Kinematics Question 455
Question: If $ \vec{A}\times \vec{B}=(\vec{C}+\vec{D}), $ then select the correct alternative
Options:
A) $ \vec{B} $ it’s parallel to $ \vec{C}+\vec{D} $
B) $ \vec{A} $ is perpendicular to $ \vec{C} $
C) component of $ \vec{C} $ along $ \vec{A}= $ component of $ \vec{D} $ along $ \vec{A} $
D) component of $ \vec{C} $ along $ \vec{A}=- $ component of $ \vec{D} $ along $ \vec{A} $
Correct Answer: D [d] $ ( \vec{C}+\vec{D} ) $ is perpendicular to $ \vec{A} $ $ \therefore $ $ \overrightarrow{A}.( \vec{C}+\vec{D} )=0 $
or $ \vec{A}.\vec{C}+\vec{A}.\vec{D}=0 $
or A (component of $ \vec{C} $ along A)
+A (component of $ \vec{D} $ along A) = 0
or Component of $ \vec{C} $ along A
= ? component of $ \vec{D} $ along A.
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Answer:
Solution:
