The resultant of these forces OP, OQ, OR, OS and OT is approximately.......... N. [Take √(3) = 1.7, √(2) = 1.4 Given i and j unit vectors along x, y axis] (a) 9.25i + 5j (b) 3i + 15j (c) 2.5i - 14.5j (d) -1.5i - 15.5j

Answer: (a). Resolve each force (angles measured from +x): P at 60°, Q at 30°, R at -45°, S at 225°, T at 150°. x: 20cos 60° + 10cos 30° + 20cos 45° - 15cos 45° - 15cos 30° = 10 + 8.5 + 14 - 10.5 - 12.75 = 9.25 y: 20sin 60° + 10sin 30° - 20sin 45° - 15sin 45° + 15sin 30° = 17 + 5 - 14 - 10.5 + 7.5 = 5 So the answer is (a) 9.25i + 5j.

The resultant of these forces OP, OQ, OR, OS and OT is… | JEE Main 2021 (27 Aug, Shift 1)

Q14JEE Main 2021 · 27 AUG · SHIFT 1Medium

Vector Basics, Unit Vectors & Components

The resultant of these forces $\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}, \overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately .......... N. [Take $\sqrt{3} = 1.7,\ \sqrt{2} = 1.4$ Given $\hat{i}$ and $\hat{j}$ unit vectors along x, y axis]

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Answer: (a)

Resolve each force (angles measured from $+x$ ): P at $60^\circ$ , Q at $30^\circ$ , R at $-45^\circ$ , S at $225^\circ$ , T at $150^\circ$ .
$x$ : $20\cos60^\circ + 10\cos30^\circ + 20\cos45^\circ - 15\cos45^\circ - 15\cos30^\circ = 10 + 8.5 + 14 - 10.5 - 12.75 = 9.25$
$y$ : $20\sin60^\circ + 10\sin30^\circ - 20\sin45^\circ - 15\sin45^\circ + 15\sin30^\circ = 17 + 5 - 14 - 10.5 + 7.5 = 5$
So the answer is (a) $9.25\hat{i} + 5\hat{j}$ .

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