(au) × v + (bu) × v = [(a + b)u] × v
u = ( x , y , z )
v = ( x₂, y₂,z₂)
LHS:
= (au) × v + (bu) × v
= [ ay z₂- y₂az , - (axz₂- x₂az) , axy₂- x₂ay ] + [ by z₂- y₂bz , - (bxz₂- x₂bz) , bxy₂- x₂by ]
RHS:
= [(a + b)u] × v
= [ (a + b)x , (a + b)y , (a + b)z ] × ( x₂, y₂,z₂)
= [ (a + b)y z₂- y₂(a + b)z , -((a + b)x z₂- x₂(a + b)z) , (a + b)x y₂- x₂(a + b)y]
This is how far I got to prove left side equals right side. . .now I don't know what to do.
To prove that the left-hand side (LHS) equals the right-hand side (RHS) of the equation, we need to simplify and compare the expressions.
Let's expand the LHS expression first:
LHS = [ ay z₂ - y₂az , - (axz₂ - x₂az) , axy₂ - x₂ay ] + [ by z₂ - y₂bz , - (bxz₂ - x₂bz) , bxy₂ - x₂by ]
Simplifying this further, we can combine like terms:
LHS = [ (ay z₂ - y₂az) + (by z₂ - y₂bz) , - ((axz₂ - x₂az) + (bxz₂ - x₂bz)) , (axy₂ - x₂ay) + (bxy₂ - x₂by) ]
= [ (ay z₂ - y₂az + by z₂ - y₂bz) , - (axz₂ - x₂az + bxz₂ - x₂bz) , (axy₂ - x₂ay + bxy₂ - x₂by) ]
= [ (ay + by)z₂ - y₂(a + b)z , - (ax + bx)z₂ + x₂(a + b)z , (ax + bx)y₂ - x₂(a + b)y ]
Now let's compare this to the RHS expression:
RHS = [ (a + b)y z₂ - y₂(a + b)z , -((a + b)x z₂ - x₂(a + b)z) , (a + b)x y₂ - x₂(a + b)y ]
The expressions on both sides are now in a similar form. To prove the equality, we'll show that each component in the LHS is equal to the corresponding component in the RHS:
1. Comparing the x-components:
(ay + by)z₂ - y₂(a + b)z = (a + b)y z₂ - y₂(a + b)z
Distributing and simplifying:
ayz₂ + byz₂ - y₂az - y₂bz = ayz₂ + byz₂ - y₂az - y₂bz
Both sides are equal.
2. Comparing the y-components:
- (ax + bx)z₂ + x₂(a + b)z = -((a + b)x z₂ - x₂(a + b)z)
Distributing and simplifying:
-axz₂ - bxz₂ + x₂az + x₂bz = -axz₂ - bxz₂ + x₂az + x₂bz
Both sides are equal.
3. Comparing the z-components:
(ax + bx)y₂ - x₂(a + b)y = (a + b)x y₂ - x₂(a + b)y
Distributing and simplifying:
axy₂ + bxy₂ - x₂ay - x₂by = axy₂ + bxy₂ - x₂ay - x₂by
Both sides are equal.
Since each component of the LHS is equal to the corresponding component of the RHS, we can conclude that the LHS equals the RHS.
Therefore, the equation (au) × v + (bu) × v = [(a + b)u] × v holds true for vector u = (x, y, z) and vector v = (x₂, y₂, z₂).