IIT JEE , AIEEE Forum - Mathematics , Vectors - Radon challenger series Part-5

Radon222

$Assertion(A): \text{If in a triangle ABC, }\\ \\ \vec{BC}=\frac{\vec{e}}{|\vec{e}|}-\frac{\vec{f}}{|\vec{f}|} \text{ and } \vec{AC}\frac{2\vec{e}}{|\vec{e}|};|\vec{e}| \not= |\vec{f}|,\\ \\ \text{ then the value of cos2A+cos2B+cos2C is -1 }\\ \\ Reason(R):\text{If in a triangle ABC, angle C =90, }\\ \text{then cos2A+cos2B+cos2C is -1 }\\ \\$

A) A and R both correct and R is correct explanation

B) A and R both correct and R is not correct explanation

C) A is true R is false

D) A is false R is true

akhil_o

Radon222

Sorry guys there is a correction, this one is correct

$Assertion(A): \text{If in a triangle ABC, }\\ \\ \vec{BC}=\frac{\vec{e}}{|\vec{e}|}-\frac{\vec{f}}{|\vec{f}|} \text{ and } \vec{AC}=\frac{2\vec{e}}{|\vec{e}|};|\vec{e}| \not= |\vec{f}|,\\ \\ \text{ then the value of cos2A+cos2B+cos2C is -1 }\\ \\ Reason(R):\text{If in a triangle ABC, angle C =90, }\\ \text{then cos2A+cos2B+cos2C is -1 }\\ \\$

akhil_o

its the same na?
D)
Assertion is true only for perpendicular vectors i think
Reason is true

elastiboysai

Radon222

Wrong !

Radon222

$\vec{BA}=\vec{BC}-\vec{AC} \\ \\ =\left( \frac{\vec{e}}{|\vec{e}|}-\frac{\vec{f}}{|\vec{f}|}\right)-\left(\frac{2\vec{e}}{|\vec{e}|}\right)\\ \\ -\left( \frac{\vec{e}}{|\vec{e}|}+\frac{\vec{f}}{|\vec{f}|}\right)\\ \\ Now.\vec{BA}.\vec{BC}=-\left( \frac{\vec{e}}{|\vec{e}|}+\frac{\vec{f}}{|\vec{f}|}\right)\left( \frac{\vec{e}}{|\vec{e}|}-\frac{\vec{f}}{|\vec{f}|}\right) \\ \\ =-\left(\frac{e^2}{e^2}-\frac{f^2}{f^2}\right)=-(1-1)=0 \\ \\ \rightarrow \angle B =90 \\ \\ \rightarrow cos2A+cos2C=2cos(A+C)cos(A-C)=0\\ \\ \rightarrow cos2A+cos2C+cos2B=-1 \\ \\ \text{This condition also holds true when angle C=90}\\$

Thus both assertion and reason are correct but the reason is not correct explanation of assertion.Therefore the answer is B

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