If $ \vec{a}, \vec{b}, \vec{c} $ are 3 vectors such that $ |\vec{a}| = 5, |\vec{b}| = 8, |\vec{c}| = 11 $ and $ \vec{a} + \vec{b} + \vec{c} = 0 $, then the angle between the vectors $ \vec{a} $ and $ \vec{b} $ is:

Question

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We are given the following conditions: $$ |\vec{a}| = 5, \quad |\vec{b}| = 8, \quad |\vec{c}| = 11, \quad \vec{a} + \vec{b} + \vec{c} = 0. $$ Using the property $ \vec{a} + \vec{b} + \vec{c} = 0 $, we can write $ \vec{c} = -(\vec{a} + \vec{b}) $. Now, to find the angle $ \theta $ between $ \vec{a} $ and $ \vec{b} $, we use the formula for the dot product: $$ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta. $$ Substitute the known values and solve for $ \cos \theta $, we find that: $$ \cos \theta = \frac{-2}{5}. $$ Thus, the correct answer is $ \cos^{-1}\left( \frac{-2}{5} \right) $.

Answer

$ \cos^{-1}\left( \frac{-2}{5} \right) $

Answer

$ \cos^{-1}\left( \frac{10}{11} \right) $

Answer

$ \cos^{-1}\left( \frac{41}{55} \right) $

Answer

$ \frac{\pi}{3} $

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 \( \vec{a}, \vec{b}, \vec{c} \) are 3 vectors such that \( |\vec{a}| = 5, |\vec{b}| = 8, |\vec{c}| = 11 \) and \( \vec{a} + \vec{b} + \vec{c} = 0 \), then the angle between the vectors \( \vec{a} \) and \( \vec{b} \) is:

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The Correct Option is A

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