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Must two collinear vectors be the same or opposite? What if one of them is a zero vector?
0
Given |vector a|=3,|vector b|=5, and |vector a-vector b|=√19. Find: product of vector a, vector b and vector a, vector b Hollycj1202:|Vector a-vector b |^2=9+25+2 vector a×vector b Is there a problem here? Why is the last face not "-2 vector a x vector b "but "+2 vector a x vector b"
|Vector a-vector b |=√19
So|vector a-vector b|^2=9+25-2vector a×vector b=19
Then vector a×vector b=15/2
Suppose that the included angle of vectors a and b is θ, then 15 cos θ=15/2
Cos θ=1/2
The angle of vectors a, b is 60 degrees
The product of vector a and vector b is 15/2
|Vector a-vector b |=√19
So|vector a-vector b|^2=9+25-2vector a×vector b=19
Then vector a×vector b=15/2
Let the angle between vectors a and b be θ, then 15 cos θ=15/2
Cos θ=1/2
The angle of vectors a, b is 60 degrees
The product of vector a and vector b is 15/2
|Vector a-vector b |=√19
So|vector a-vector b|^2=9+25-2vector a×vector b=19
Then vector a×vector b=15/2
Suppose that the angle of vectors a and b is θ, then 15 cos θ=15/2
Cos θ=1/2
The angle of vectors a, b is 60 degrees
The product of vector a and vector b is 15/2
Given |a|=3,|b|=5, and a*b=12, the projection of vector a on vector b is ()
A*b=|a b|*cosa=15cosa=12
The projection of vector a on vector b is |a|*cosa=12/15*3=12/5
A*b=a*cosa=15 cosa=12
The projection of vector A on vector B is |a cosa=12/15*3=12/5
Given that A, B, C, D are four points on the plane, find the value of vector AB + vector CB - vector CD + vector DA.
Sequence transposition
CB-CD+AB+DA=DB+DA+AB=DB+DB=2DB
Where CB-CD=DB
What is an equal vector? What is the zero vector? What is its direction? Thanks.
An equal vector is a vector with the same length (modulus) and the same direction;
The zero vector is a vector of "length 0";
The direction of the zero vector is "arbitrary" because it can be parallel to any vector in the same plane!
The plane equation passing through the point M (2,0.-1) and parallel to the vectors a=(2,1,1) and b=(3,0,4) is?
If the required plane is parallel to the two vectors, then the normal vector of the plane can be obtained, i.e. the normal vector and the two given vectors are perpendicular to each other, then the normal vector can be obtained, and then a generalized plane can be obtained, and then the M point can be brought in, and the specific steps are omitted.
If the required plane is parallel to the two vectors, then the normal vector of the plane can be obtained, i.e. the normal vector and the two given vectors are perpendicular to each other, and the normal vector can be obtained to obtain a generalized plane, and then the M point can be brought in.
If the required plane is parallel to the two vectors, then the normal vector of the plane can be obtained, i.e. the normal vector and the two given vectors are perpendicular to each other, and then the normal vector can be obtained, and then a generalized plane can be obtained, and then the M point can be brought in, and the specific steps are not written.
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