Explain in detail the triangle law of addition.

III. Long Answer Questions · Q1

Q. Explain in detail the triangle law of addition.

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Step 1. Statement of the law. To add two vectors $\vec A$ and $\vec B$ inclined at angle $\theta$ : represent $\vec A$ and $\vec B$ as two sides of a triangle taken in the same order — draw $\vec A$ from O to P, then draw $\vec B$ starting from P (the head of $\vec A$ ) to a point Q. The resultant $\vec R=\vec A+\vec B$ is then the third side of the triangle, drawn directly from O to Q.

Step 2. Magnitude. Extend OP to a point N so that PN is along OP extended, and drop a perpendicular from Q onto this extended line, meeting it at N. In right triangle PQN (with $\angle$ QPN $=180°-\theta$ 's supplement, giving the angle at P equal to $\theta$ measured appropriately), $PN=B\cos\theta$ and $QN=B\sin\theta$ . Then $ON=OP+PN=A+B\cos\theta$ , and applying Pythagoras to right triangle OQN:

OQ^2=ON^2+QN^2=(A+B\cos\theta)^2+(B\sin\theta)^2

Expanding, $R^2=A^2+2AB\cos\theta+B^2\cos^2\theta+B^2\sin^2\theta=A^2+B^2+2AB\cos\theta$ (using $\cos^2\theta+\sin^2\theta=1$ ), so

R=\sqrt{A^2+B^2+2AB\cos\theta}

Step 3. Direction. If $\vec R$ makes angle $\alpha$ with $\vec A$ , then in right triangle OQN, $\tan\alpha=\dfrac{QN}{ON}=\dfrac{B\sin\theta}{A+B\cos\theta}$ .

Step 4. Special cases (checks). For $\theta=0$ (parallel vectors), $R=A+B$ , the ordinary sum. For $\theta=180°$ (anti-parallel), $R=|A-B|$ . For $\theta=90°$ , $R=\sqrt{A^2+B^2}$ , ordinary Pythagoras — all consistent with simple intuition.

✓Final answer

\vec R=\vec A+\vec B

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL5 marks
$Q. (a) Explain the Triangle Law of vector addition. OR$
(b) Derive the expression for the terminal velocity of a sphere moving in a high-viscous fluid using Stokes force.
›Reveal solutionSolution

The triangle law says that placing two vectors head-to-tail along two sides of a triangle gives their resultant as the closing third side, with magnitude R = square root of (A^2+B^2+2AB*cos(theta)).

Statement: If two vectors acting on a particle can be represented, both in magnitude and direction, by the two sides of a triangle taken in order (i.e., the tail of the second vector coincides with the head of the first), then the resultant of these two vectors is represented, in magnitude and direction, by the third side of the triangle taken in the reverse order (from the tail of the first vector to the head of the second).

Derivation of the magnitude and direction:

Let vector A = OP be represented by one side, and vector B = PQ be represented by the second side (drawn from the head of A), making an angle theta with A (the angle between A and B). By the triangle law, the resultant R = OQ is the third side, closing the triangle.

Drop a perpendicular QN from Q to the extension of OP. In right triangle ONQ:

QN = Bsin(theta) (the perpendicular height) ON = A + Bcos(theta) (the extended base)

Magnitude of resultant: Using the Pythagorean theorem in triangle ONQ,

OQ^2 = ON^2 + QN^2

R^2 = (A + Bcos(theta))^2 + (Bsin(theta))^2

R^2 = A^2 + 2ABcos(theta) + B^2cos^2(theta) + B^2sin^2(theta) R^2 = A^2 + 2ABcos(theta) + B^2*(cos^2(theta) + sin^2(theta))

R^2 = A^2 + B^2 + 2ABcos(theta) (since cos^2(theta) + sin^2(theta) = 1) So R = square root of (A^2 + B^2 + 2ABcos(theta))

Direction of resultant: Let alpha be the angle the resultant R makes with vector A. In right triangle ONQ,

tan(alpha) = QN/ON = (Bsin(theta))/(A + Bcos(theta))

So alpha = arctan[(Bsin(theta))/(A + Bcos(theta))]

This gives both the magnitude and direction of the resultant of two vectors added by the triangle law — the same law used repeatedly (as the 'polygon law') to add more than two vectors, by chaining them head-to-tail in sequence.

✓Final answer
$By the triangle law, placing B's tail at A's head, the resultant R (from A's tail to B's head) has magnitude R = square root of (A^2 + B^2 + 2AB cos(theta)) and makes an angle alpha = arctan[(B sin(theta))/(A + B*cos(theta))] with A.$

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Q.Explain in detail the triangle law of addition.

Concept understanding — Triangle Law of Vector Addition

Q.Explain in detail the triangle law of addition.

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