Minimum distance between a point and a line in 3D

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Bill Foster 6 years, 5 months ago

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Newcastle University Mathematics and Statistics 9 years, 2 months ago

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Question statement

In $\mathbb{R}^3$ find the distance between the point $\boldsymbol{p} = \var{p}$ and the line through the point $\boldsymbol{x_0} = \var{x_0}$ that is parallel to the vector $\boldsymbol{v} = \var{v}$ .

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			Name	Type	Generated Value

			directions	list	List of 3 items
			p	vector	vector(0,5,-1)

			Name	Type	Generated Value

s1	integer	-1
a	integer	-4
b	integer	-3
g	integer	-7
x_0	vector	vector(-7,-5,-3)

			Name	Type	Generated Value

c	integer	-2
d	integer	5
f	integer	6
v	vector	vector(7,4,8)

			Name	Type	Generated Value

top

vector

vector(72,-42,-42)

			Name	Type	Generated Value

Automatically regenerate variables?

Name

Data type

Value

xxxxxxxxxx
 
map(id(3)[x],x,shuffle(0..2))

JME function reference

Description

Shuffled list of axis vectors

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

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Generated value: `list`

[ vector(0,0,1), vector(0,1,0), vector(1,0,0) ]

This variable doesn't seem to be used anywhere.

Parts

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Distance = Unnamed gap

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The line through $\boldsymbol{x_0} = \var{x_0}$ in the direction of $\boldsymbol{v}=\var{v}$ has equation:

$\boldsymbol{r} = \boldsymbol{x_0} + \lambda \boldsymbol{v} = \var{x_0} + \lambda \var{v}$

The minimum distance between this line and the point $\boldsymbol{p} = \var{p}$ is given by

$\frac{\left\lVert(\boldsymbol{p}-\boldsymbol{x_0}) \times \boldsymbol{v} \right\rVert}{\lVert \boldsymbol{v} \rVert}$

Now,

$\begin{align} \boldsymbol{p} - \boldsymbol{x_0} &= \var{p-x_0} \Rightarrow \\[1em] (\boldsymbol{p}-\boldsymbol{x_0}) \times \boldsymbol{v} &= \var{p-x_0} \times \var{v} \\[1em] &= \var{top} \end{align}$

Since

$\begin{align} \left\lVert \begin{matrix} \var{top[0]} \\ \var{top[1]} \\ \var{top[2]} \end{matrix} \right \rVert &= \simplify[]{sqrt({top[0]}^2 + {top[1]}^2 + {top[2]}^2)} \\ &= \sqrt{\var{abs(top)^2}} \end{align}$

and $\lVert \boldsymbol{v} \rVert = \simplify[]{sqrt({v[0]}^2 + {v[1]}^2 + {v[2]}^2)} = \sqrt{\var{abs(v)^2}}$ , the distance is then:

$\simplify{sqrt({abs(top)^2}/{abs(v)^2})}$

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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.
David's copy of Vectors by David Rickard in PHYS1010.
Eryk's copy of MATH6005 Vectors by Eryk Wilczy?ski in Eryk's workspace.
Maria's copy of Vectors by Maria Aneiros in Maria's workspace.
Vectors by Jeremie Wenger in Jeremie's workspace.
Newcastle U vector questions (3D points, lines and planes) by Jorj Kowszun in Jorj's workspace.
Nicolas's copy of Vectors by Nicolas Sawyer in Nicolas's workspace.
Maher's copy of Vectors by Maher Elbohouty in Vectors.
Peter's copy of Vectors QUIZ by Peter Keys in Peter's workspace.
All the good stuff by Chris Templet in Chris's workspace.

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Minimum distance between a point and a line in 3D

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Bill Foster 6 years, 5 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

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Generated value: `list`

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Minimum distance between a point and a line in 3D

Metadata

Taxonomy: mathcentre

Taxonomy: Kind of activity

Taxonomy: Context

Contributors

Feedback

History

Comment

Checkpoint description

Bill Foster 6 years, 5 months ago

Newcastle University Mathematics and Statistics 9 years, 2 months ago

Error in variable testing condition

Error

Warning

Generated value: list

Parts

Gap-fill

Generated value: `list`