Can someone show me, step by step, how to prove
v*v = ||v||^2
v*v = ||v||^2
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Suppose v=(a, b, c)
The dot product of v by v would be as follows v*v = a*a+b*b+c*c
the magnitude of v, or ||v|| is √(a²+b²+c²)
it follows that ||v||² would then be (a²+b²+c²) which equals a*a+b*b+c*c
therefore v*v=||v||²
The dot product of v by v would be as follows v*v = a*a+b*b+c*c
the magnitude of v, or ||v|| is √(a²+b²+c²)
it follows that ||v||² would then be (a²+b²+c²) which equals a*a+b*b+c*c
therefore v*v=||v||²
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Let v = xi + yj + zk where i, j and k are the unit vectors in the x, y and z directions.
v * v = (xi + yj + zk) * (xi + yi + zk)
= xi^2 + xi*yj + xi * zk
+ yj * xi + yj^2 + yj * zk
+ zk * xi + zk * yj + zk^2
and all the cross terms are 0 because i and j are perpendicular, i and k are perpendicular, j and k are perpendicular. So
v * v = xi^2 + yj^2 + zk^2
v * v = (xi + yj + zk) * (xi + yi + zk)
= xi^2 + xi*yj + xi * zk
+ yj * xi + yj^2 + yj * zk
+ zk * xi + zk * yj + zk^2
and all the cross terms are 0 because i and j are perpendicular, i and k are perpendicular, j and k are perpendicular. So
v * v = xi^2 + yj^2 + zk^2
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I'll prove this in the case that v is in R3; this proof can be trivially generalized.
Suppose that v =. Dotting v with itself gives:
v · v = ·
= (v₁)(v₁) + (v₂)(v₂) + (v₂)(v₂), by the definition of the dot product
= (v₁)^2 + (v₂)^2 + (v₃)^2.
On the other hand:
||v|| = ||||
= √[(v₁)^2 + (v₂)^2 + (v₃)^2]
==> ||v||^2 = (v₁)^2 + (v₂)^2 + (v₃)^2.
Thus in R3, v · v = ||v||^2. This proof can be easily generalized to all of Rn.
I hope this helps!
Suppose that v =
v · v =
= (v₁)(v₁) + (v₂)(v₂) + (v₂)(v₂), by the definition of the dot product
= (v₁)^2 + (v₂)^2 + (v₃)^2.
On the other hand:
||v|| = ||
= √[(v₁)^2 + (v₂)^2 + (v₃)^2]
==> ||v||^2 = (v₁)^2 + (v₂)^2 + (v₃)^2.
Thus in R3, v · v = ||v||^2. This proof can be easily generalized to all of Rn.
I hope this helps!