- #1
newtomath
- 37
- 0
I am working on 2 problems and wanted to know your thoughts on them:
Problem 1
Given the vectors x = <3,2,-4>, y = <-3/2,1,-2>, and z = <0,2,1>, select all statements below that apply.
A.
the vectors x and y are orthogonal.
B.
the vectors x and y are in opposite directions.
C.
the vectors x and z are orthogonal
D.
the vectors x and z are in opposite directions
2 Vectors are orthogonal if their dot product is 0.
2 vectors are orthogonal is the angle between them is 180.
To determine the angle between 2 vectors we can use the equation Cos ⊖= The dot product of the 2 vectors/ Distance of vector1 * distance of vector 2
A.
the vectors x and y are orthogonal.
The dot product of x and y is ( 3* -3/2 , 2*1, -4*-2) = (-9/2 + 2 + 8) = 11/2. hence x and y are not orthogonal
B.
the vectors x and y are in opposite directions.
Cos ⊖= 11/2 / (sqrt ( 29)* (sqrt(29)/ 2)) = 11/2 / 29/2 = 11/29
Cos ⊖ = 11/29 = approx 67.71 degrees. x and y are not in opposite directions.
C.
the vectors x and z are orthogonal
The dot product of x and z is ( 3* 0 , 2*2 , -4*-1) = (0 + 4 - 4). x and z are orthogonal.
D.
the vectors x and z are in opposite directions
Cos ⊖= 0 = 90 degrees. This confirms our answer in part C that x and z are orthogonal and create a right angle. they are not in opposite directions.
Problem 2
Given the vector x = <3,2,-4>,
1. find the length of the vector x (write sqrt(#) for square root to write your answer exactly).
2. find the unit vector in the same direction as x. (Write your answer exactly using the form <v1,v2,v3,...> , rationalize all denominators)
1) The length of a vector is the square root of the sum of squares. The length is sqrt ( 3^2 + 2^2 +(-4)^2 = sqrt ( 29)
2) The unit vector in the same direction of x is given by x/ length of x. We have ( 3, 2, -4) / sqrt(29) from part A. So we have ( 3*sqrt(29)/ 29 , 2*sqrt(29), -4*sqrt(29)/ 29)
Problem 1
Given the vectors x = <3,2,-4>, y = <-3/2,1,-2>, and z = <0,2,1>, select all statements below that apply.
A.
the vectors x and y are orthogonal.
B.
the vectors x and y are in opposite directions.
C.
the vectors x and z are orthogonal
D.
the vectors x and z are in opposite directions
2 Vectors are orthogonal if their dot product is 0.
2 vectors are orthogonal is the angle between them is 180.
To determine the angle between 2 vectors we can use the equation Cos ⊖= The dot product of the 2 vectors/ Distance of vector1 * distance of vector 2
A.
the vectors x and y are orthogonal.
The dot product of x and y is ( 3* -3/2 , 2*1, -4*-2) = (-9/2 + 2 + 8) = 11/2. hence x and y are not orthogonal
B.
the vectors x and y are in opposite directions.
Cos ⊖= 11/2 / (sqrt ( 29)* (sqrt(29)/ 2)) = 11/2 / 29/2 = 11/29
Cos ⊖ = 11/29 = approx 67.71 degrees. x and y are not in opposite directions.
C.
the vectors x and z are orthogonal
The dot product of x and z is ( 3* 0 , 2*2 , -4*-1) = (0 + 4 - 4). x and z are orthogonal.
D.
the vectors x and z are in opposite directions
Cos ⊖= 0 = 90 degrees. This confirms our answer in part C that x and z are orthogonal and create a right angle. they are not in opposite directions.
Problem 2
Given the vector x = <3,2,-4>,
1. find the length of the vector x (write sqrt(#) for square root to write your answer exactly).
2. find the unit vector in the same direction as x. (Write your answer exactly using the form <v1,v2,v3,...> , rationalize all denominators)
1) The length of a vector is the square root of the sum of squares. The length is sqrt ( 3^2 + 2^2 +(-4)^2 = sqrt ( 29)
2) The unit vector in the same direction of x is given by x/ length of x. We have ( 3, 2, -4) / sqrt(29) from part A. So we have ( 3*sqrt(29)/ 29 , 2*sqrt(29), -4*sqrt(29)/ 29)