + \frac 1 7 t \hat j + t \hat k$.

In vector form, the above is $L = \begin{bmatrix} \frac 3 7 \\ \frac 1 7 \\ 1 \end{bmatrix}t + \begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix}$

Geometrically, to get points on this line, go out 3 units on the x-axis, and then go off in the direction of <3t/7, t/7, t>.

Thus equation of line is $x\hat i + y\hat j + z\hat k = 3\hat i + (\frac{3}{7}\hat i + \frac{1}{7}\hat j)t$

Your equation doesn't follow from your work that preceded it.