question_answer

If in a triangle \[ABC,AB=\frac{u}{|u|}-\frac{v}{|v|}\] and \[AC=\frac{2u}{|u|},\] where \[|u|\ne |v|,\] then

A)  \[1+\cos 2A+\cos 2B+\cos 2C=0\]

B)  \[1+\cos 2A+\cos 2B=0\]

C)  \[1+\cos 2A+\cos 2B+\cos 2C=3\]

D)  \[1+\cos 2A+\cos 2B=4\]

Correct Answer: A

Solution :

 Idea Use triangle law of addition of vectors \[\therefore \]In \[\Delta ABC\] Then, CB = CA + AB= a + b For a unit vector \[(\hat{a})\]\[\hat{a}=\frac{a}{|a|}\]Here, AB + BC = AC \[BC=\frac{2u}{|u|}-\frac{u}{|u|}+\frac{v}{|v|}\]                            \[=\frac{u}{|u|}+\frac{v}{|v|}\] \[AB\cdot BC=\left( \frac{u}{|u|}-\frac{v}{|v|} \right)\left( \frac{u}{|u|}+\frac{v}{|v|} \right)\] \[=(\hat{u}-\hat{v})\cdot (\hat{u}+\hat{v})\]     \[=1-1=0\] \[\angle B={{90}^{o}}\] \[\Rightarrow 1+\cos 2A+\cos 2B+\cos 2C=0\] TEST Edge Two vectors are perpendicular, parallel and points A, B and C are collinear. Related questions are asked to solve such types of questions to understand the product of vectors and using trigonometry?s identities.