Vector addition and subtraction (graphical method)

This is core, board-examinable content in CBSE Class 11 Physics, chapter Motion in a Plane: the prescribed scope explicitly lists "addition/subtraction (triangle and parallelogram law)" alongside scalars versus vectors, position and displacement vectors, and multiplication by a real number. It sits between "multiplication of vectors by a real number" (the previous article, which fixes what stretching or reversing a single vector means) and "resolution of vectors" (the next article, which replaces this graphical picture with the faster algebraic component method). Everything here stays within the geometric, angle-and-triangle toolkit — no components, no unit vectors $\hat i,\hat j$, which the next node handles.

A vector is drawn as an arrow: its length represents magnitude and its direction represents, well, direction. Two vectors of the same physical kind — two forces, two velocities, two displacements — can be combined into a single equivalent vector, the resultant, and the graphical method finds that resultant purely by drawing, with the algebra entering only at the very last step to read off numbers.

Triangle law. Draw $\vec A$ as an arrow. From the tip (arrowhead) of $\vec A$, draw $\vec B$ — keeping $\vec B$'s own length and direction unchanged, just relocating its tail to sit at $\vec A$'s tip. The resultant $\vec R$ is the vector from the tail of $\vec A$ straight to the tip of $\vec B$, closing the triangle:

where: $\vec A$ and $\vec B$ are the two vectors being added (drawn to scale, at whatever angle they actually act), and $\vec R$ is the single equivalent vector — same physical effect as $\vec A$ and $\vec B$ acting together.

Parallelogram law. Draw $\vec A$ and $\vec B$ from the same starting point, as two adjacent sides of a parallelogram, and complete the figure. The resultant is the diagonal of the parallelogram drawn from that shared starting point. This is not a second, different rule — it is the triangle law seen from a different corner of the same figure, because the far side of the parallelogram is just $\vec B$ relocated to $\vec A$'s tip, which is exactly the triangle construction.

The angle $\theta$ between $\vec A$ and $\vec B$ decides everything about the resultant. Applying the law of cosines to the triangle formed gives the magnitude of $\vec R$, and applying the law of sines gives its direction relative to $\vec A$:

where: $A$ and $B$ are the magnitudes of the two vectors, $\theta$ is the angle between them (measured as they are drawn tail-to-tail), $R$ is the resultant's magnitude, and $\alpha$ is the angle the resultant makes with $\vec A$. Physically, this says a bigger angle between two pushes wastes more of their combined effort — at $\theta=0^\circ$ ($\cos\theta=1$) the vectors line up and $R=A+B$, the biggest possible resultant; at $\theta=180^\circ$ ($\cos\theta=-1$) they oppose and $R=|A-B|$, the smallest. Vector addition obeys two algebraic properties, both visible directly in the diagrams above:

• Commutative: $\vec A + \vec B = \vec B + \vec A$ — draw $B$ first then $A$ in the triangle and you land on the same diagonal of the parallelogram.
• Associative: $(\vec A + \vec B) + \vec C = \vec A + (\vec B + \vec C)$ — grouping three vectors differently before adding still closes the same overall polygon.

Vector subtraction is defined as addition of the negative vector, where $-\vec B$ has the same magnitude as $\vec B$ but points in exactly the opposite direction:

Graphically: reverse $\vec B$'s arrow (flip it $180^\circ$, keep the length), then apply the ordinary triangle or parallelogram law to add $\vec A$ and $(-\vec B)$. Because the angle between $\vec A$ and $-\vec B$ is $(180^\circ-\theta)$, and $\cos(180^\circ-\theta) = -\cos\theta$, the magnitude formula becomes:

where: the symbols mean the same as before, and the only change from the addition formula is the sign in front of the $2AB\cos\theta$ term — a direct consequence of reversing one vector before adding.

The goal of this derivation is to justify, from plain triangle geometry, why the resultant of two vectors drawn tip-to-tail has magnitude $R=\sqrt{A^2+B^2+2AB\cos\theta}$ and makes an angle $\alpha=\tan^{-1}\!\left[B\sin\theta/(A+B\cos\theta)\right]$ with $\vec A$ — the two formulas that turn a hand-drawn triangle into exam-ready numbers, so that a student who has only sketched the construction can still extract exact answers without measuring the page with a ruler and protractor.

Board level - resultant of two forces at 60 degrees

Two forces of magnitude $A = 6$ N and $B = 8$ N act on a body at an angle of $60^\circ$ to each other. Find the magnitude and direction of the resultant force.

