JEE Advanced 2015 · Paper 1 · Mathematics Q58

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MathematicsMultiple correct+4 / -2Question 58 of 60

Let $\Delta PQR$ be a triangle. Let $\vec a=\overrightarrow{QR}$ , $\vec b=\overrightarrow{RP}$ and $\vec c=\overrightarrow{PQ}$ . If $|\vec a|=12$ , $|\vec b|=4\sqrt3$ and $\vec b\cdot\vec c=24$ , then which of the following is (are) true?

\dfrac{|\vec c|^2}2-|\vec a|=12

\dfrac{|\vec c|^2}2+|\vec a|=30

|\vec a\times\vec b+\vec c\times\vec a|=48\sqrt3

\vec a\cdot\vec b=-72

Answer and worked solution

Official answer: A, C, D

Basic relation. $\vec a+\vec b+\vec c=\vec0$ .

Dot products. $\vec b\cdot\vec c=-\vec a\cdot\vec b-|\vec b|^2=24$ , so $\vec a\cdot\vec b=-72$ . (D) is true.

$|\vec c|^2$ . $|\vec a+\vec b|^2=144+48-144=48$ .

(A) $24-12=12$ . True.
(B) $24+12=36$ , not 30. False.

Cross products. $\vec a\times\vec b+\vec c\times\vec a=\vec a\times\vec b-(\vec a+\vec b)\times\vec a=2\,\vec a\times\vec b$ . And
$|\vec a\times\vec b|=\sqrt{144\cdot48-72^2}=24\sqrt3,$
so the magnitude is $48\sqrt3$ . (C) is true.

(A), (C), (D)

Topics:Dot product Cross product

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JEE Advanced 2015 · Paper 1, Mathematics question 58