Vector Algebra Question 38
Question: The components of a vector $ \vec{a} $ along and perpendicular to a non-zero vector $ \vec{b} $ are
Options:
A) $ ( \frac{\vec{a}.\vec{b}}{{{| {\vec{b}} |}^{2}}} )\vec{b}\And \vec{a}-( \frac{\vec{a}.\vec{b}}{{{| {\vec{b}} |}^{2}}} )\vec{b} $
B) $ ( \frac{\vec{a}.\vec{b}}{{{| {\vec{a}} |}^{2}}} )\vec{b}\And \vec{a}+( \frac{\vec{a}.\vec{b}}{{{| {\vec{a}} |}^{2}}} )\vec{b} $
C) $ ( \frac{\vec{a}.\vec{b}}{{{| {\vec{a}} |}^{2}}} )\vec{a}-( \frac{\vec{a}.\vec{b}}{{{| {\vec{b}} |}^{2}}} )\vec{a} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ \overrightarrow{OM}= $ component
of $ \vec{a} $ along $ \vec{b} $
$ \overrightarrow{MA}= $ component of $ \vec{a} $
Perpendicular to $ \vec{b} $
$ \Delta ,OMA\Rightarrow \cos \theta =\frac{OM}{OA} $
$ \Rightarrow OM=|\overrightarrow{OM}|=|\overrightarrow{OA}|cos\theta =|\vec{a}|cos\theta $
$ \because \vec{a}.\vec{b}=|\vec{a}||\vec{b}|cos\theta =|\vec{b}|(OM) $
$ \therefore ,\overrightarrow{OM}=|\overrightarrow{OM}|\hat{b}=( \frac{\vec{a}.\vec{b}}{|\vec{b}|} )\frac{{\vec{b}}}{|\vec{b}|}=( \frac{\vec{a}.\vec{b}}{|\vec{b}{{|}^{2}}} )\vec{b} $
$ \overrightarrow{OM}+\overrightarrow{MA}=\overrightarrow{OA}\therefore \overrightarrow{MA}=\overrightarrow{OA}-\overrightarrow{OM} $
$ =\vec{a}-( \frac{\vec{a}\cdot \vec{b}}{|\vec{b}{{|}^{2}}} )\vec{b} $
BETA
