Chapter 1: Vectors, Lines and Planes
Section 1.3: Dot Product
Example 1.3.10
Find a vector of unit length that is orthogonal to the vectors A=i+j+k and B=2 i+3 j−k.
Solution
Mathematical Solution
Let the unknown vector be V=a i+b j+c k.
The condition that it be orthogonal to A is A·V=a+b+c=0.
The condition that it be orthogonal to B is B·V=2 a+3 b−c=0.
The condition that it have unit length is V·V=a2+b2+c2=1.
Solving these three equations for a,b,c leads to the vectors V=±1264−3−1.
Maple Solution - Interactive
Within the confines of the Student MultivariateCalculus package, the dot product is taken over the real numbers.
Tools≻Load Package: Student Multivariate Calculus
Loading Student:-MultivariateCalculus
Write the vector A as per Table 1.1.1.
Context Panel: Assign Name
A=1,1,1→assign
Write the vector B as per Table 1.1.1.
B=2,3,−1→assign
Write the vector V as per Table 1.1.1.
V=a,b,c→assign
Form and solve the three equations A·V=0,B·V=0,V2=1
Common Symbols palette: Dot-product operator
Press the enter key.
Context Panel: Solve≻Solve (explicit)
A·V=0,B·V=0,V·V=1
a+b+c=0,2a+3b−c=0,a2+b2+c2=1
→solve
a=−22613,b=32626,c=2626,a=22613,b=−32626,c=−2626
Transfer the values of a,b,c to the vector V
Expression palette: Evaluation template Reference the solutions via its equation label. Append [1] to obtain the first solution.
Expression palette: Evaluation template Reference the solutions via its equation label. Append [2] to obtain the second solution.
Vx=a|f(x)1 = −22613326262626
Vx=a|f(x)2 = 22613−32626−2626
Note that the two solutions for vector V are collinear, but point in opposite directions.
Maple Solution - Coded
Initialize
Define the vectors A, B, and V.
A,B,V≔1,1,1,2,3,−1,a,b,c:
Obtain and solve the three equations A·V=0,B·V=0,V2=1
S≔solveDotProductA,V,DotProductB,V,NormV2=1,a,b,c,explicit
S≔a=−22613,b=32626,c=2626,a=22613,b=−32626,c=−2626
Transfer the data in each of these two solutions to the generic vector V
evalV,S1
−22613326262626
evalV,S2
22613−32626−2626
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