Find the angle between planes

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Bill Foster

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Bill Foster 6 years, 4 months ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

Calculate the angle $\alpha,\;0\leq\alpha \leq \frac{\pi}{2}$ , between the plane $\Pi_1$ , passing through the points

$\begin{align} \boldsymbol{r_1} &= \var{r1}, & \boldsymbol{r_2} &= \var{r2}, & \boldsymbol{r_3} &= \var{r3} \end{align}$

and the plane, $\Pi_2$ , whose equation is

$\simplify[std]{{a1}x+{b1}y+{c1}z={d1}}$

Give any introductory information the student needs.

			Name	Type	Generated Value

tn1	vector	vector(-3,8,-4)
r1	vector	vector(6,-5,-8)
r2	vector	vector(2,-3,-1)
r3	vector	vector(6,-4,-6)
a1	integer	2
b1	integer	3
tol	number	0.001
alpha	number	0.9222985256
ans	number	0.9222985256
n1	vector	vector(-3,8,-4)
c1	integer	-8
tr3	vector	vector(6,-4,-6)
d1	integer	-3

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
cross(r2-r1,tr3-r1)

JME function reference

Description

Describe what this variable represents, and list any assumptions made about its value.

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Generated value: `vector`

vector(-3,8,-4)

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This variable doesn't seem to be used anywhere.

Parts

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1. Gap Number entry
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$\alpha =$ Unnamed gap radians

(Enter your answer in radians, to 3 decimal places)

Advice

The angle between two planes is given by the angle between their normals.

The plane $\Pi_1$ can be written in the form

$\boldsymbol{r} = \boldsymbol{r}_1+\lambda( \boldsymbol{r}_2 - \boldsymbol{r}_1)+\mu( \boldsymbol{r}_3- \boldsymbol{r}_1)$

and the normal $\boldsymbol{n}_1$ to this plane is given by:

$\boldsymbol{n}_1 = (\boldsymbol{r}_2 - \boldsymbol{r}_1)\times (\boldsymbol{r}_3- \boldsymbol{r}_1)$

For $\boldsymbol{r}_1$ , $\boldsymbol{r}_2$ , $\boldsymbol{r}_3$ as given.

$\boldsymbol{n}_1 = \var{n1}$

The normal to the plane $\Pi_2$ is given by

$\boldsymbol{n}_2 = \var{vector(a1,b1,c1)}$

The angle between the two normals (and hence the two planes) can be found using:

$\cos(\alpha) = \frac{\boldsymbol{n}_1 \cdot \boldsymbol{n}_2}{\lVert\boldsymbol{n}_1\rVert \lVert\boldsymbol{n}_2\rVert}$

On calculating this, we obtain

$\begin{align} \boldsymbol{n}_1 \boldsymbol{\cdot} \boldsymbol{n}_2 &= \var{a1*n1[0]+b1*n1[1]+c1*n1[2]} \\ \lVert\boldsymbol{n}_1\rVert &= \simplify[std]{sqrt({n1[0]^2+n1[1]^2+n1[2]^2})} \\ \lVert\boldsymbol{n}_2\rVert &= \simplify[std]{sqrt({a1^2+b1^2+c1^2})}\\ \cos(\alpha) &= \simplify[std]{{(a1*n1[0]+b1*n1[1]+c1*n1[2])/sqrt((n1[0]^2+n1[1]^2+n1[2]^2)*(a1^2+b1^2+c1^2))}} \end{align}$

Now calculate $\arccos(\alpha) = \var{precround(alpha,3)}$ . The angle returned by your calculator will give a value between $0$ and $\pi$ . If it's bigger than $\frac{\pi}{2}$ , subtract the calculated value from $\pi$ to obtain an acute angle. So the angle between the two planes is

$\alpha' = \pi - \var{precround(alpha,3)} = \var{precround(ans,3)} \text{ radians}$

$\alpha = \var{precround(ans,3)} \text{ radians}$

to 3 decimal places.

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This question is used in the following exams:

Vectors by Newcastle University Mathematics and Statistics in Content created by Newcastle University.
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All the good stuff by Chris Templet in Chris's workspace.

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Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Find the angle between planes

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 4 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: vector

Parts

Gap-fill

Generated value: `vector`