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
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
\(\frac {2π}{3}\)
\(\frac {5π}{6}\)
\(\frac {5π}{9}\)
\(\frac {3π}{4}\)
The Correct Option is A
Solution and Explanation
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 \(\frac {2π}{3}\)
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