Vector Algebra Question 241
Question: Vectors $ 3\vec{a}-5\vec{b} $ and $ 2\vec{a}+\vec{b} $ are mutually perpendicular. If $ \vec{a}+4\vec{b} $ and $ \vec{b}-\vec{a} $ are also mutually perpendicular, then the cosine of the angle between $ \vec{a} $ and $ \vec{b} $ is
Options:
A) $ \frac{19}{5\sqrt{43}} $
B) $ \frac{19}{3\sqrt{43}} $
C) $ \frac{19}{2\sqrt{45}} $
D) $ \frac{19}{6\sqrt{43}} $
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Answer:
Correct Answer: A
Solution:
- [a] $ (3\vec{a}-5\vec{b}).(2\vec{a}+\vec{b})=0 $ Or $ 6{{| {\vec{a}} |}^{2}}-5{{| {\vec{b}} |}^{2}}=7\vec{a}\cdot \vec{b} $ Also, $ (\vec{a}+4\vec{b}).(\vec{b}-\vec{a})=0 $ Or $ -{{| {\vec{a}} |}^{2}}+4{{| {\vec{b}} |}^{2}}=3\vec{a}\cdot \vec{b} $ Or $ \frac{6}{7}{{| {\vec{a}} |}^{2}}-\frac{5}{7}{{| {\vec{b}} |}^{2}}=-\frac{1}{3}{{| {\vec{a}} |}^{2}}+\frac{4}{3}{{| {\vec{b}} |}^{2}} $ Or $ 25{{| {\vec{a}} |}^{2}}=43{{| {\vec{b}} |}^{2}} $
$ \Rightarrow 3\vec{a}\cdot \vec{b}=-{{| {\vec{a}} |}^{2}}+4{{| {\vec{b}} |}^{2}}=\frac{57}{25}{{| {\vec{b}} |}^{2}} $ Or $ 3| {\vec{a}} || {\vec{b}} |\cos \theta =\frac{57}{25}{{| {\vec{b}} |}^{2}} $ Or $ \sqrt[3]{\frac{43}{25}}{{| {\vec{b}} |}^{2}}\cos \theta =\frac{57}{25}{{| {\vec{b}} |}^{2}} $ Or $ \cos \theta =\frac{19}{5\sqrt{43}} $
BETA
