Question

If a and b are two vectors such that I$\vec {a}$I + I$\vec {b}$I = $\sqrt 2$ with $\vec {a}$.$\vec {b}$ = –1, then the angle between $\vec {a}$ and $\vec {b}$ is

• A. $\frac {2π}{3}$
• B. $\frac {5π}{6}$
• C. $\frac {5π}{9}$
• D. $\frac {3π}{4}$

Answer 1

Answer: (A)

$\frac {2π}{3}$

Answer 2

To find the angle between vectors $\vec {a}$ and $\vec {b}$, we can use the dot product formula:
$\vec {a}$·$\vec {b}$ = |$\vec {a}$| |$\vec {b}$| cosθ
Given $\vec {a}$ · $\vec {b}$ = -1 and |$\vec {a}$| + |$\vec {b}$| = $\sqrt {2}$
Now by substituting
-1 = $\sqrt {2}$ $\sqrt {2}$ cosθ
-1 = 2 cosθ
cosθ = -$\frac {1}{2}$
To find the angle theta, we need to determine the inverse cosine (arccos) of -$\frac {1}{2}$. The angle theta will have two possible values, as cosine is negative in the second and third quadrants.
θ = 2π/3
Hence the angle between $\vec {a}$ and $\vec {b}$ is $\frac {2π}{3}$