JEE Advanced 2017 · Paper 2 · Physics Q5

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PhysicsSingle correct+3 / -1Question 5 of 54

Three vectors $\vec{P}$ , $\vec{Q}$ and $\vec{R}$ are shown in the figure. Let $S$ be any point on the vector $\vec{R}$ . The distance between the points $P$ and $S$ is $b|\vec{R}|$ . The general relation among vectors $\vec{P}$ , $\vec{Q}$ and $\vec{S}$ is

From the origin O, vectors P⃗ (to point P, steeper) and Q⃗ (to point Q, shallower) are drawn; R⃗ = Q⃗ − P⃗ runs from P to Q. A point S on R⃗ is at distance b|R⃗| from P, and S⃗ runs from O to S.

\vec{S} = (1 - b)\vec{P} + b\vec{Q}

\vec{S} = (b - 1)\vec{P} + b\vec{Q}

\vec{S} = (1 - b^2)\vec{P} + b\vec{Q}

\vec{S} = (1 - b)\vec{P} + b^2\vec{Q}

Answer and worked solution

Official answer: A

$S$ lies on $\vec R=\vec Q-\vec P$ , a distance $b|\vec R|$ from $P$ . So
$\vec S=\vec P+b\vec R=\vec P+b(\vec Q-\vec P)=(1-b)\vec P+b\vec Q.$

(A)

Topics:Vectors

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JEE Advanced 2017 · Paper 2, Physics question 5