1.A given line is described by the system of equations x=2t+7; y=-3t+4; z=5-t; find the equation of a line parallel to the given line and which contains the point (1,1,0).
2.Let a=<0,6,2> and b=<0,2,4>, then the component of a parallel to b is given by?
- the answer is <0,2,4> but I do not understand why
Also, I do not understand how to add magnitudes.
Please help.
Let u and w be vectors in R3 such that ||u||=3 and ||w||=2 and the angle between the vectors is pi/2; what is the value of ||3u-4w||?
- answer is sqrt(145)
1. To find the equation of a line parallel to the given line and passing through the point (1, 1, 0), we need to determine the direction of the line first. Since we want it to be parallel to the given line, it will have the same direction vector.
The direction vector of the given line can be obtained by taking the coefficients of t from the three equations: (2, -3, -1). This direction vector represents the direction and magnitude of the line.
Now, we know that any parallel line to the given line will have the same direction vector. So, the equation of the parallel line will have the form:
x = 2t + a (replacing 'a' with the x-coordinate of the given point)
y = -3t + b (replacing 'b' with the y-coordinate of the given point)
z = -t + c (replacing 'c' with the z-coordinate of the given point)
Substituting the coordinates of the given point (1, 1, 0) into the equations, we get:
x = 2t + 1
y = -3t + 1
z = -t + 0
Therefore, the equation of the line parallel to the given line and passing through (1, 1, 0) is:
x = 2t + 1
y = -3t + 1
z = -t
2. To find the component of vector a parallel to vector b, we need to project vector a onto vector b. The formula for projecting one vector onto another is given by:
Projection of a onto b = (a · b) / |b|
Where "a · b" represents the dot product of vectors a and b, and "|b|" represents the magnitude (or length) of vector b.
Given the vectors a = <0, 6, 2> and b = <0, 2, 4>, we can find the component of a parallel to b using the projection formula:
Projection of a onto b = (a · b) / |b|
= (<0, 6, 2> · <0, 2, 4>) / |<0, 2, 4>|
Calculating the dot product:
(0 * 0) + (6 * 2) + (2 * 4) = 0 + 12 + 8 = 20
Calculating the magnitude of b:
|<0, 2, 4>| = sqrt(0^2 + 2^2 + 4^2) = sqrt(0 + 4 + 16) = sqrt(20) = 2sqrt(5)
Therefore, the component of vector a parallel to vector b is given by:
(20 / (2sqrt(5))) * b = (10 / sqrt(5)) * <0, 2, 4> = <0, 2sqrt(5), 4sqrt(5)>