Paragraph: Let O be the origin, and O X , O Y , O Z be three unit vectors in the directions of the sides Q R , R P , P Q , respectively, of a triangle P Q R . Question: If the triangle P Q R varies, then the minimum

Correct answer: A. - (3)/(2) (P + Q) + (Q + R) + (R + P) = (180 - R) + (180 - P) + (180 - Q) = - [cosP + cosQ + cosR] In any PQR , cosP + cosQ + cosR ≤ (3)/(2) ⇒ - (cosP + cosQ + cosR) ≥ - (3)/(2)

JEE Advanced 2017Paper 2

Exam: JEE Advanced
Year: 2017
Paper: Paper 2
Subject: Mathematics
Chapter: Properties of Triangles

JEE Advanced 2017 · Paper 2 · Mathematics · Properties of Triangles

Paragraph: Let O be the origin, and O X , O Y , O Z be three unit...

Paragraph: Let

O

be the origin, and

\overrightarrow{O X}, \overrightarrow{O Y}, \overrightarrow{O Z}

be three unit vectors in the directions of the sides

\overrightarrow{Q R}, \overrightarrow{R P}

\overrightarrow{P Q}

, respectively, of a triangle

P Q R

.
Question: If the triangle

P Q R

varies, then the minimum value of

\cos (P+Q)+\cos (Q+R)+\cos (R+P)

Options

A $- \dfrac{3}{2}$
B $\dfrac{3}{2}$
C $\dfrac{5}{3}$
D $- \dfrac{5}{3}$

Correct answer

A. $- \dfrac{3}{2}$

Step-by-step solution

$\cos \left(P + Q\right) + \cos \left(Q + R\right) + \cos \left(R + P\right)$

$= \cos \left(180 - R\right) + \cos \left(180 - P\right) + \cos \left(180 - Q\right)$

$= - \left[cosP + cosQ + cosR\right]$

In any $\delta PQR$ , $cosP + cosQ + cosR \le \dfrac{3}{2}$

$⇒ - \left(cosP + cosQ + cosR\right) \ge - \dfrac{3}{2}$

Practice Properties of Triangles on Quantrex Academy →

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