Why vector AB+vector BA=0 is wrong Equal to 0 is not vector 0 If it is equal to vector 0 but not 0, give a reason. Is it 0 without direction? I'm just not sure. Why is it in the book? Vector PB+vector OP-vector OB=0 is correct Instead of writing the 0 vector (i.e. no arrow above) Why vector AB+vector BA=0 is wrong Equal to 0 is not vector 0 If it is equal to vector 0, not 0, give a reason,0, no direction I'm just not sure. Why is it in the book? Vector PB+vector OP-vector OB=0 is correct Instead of writing the 0 vector (i.e. no arrow above)
This zero is not zero
0 Is only a real number, but vector 0 is a vector with direction and size
You know the vector! Vector AB+vector BA=vector 0
This vector 0 represents an arbitrary size of 0 in its direction
Now you understand!
This zero is not zero
0 Is only a real number, but vector 0 is a vector with direction and size
You know the vector! Vector AB + Vector BA = Vector 0
This vector 0 represents an arbitrary size of 0 in its direction
Now you understand!
Given that in a triangle ABC, AB=2, AC=1, if D is the midpoint of the BC, what is the AD vector point multiplied by the BC vector?
AD vector =(AB vector + AC vector)/2
BC vector = AC vector - AB vector
AD vector * BC vector =(AB vector + AC vector)*(AC vector - AB vector)/2
=(AC vector * AC vector - AB vector * AB vector)/2
=(|AC|^2-||AB|^2/2
=(1-4)/2
=-3/2
A =60 degrees in triangle ABC A B =2 AC =3 D is the midpoint of BC What is the vector AD*AB equal to In triangle ABC, A=60 degrees A B=2 AC=3 D is the midpoint of BC, then what is the vector AD*AB equal to
Vector AD = Vector AB + Vector BD
Vector AD = Vector AC + Vector CD
Because vector BD+vector CD=0 vector
So 2 vector AD = vector AB + vector AC
Vector AD* Vector AB
=[(Vector AB + vector AC)- Vector AB ]/2
=(Vector AB.vector AB + vector AB.vector AC)/2
=(2*2+2*3*Cos60°)/2
=(4+3)/2
=7/2
D is the midpoint on edge AB of triangle ABC, then vector CD = A-BC+1/2BA B-BC-1/BA CBC-1/2BA DBC +1/2BA
Solution has a graph
Vector CD = Vector CB + Vector BD
=- Vector BC +1/2 vector BA
Therefore, A.
In triangle ABC, point D is the midpoint of AB and satisfies |vector CD|=1/2|vector AB|, then vector CA·CB=? The title is written here. In triangle ABC, point D is the midpoint of AB, and satisfies |vector CD|=1/2|vector AB|, then vector CA·CB=? The title is written here.
This is very simple... You didn't think of that for a while...
The centre line of the hypotenuse is half the hypotenuse, this is a right triangle, the angle C is 90°,
Then vector CA·CB equals 0.
It looks like there's no condition.
This is very simple... You didn't think of that for a while...
The centre line of the hypotenuse is half the hypotenuse, this is a right triangle, the angle C is 90°,
Then vector CA·CB equals zero.
It looks like there's no condition.
As shown in the figure, it is known that angle B + angle D + angle BED=360, verify AB parallel CD As shown in the figure, known: angle B + angle D + angle BED=360, verify AB parallel CD
EF//AB through E
∥AB
BEF°∠B=180
D°∠FED=180
EF∥CD
CD//AB