Prove that for any vector A and B the following is true
llAxBll^2 +lA·Bl^2 =llAll^2 llBll^2
A×B = |A|*|B|*sinθ
A•B = |A|*|B|*cosθ
now, what is your most basic trig identity?
vector identity :
|A×B| = |A|*|B|*sinθ
To prove the given statement, we will use vector properties and algebraic manipulation.
First, let's recall the properties of the vector cross product (A x B):
1. The magnitude of the cross product is given by: |A x B| = |A| * |B| * sin(θ), where θ is the angle between vectors A and B.
2. The cross product is perpendicular to both A and B.
Next, we'll start by expressing the left-hand side (LHS) of the equation:
LHS = |A x B|^2 + |A · B|^2
Now, let's expand each term of the LHS separately:
1. |A x B|^2:
Using the properties of the cross product, we can rewrite this as:
|A x B|^2 = (|A| * |B| * sin(θ))^2 = |A|^2 * |B|^2 * sin^2(θ)
2. |A · B|^2:
Using the dot product property, we have:
|A · B|^2 = (|A| * |B| * cos(θ))^2 = |A|^2 * |B|^2 * cos^2(θ)
Now, let's rewrite the equation with the expanded terms:
LHS = |A|^2 * |B|^2 * sin^2(θ) + |A|^2 * |B|^2 * cos^2(θ)
Taking the common factor of |A|^2 * |B|^2, we have:
LHS = |A|^2 * |B|^2 * (sin^2(θ) + cos^2(θ))
By the trigonometric identity sin^2(θ) + cos^2(θ) = 1, we can simplify this further:
LHS = |A|^2 * |B|^2 * 1
Simplifying, we get:
LHS = |A|^2 * |B|^2
Notice that the right-hand side (RHS) of the equation is |A|^2 * |B|^2, which matches the result we obtained for the LHS. Therefore, we have proved that:
|A x B|^2 + |A · B|^2 = |A|^2 * |B|^2
Hence, we have proven that for any vectors A and B, the given equation holds true.