Vector Algebra Question 37
Question: Resolved part of vector $ \vec{a} $ along vector $ \vec{b} $ is $ {{\vec{a}}_1} $ and that perpendicular to $ \vec{b} $ is $ {{\vec{a}}_2} $ then $ {{\vec{a}}_1}\times {{\vec{a}}_2} $ is equal to
Options:
A) $ \frac{(\vec{a}\times \vec{b})\cdot \vec{b}}{{{| {\vec{b}} |}^{2}}} $
B) $ \frac{(\vec{a}\cdot \vec{b})\vec{a}}{{{| {\vec{a}} |}^{2}}} $
C) $ \frac{(\vec{a}\cdot \vec{b})(\vec{b}\times \vec{a})}{{{| {\vec{b}} |}^{2}}} $
D) $ \frac{(\vec{a}\cdot \vec{b})(\vec{b}\times \vec{a})}{| \vec{b}\times \vec{a} |} $
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Answer:
Correct Answer: C
Solution:
- [c] $ {{\vec{a}}_1}=(\vec{a}\cdot \hat{b})\hat{b}=\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}} $
$ \Rightarrow {{\vec{a}}_2}=\vec{a}-{{\vec{a}}_1}=\vec{a}-\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}} $ Thus, $ {{\vec{a}}_1}\times {{\vec{a}}_2}=\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}}\times ( \vec{a}-\frac{(\vec{a}\cdot \vec{b})\vec{b}}{|\vec{b}{{|}^{2}}} ) $ $ =\frac{(\vec{a}\cdot \vec{b})(\vec{b}\times a)}{|\vec{b}{{|}^{2}}} $
BETA
