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Let V,W be the points with coordinates (1,1,1),(0,2,3), respectively, let O be the origin of coordinates and let all measurements be in metres. Calculate the distance between the points V and W. (5 marks) (b) Let

V
and
W
be the vectors corresponding to the displacements OV and OW in part (a), respectively. Calculate the angle between
V
and
W
. (5 marks) (c) Find the area of the triangle O,V,W. (5 marks) (d) Find the coordinates of a point R that is 1 metre from the origin such that the vector
OR
is perpendicular to the surface of the triangle O,V,W. (5 marks) (e) If the Cartesian coordinate system in part (a) is rotated in an anticlockwise (i.e. positive direction) around the z-axis by 10

, calculate the new coordinates of point V. (5 marks) Question 3 is on the next page....

User Prdatur
by
8.2k points

1 Answer

6 votes

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)

__

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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Let V,W be the points with coordinates (1,1,1),(0,2,3), respectively, let O be the-example-1
Let V,W be the points with coordinates (1,1,1),(0,2,3), respectively, let O be the-example-2
User Justin Kramer
by
7.4k points

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