MATH 127
Vector Analysis
• Find the angle between two vectors where
→A=2ˆi+2ˆj and →B=6ˆi−3ˆj
θ=cos−1→A.→B|→A|.|→B|=cos−112−62√2.√45=71.56∘
• Important properties:
1. The area of the parallelogram is given, A = |→A×→B|
2. The volume of the parallelepiped is given, V = |→A.(→B×→C|)
3. Are of Triangles = 12|→A×→B|
4. If the three vectors a co-planar then (→A×→B).→C=0 or →A(→B×→C=0)
5. (→A×→B).→C=0 implies two of the vectors parallel.
6. (→A×→B).→C=→A.(→B×→C)=→B.(→C×→A) = same volume.
• Position vector:
OP =xˆi+yˆj+zˆkPQ =Q−P=(x1−x)ˆi+(y1−y)ˆj+(z1−z)ˆkOQ =x1ˆi+y1ˆj+z1ˆk
Example: A plane is defined by any three points that are in the plane. If a plane contains the point
P=(1,0,0),Q(1,1,1) and R(2,−1,3)
then find a vector that is orthogonal to the plane.
Solution: →PQ=ˆj+ˆk→PR=ˆi−ˆj+3ˆk|→PQ×→PR=|ˆiˆjˆk0111−13|
Example: Find the area of the parallelogram with vectors
P(2,2,0),Q(9,2,0),R(10,3,0) and S(3,3,0)
Solution: →QP=7ˆi→QR=ˆi+ˆj|→QP×→QR=|ˆiˆjˆk700110|=|7ˆk|=7sqr unit
Example: Find the area of the triangle with vectors
A(1,0,0),B(0,1,0),C(0,0,1)
.
Solution: △ABC=12|→AB×→AC=12|ˆiˆjˆk−110−101|=12|ˆi+ˆj+ˆk|=12√12+12+12=√32 sqr unit
Example: find the volume of the parallelepiped, determined by the vectors (1,1,1),(4,7,2),(3,2,1), and (5,4,3)
Example: Determine, whether the three vectors
→A=ˆi+4ˆj−7ˆk,→B=2ˆi−ˆj+4ˆk, and →C=−9ˆj+18ˆk
lie in the same plane or not.
Solution: →C(→A×→B)=|0−91814−72−14|=9(4+14)+18(−1−8)=0
∴ these vectors exists in the same plane.
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