Chapter 1: Vectors, Lines and Planes
Section 1.6: Lines
Find the distance from the point P:2,1,−3 to the line R=A+t V, where A=i−j+2 k and V=7 i−9 j+5 k.
Figure 1.6.4(a) shows the line R, the direction vector V (in red), the points A and P, the vector U=P−A (in black), and the two components of the projection of U onto V, one in green (UV) and one in blue (U⊥V).
The distance from point P to the line R is
U⊥V = U−UV = U−U·VV·V V
where U=P−A=21−3−1−12 = 12−5.
Figure 1.6.4(a) Projecting U=P−A onto line
U⊥V = U−U·VV·V V = 12−5−−361557−95 = 407155−14155−11931
and d=U−U·VV·V V = 1155⁢519870 = 1155⁢3354⁢155 ≐ 4.65.
Alternatively, Table 1.5.1 lists a formula for finding the distance from point P to the line through points Q and R. This formula is not immediately applicable, but if a second point is found on the line, then the requisite distance is given by A×B / A, where A=R−Q and B=P−Q. Taking t=0 and t=1 on the given line provides the two points Q:1,−1,2 and R:8,−10,7, respectively. Hence,
A=R−Q=8−107−1−12=7−95 and B=P−Q=21−3−1−12=12−5
and A×B= |ijk7−9512−5|=354023, so d=A×B∥A∥ = 1225+1600+52949+81+25 = 519870155
Maple Solution - Interactive
Tools≻Load Package: Student Multivariate Calculus
Define the line R=A+t V
Write the sequence of names A, V.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Line
Context Panel: Assign to a Name≻R
1,−1,2,7,−9,5→make lineStudent:-MultivariateCalculus:-Line⁡1,−1,2,7−95,variables=x,y,z,parameter=t,id=1→assign to a nameR
Obtain the distance from point P to line R
Write the sequence of point P and the name R.
Context Panel: Student Multivariate Calculus≻Lines & Planes≻Distance
Context Panel: Combine≻radical
Context Panel: Approximate≻5 (digits)
2,1,−3,R→distance3354⁢155155= combine 519870155→at 5 digits4.6517
The traditional approach to this calculation is vector-based. The distance is the magnitude of the orthogonal component of a vector U, to point P from an arbitrary point on the line, projected onto V. The arbitrary point on the line is generally taken as A.
Implement a vector-based solution
Define the position vector A.
Context Panel: Assign to a Name≻A
1,−1,2→assign to a nameA
Define the position vector P.
Context Panel: Assign to a Name≻P
2,1,−3→assign to a nameP
Define the direction vector V.
Context Panel: Assign to a Name≻V
7,−9,5→assign to a nameV
Context Panel: Assign Name
Obtain the norm of the component of U orthogonal to V in the plane determined by U and V
Keyboard the norm bars.
Common Symbols palette: Dot product operator
Context Panel: Evaluate and Display Inline
U−U·VV·V V = 519870155
task template provides the vector U⊥V , the component of U orthogonal to V.
Linear Algebra≻Visualizations≻Projection Plot onto 1-D
Projection onto 1-D Subspace
Vector to be projected
The distance is then the magnitude of the orthogonal component.
Extract the orthogonal component via copy/paste.
Context Panel: Student Multivariate Calculus≻Norm
Note that the ProjectionPlot command upon which the task template is based, really does select an arbitrary point on the line. Hence, the figure appearing in the task template differs slightly from Figure 1.6.4(a) where the "arbitrary" point is taken as point A.
Maple Solution - Coded
A solution with the "Lines & Planes" tools in the Student MultivariateCalculus package:
Install the Student MultivariateCalculus package.
Apply the Line command to define the line R.
Apply the Distance command.
d≔Distance2,1,−3,R = 1155⁢3354⁢155
Apply the evalf command.
evalfd,5 = 4.6518
A traditional vector-based solution:
Define the position vectors A and P.
Obtain the vector U=P−A.
Apply the Norm and DotProduct commands to obtain the norm of U⊥V=U−U·VV·V V
NormU−DotProductU,VDotProductV,V V = 519870155
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