Answer:
a) √6 ≈ 2.449 meters
b) 36.8°
c) √3.5 ≈ 1.871 square meters
d) R(√14/14, -3√14/14, √14/7)
e) V' ≈ (1.158, 0.811, 1)
Explanation:
Given vectors V(1, 1, 1) and W(0, 2, 3), you want the values of various functions of these vectors.
a) Distance
The distance between V and W is |W-V| = |(0-1, 2-1, 3-1)| = √(1²+1²+2²) = √6
The distance between V and W is √6 ≈ 2.449 meters.
b) Angle
The angle between the vectors can be found several ways. Here, we choose to use the dot product:
V•W = |V|·|W|·cos(θ) . . . . . where θ is the angle between the vectors
θ = arccos(V•W/(|V|·|W|)) ≈ 36.8°
The angle between V and W is about 36.8°.
c) Area
The area enclosed by the triangle whose sides are V and W is given by half the magnitude of the cross product.
Area = 1/2|V×W| = 1/2|V|·|W|·sin(θ) = √14/2 ≈ 1.8708
The area of the triangle is √3.5 ≈ 1.8708 square meters.
d) Normal
The normal vector is given by the cross product. The point R that is one unit from the origin in that direction is the normalized cross product:
V×W = (1, -3, 2)
R = (1, -3, 2)/√14
Point R is (√14/14, -3√14/14, √14/7)
e) Rotated
Rotating the coordinate system CCW by 10° is the same as rotating the vector CW by 10°. Since we're rotating about the z-axis, the rotation affects only the x- and y-coordinates. The rotation matrix for the vector is shown in the second attachment. The new coordinates of point V are ...
V' ≈ (1.15845593068, 0.81115957535, 1)
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Additional comment
The angle between V and W can also be found using the law of cosines. That will tell you the angle is arccos(5/√39) ≈ 36.8°.
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