Final answer:
To compute the magnitude of the vector subtraction e−f using the dot product, we can use the formula ∥e−f∥ = √(∥e∥² + ∥f∥² - 2(e · f)). Given the value of ∥e+f∥ = 8/7, we can solve for (e · f) and substitute it back into the formula to find ∥e−f∥, which simplifies to 8/7.
Step-by-step explanation:
To compute the magnitude of the vector subtraction e−f using the dot product, we can use the following formula: ∥e−f∥ = √(∥e∥² + ∥f∥² - 2(e · f)). We can substitute the given value of ∥e+f∥ = 8/7 into this formula and solve for ∥e−f∥. Since e and f are unit vectors, their magnitudes are 1, so the formula simplifies to ∥e−f∥ = √(1 + 1 - 2(e · f)). Using the given value of ∥e+f∥ = 8/7, we can solve for (e · f) and substitute it back into the formula to find ∥e−f∥.
Given: ∥e+f∥ = 8/7
∥e−f∥ = √(1 + 1 - 2(e · f))
Substituting the value of ∥e+f∥:
8/7 = √(1 + 1 - 2(e · f))
Squaring both sides:
(8/7)² = 1 + 1 - 2(e · f)
64/49 = 2 - 2(e · f)
(e · f) = (2 - 64/49)/2
(e · f) = (98/49 - 64/49)/2
(e · f) = 34/49/2
(e · f) = 34/98
Substituting back into the formula:
∥e−f∥ = √(1 + 1 - 2(34/98))
∥e−f∥ = √(1 + 1 - (68/98))
∥e−f∥ = √(1 + 1 - (34/49))
∥e−f∥ = √(49/49 + 49/49 - 34/49)
∥e−f∥ = √(64/49)
∥e−f∥ = 8/7