# Finding acute angle between lines/planes and between planes.

1. Mar 29, 2016

### chwala

1. The problem statement, all variables and given/known data
a) Find the acute angle between the line $r = i+3j+5k+t (2i+4j+k)$ and the plane x-y+z=0
b)Two planes are defined by the equations $x+2y+z=4$ and $2x-3y=6$
i. find the acute angle between them.
ii. find a vector equation of their line of intersection.

2. Relevant equations

3. The attempt at a solution
a)
I considered the directional vector $p= 2i+4j+k$ and the normal to plane $q= i-j+k$
$p.q=|p||q| cos θ$ using this i got $θ=97.2^0$degrees. The text book answer is given as $θ-0.5π= 7.2$degrees
b)
i.
I let $m=i+2j+k$ and $k=2i-3j$ using $m.k=|m||k| cos Φ$ i got $θ=116.9$ degrees. The textbook answer is $63.1$ degrees, now my question is can we conclude that to get an acute angle α between a line and a plane we can use $α=θ-0.5π$ and to get an acute angle α between two planes then we use $α=180-θ$ is that correct?

2. Mar 29, 2016

### BvU

a) it says it wants the acute angle between the plane and the line. Make a drawing !
b i) what is the relationship between your two ponts and the planes ?
b ii ) ?

3. Mar 29, 2016

### chwala

I am unable to make this drawing in 3d kindly assist..

4. Mar 29, 2016

### BvU

5. Mar 29, 2016

### chwala

bii)
$x+2y+z=4$ , $2x-3y=6$ on solving we get $(x,y,z)= (3,0,1)$ also another solution is $(x,y,z)= (0,-2,8)$ thus expressing this in the form of $r$ = $a + tp$ we can have $(3,0,1)=$(0,-2,8)$+$t(m,n,p)$, where$(m,n,p)= (3,-2,7) i.e for $t=1$ thus in form of a vector equation, we now have $r(x,y,z)= (0,-2,8) + t(3,2,-7)$

6. Mar 29, 2016

### BvU

Agree. Well done.
Re a): is it clear your answer was 90 degree off because you calculated the angle between the normal and the line ?
So that in both cases you are allowed to use $\min (\alpha, \pi -\alpha)$ ?

7. Mar 29, 2016

### chwala

Thanks a lot Bvu

8. Mar 29, 2016

### chwala

can you also draw the two planes in part 1b) above please..........

9. Mar 29, 2016

### BvU

That's a bit more awkward...A sensible representation of the first plane is a bit beyond my visio capabilities. Best I can offer is the second plane: a vertical (in the z direction) plane through (3,0,0) and (0,-2,0) (blueish) and the line (magenta) intersecting the two planes. Plus the intersection of the z=0 plane and the first plane: $x + 2y = 4$ (green). Picture doesn't show the angle between the two planes, though (blue and green line are not perpendicular) !

For the two planes, you calculated the angle between the two normal vector directions. Since the angle between those two vectors is equal to the angle between the two planes, the (acute) angle = $\min (\alpha, \pi -\alpha)$ is valid.

For the intersection between $x+2y+z=4$ and $2x−3y=6$ you used the words "on solving we get". What you did was that you found two points for both of which the two equations are satisfied. To me, 'solving' $\ x+2y+z=4 \ \& \ 2x−3y=6 \$ means you have to find all such points .

It isn't wise to use the same symbol for two different things: the choice of k in $k=2i-3j$ is confusing...

#### Attached Files:

• ###### planes1.jpg
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10. Mar 29, 2016

### chwala

This looks good, you must have a good computer to do this..........

11. Mar 29, 2016

### chwala

I agree i should have used a different character for $k$

12. Mar 31, 2016

### chwala

Can we also say that $r(x,y,z)= (3,0,1) + t(-3,-2,7)$ is a Vector equation for the two planes?

Kindly note that $(3,0,1)$ and $(0,-2,8)$ are both points that satisfy the plane equations. The two points can be expressed as a linear combination of each other.

How would these planes look like in a 3D plane?

13. Mar 31, 2016

### BvU

No. It is a vector equation for the intersection of the two planes. The two planes each have a separate equation ($\ x+2y+z=4 \$ and $\ 2x−3y=6 \$). Points that satisfy $r(x,y,z)= (3,0,1) + t(-3,-2,7)$ are the points that satisfy $\ x+2y+z=4 \ \& \ 2x−3y=6 \$ (i.e. both equations).

No. Clearly $(3,0,1)$ is not a multiple of $(0,-2,8)$
See post #9. One is the blueish plane. The other is the plane that contains both the green and the magenta line.

A further comment (on second thoughts, more like a challenge):
You found $(3,0,1)$ by filling in $y = 0 \Rightarrow x = 3$ (from the second plane equation) and picking a point on the line $x+z =4$ (from the first plane equation). And you found $(0,-2,8)$ likewise with $y = 0$. Then you worked out the equation for a line through both points. Well done.

But there is a more direct way to solve $\ x+2y+z=4 \ \ \& \ \ 2x−3y=6 \$ and end up with the equation for a straight line (the same line, of course). Do you see what I mean ?

--

14. Mar 31, 2016

### chwala

If $r=a +tp$ is a vector equation, then
$(3,0,1)= (0,-2,8) + 1(3,2,-7)$ also

$(0,-2,8) = (3,0,1) + 1(-3,-2,7)$ where $t=1$
remember t is just a scalar quantity

15. Mar 31, 2016

### BvU

What you write down is $\vec a = \vec b + (\vec a - \vec b)$ and $\vec b = \vec a + (\vec b - \vec a)$ which isn't extremely shocking. It does not mean that $\vec a$ and $\vec b$ are linear combinations of each other.

With
I invite you to solve directly for the line that lies in both planes.

16. Mar 31, 2016

### chwala

Ok Bvu i can see you are conversant with this...many people don't like to work on vectors in 3D

17. Mar 31, 2016

### chwala

what i meant by linear combination is that they can be expressed in an equation , i am using the language from linear algebra, specifically on the definition of a basis of a vector. A basis is basically a span of vectors in space V for e.g in a field of real numbers, that are also linearly indepedent. sorry for confusion

18. Apr 5, 2016

### chwala

From my research there are many approaches in finding a vector equation of a line from a set of plane equations in 3D, the more obvious method to me is to get points that satisfy the equations for instance if the points satisfying a set of equations is $(a,b,c)$ and $(m,n,q)$ then it follows that the vector equation is $r= ai+bj+ck + η((a-m)i+(b-n)j+(c-q)k)$ this can also be expressed as $r= mi+nj+qk+((m-a)i+(n-b)j+(q-c)k$ where η is a parameter. Note that in this form of $r=a +ηp, a$is the position vector and p is the directional vector

19. Apr 6, 2016

### BvU

With
I mean: $$x={3\over 2}y + 3 \ \ \& \ \ z = -{ 3\over 2}y - 3 - 2y + 4 \ \ \Leftrightarrow \\ x={3\over 2}y + 3 \ \ \& \ \ z = -{ 7\over 2}y + 1$$ so that $\left ( {3\over 2}y + 3, y, -{ 7\over 2}y + 1 \right )$, i.e. $(3,0,1) + y({3\over 2}, 1, -{ 7\over 2})$ is a line in both planes (it satisfies both equations for any value of the parameter y).

20. Apr 6, 2016

### chwala

Thanks Bvu..............