Vector-Algebra Question 413
Question: If $ \vec{a}.,\vec{b}=0 $ and $ \vec{a}+\vec{b} $ makes an angle of $ 60{}^\circ $ with $ \vec{a} $ , then
Options:
A) $ |\vec{a}|=2|\vec{b}| $
B) $ 2|\vec{a}|=|\vec{b}| $
C) $ |\vec{a}|=\sqrt{3}|\vec{b}| $
D) $ |\vec{b}|=\sqrt{3}|\vec{a}| $
Correct Answer: DShow Answer
Answer:
Solution:
$ \cos ,60{}^\circ =\frac{1}{2}=\frac{(\overset{\to }{\mathop{a}},+\overset{\to }{\mathop{b}},).\overset{\to }{\mathop{a}},}{| \overrightarrow{a}+\overrightarrow{b} || \overrightarrow{a} |}=\frac{{{| a |}^{2}}}{| \overrightarrow{a}+\overrightarrow{b} || \overrightarrow{a} |} $
$ \frac{1}{2}=\frac{| \overset{\to }{\mathop{a}}, |}{| \overrightarrow{a}+\overrightarrow{b} |} $ ?(1)
$ { as\overrightarrow{a}.\overrightarrow{b}=0,\overrightarrow{a}\bot \overrightarrow{b} } $
$ ( \overrightarrow{a}+\overrightarrow{b} ).\overrightarrow{b}=| \overrightarrow{a}+\overrightarrow{b} || \overrightarrow{b} |\cos 30{}^\circ $
$ \cos 30{}^\circ =\frac{\sqrt{3}}{2}=\frac{( \overrightarrow{a}+\overrightarrow{b} ).( \overrightarrow{b} )}{| \overrightarrow{a}+\overrightarrow{b} || \overrightarrow{b} |} $
$ \frac{\sqrt{3}}{2}=\frac{{{| {\vec{b}} |}^{2}}}{| \overrightarrow{a}+\overrightarrow{b} || \overrightarrow{b} |}=\frac{| {\vec{b}} |}{| \overrightarrow{a}+\overrightarrow{b} |} $ ?(2)
$ \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{| {\vec{b}} |}{| {\vec{a}} |}\sqrt{3}\frac{| {\vec{b}} |}{| {\vec{a}} |}\Rightarrow | {\vec{b}} |=\sqrt{3}| {\vec{a}} | $
BETA
