Question

Show that when →A+→B=→C, then C²=A²+B²+2ABcosφ, where φ is the angle between vectors →A and →B.

a) C² = A² + B² - 2ABcosφ
b) C² = A² + B² - ABcosφ
c) C² = A² + B² + 2ABcosφ
d) C² = A² + B² + ABcosφ

Answer 1

To prove C² = A² + B² + 2ABcosφ for vectors →A and →B, we use vector algebra and the definition of the dot product, ultimately showing that the correct representation of the equation is option c.

Step-by-step explanation:

To show that C² = A² + B² + 2ABcosφ when →A + →B = →C, we will use vector algebra and the definition of the dot product (scalar product).

We begin by considering the magnitude of the resultant vector →C. Since →C is the sum of →A and →B, its magnitude squared, C², can be obtained by taking the dot product of →C with itself:

→C · →C = (→A + →B) · (→A + →B) = →A · →A + →A · →B + →B · →A + →B · →B

Using the property that →A · →B = →B · →A (the dot product is commutative), this simplifies to:

→C · →C = A² + 2(→A · →B) + B²

The dot product →A · →B can be expressed as ABcosφ, where φ is the angle between the vectors →A and →B. Substituting this into our equation, we get:

C² = A² + 2ABcosφ + B²

Therefore, the correct option that represents this relation is:

c) C² = A² + B² + 2ABcosφ