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

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

cdquestions Admin · Accepted Answer

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  $\frac {2π}{3}$

Answer

$\frac {2π}{3}$

Answer

$\frac {5π}{6}$

Answer

$\frac {5π}{9}$

Answer

$\frac {3π}{4}$

List-I	List-II
(A) 4î − 2ĵ − 4k̂	(I) A vector perpendicular to both î + 2ĵ + k̂ and 2î + 2ĵ + 3k̂
(B) 4î − 4ĵ + 2k̂	(II) Direction ratios are −2, 1, 2
(C) 2î − 4ĵ + 4k̂	(III) Angle with the vector î − 2ĵ − k̂ is cos⁻¹(1/√6)
(D) 4î − ĵ − 2k̂	(IV) Dot product with −2î + ĵ + 3k̂ is 10

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

The Correct Option is A

Solution and Explanation

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