Finding acute angle between lines/planes and between planes.

[chwala]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

a) [/B]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 textbook 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?

 

[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 ) ?

 

[chwala]

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

 

[BvU]

 

[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 $r(x,y,z)= (0,-2,8) + t(3,2,-7)$

 

[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 $\min (\alpha, \pi -\alpha)$ ?

 

[chwala]

Thanks a lot Bvu

 

[chwala]

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

 

[BvU]

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

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) !

I should add three further comments:

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...

 

[chwala]

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

 

[chwala]

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

 

[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?

 

[BvU]

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

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).

Kindly note that $(3,0,1)$ and $(0,-2,8)$ are both points that satisfy both the plane equations.

The two points can be expressed as a linear combination of each other.

No. Clearly $(3,0,1)$ is not a multiple of $(0,-2,8)$

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

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 ?

--

 

[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

 

[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

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).

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

 

[chwala]

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

 

[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

 

[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

 

[BvU]

With

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).
--

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).

 

[chwala]

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).

Thanks Bvu.....