Step 1. Identify the known quantities: $A=6$ N, $B=8$ N, $\theta=60^\circ$, so $\cos 60^\circ = 0.5$ and $\sin 60^\circ = 0.866$.

Step 2. Substitute into the magnitude formula: $R=\sqrt{A^2+B^2+2AB\cos\theta}=\sqrt{6^2+8^2+2(6)(8)(0.5)}=\sqrt{36+64+48}=\sqrt{148}$.

Step 3. Evaluate: $R \approx 12.2$ N.

Step 4. Find the direction relative to $\vec A$: $\tan\alpha=\dfrac{B\sin\theta}{A+B\cos\theta}=\dfrac{8(0.866)}{6+8(0.5)}=\dfrac{6.93}{10}=0.693$, so $\alpha \approx 34.7^\circ$.

The resultant force is approximately 12.2 N, directed about 34.7 degrees from the 6 N force, on the side toward the 8 N force.

Competitive level - sum and difference of two equal perpendicular vectors, and a follow-up angle check

Two vectors have equal magnitude $A=B=10$ units and act at $90^\circ$ to each other. Find (a) the magnitude of $\vec A+\vec B$, (b) the magnitude of $\vec A-\vec B$, and (c) verify that the sum and difference vectors are themselves perpendicular to each other.

Step 1. For the sum, with $\theta=90^\circ$ so $\cos 90^\circ=0$: $|\vec A+\vec B|=\sqrt{A^2+B^2+2AB\cos 90^\circ}=\sqrt{100+100+0}=\sqrt{200}\approx 14.1$ units.

Step 2. For the difference, the angle between $\vec A$ and $-\vec B$ is $180^\circ-90^\circ=90^\circ$ as well, since reversing a vector that is already perpendicular to $\vec A$ keeps it perpendicular: $|\vec A-\vec B|=\sqrt{A^2+B^2-2AB\cos 90^\circ}=\sqrt{100+100-0}=\sqrt{200}\approx 14.1$ units.

Step 3. So (a) and (b) are numerically equal — a special feature of equal-magnitude perpendicular vectors, since the $2AB\cos\theta$ term vanishes for both the sum and the difference when $\theta=90^\circ$, leaving $\sqrt{A^2+B^2}$ either way.

Step 4. Verify perpendicularity of directions in part (c) using the parallelogram picture: $\vec A+\vec B$ points along the diagonal that bisects the right angle between $\vec A$ and $\vec B$, so it makes $45^\circ$ with each. The difference $\vec A-\vec B$ points along the other diagonal of the same parallelogram (rectangle, here, since $A=B$ and $\theta=90^\circ$), which is always perpendicular to the first diagonal in a rectangle. Hence the sum and the difference, while equal in magnitude (14.1 units each), point in directions $90^\circ$ apart.

Both the sum and the difference have magnitude approximately 14.1 units, and they point in mutually perpendicular directions. Timing note: in a competitive exam, recognize $A=B$ with $\theta=90^\circ$ instantly as $\sqrt{2}A$ for both sum and difference — that shortcut ($\sqrt{2}\times 10 = 14.14$) should take under 20 seconds, leaving the remaining time for the direction/perpendicularity reasoning rather than re-deriving the law of cosines from scratch.

• Placing both vectors tail-to-tail and then drawing the resultant as one of the parallelogram's sides instead of the diagonal — the sides are just $\vec A$ and $\vec B$ themselves, not the resultant.
• In the triangle method, starting $\vec B$ from the same point as $\vec A$ instead of from $\vec A$'s tip — that mistake actually produces the parallelogram picture's two sides, not a closed triangle.
• For subtraction, forgetting to reverse $\vec B$'s direction before applying the addition rule; $\vec A-\vec B$ is not found by connecting the tip of $B$ to the tip of $A$ without that reversal step.
• Assuming the plain-number rules $R=A+B$ or $R=A-B$ hold for any angle — they are only the special cases $\theta=0^\circ$ and $\theta=180^\circ$; every other angle needs the full $\sqrt{A^2+B^2+2AB\cos\theta}$ expression.
• Mixing up which angle goes into the formula: $\theta$ is the angle between the vectors as drawn from a common point, not the interior angle of the triangle at the join (which is $180^\circ-\theta$) — a very common sign slip that flips $+$ to $-$ in the cosine term.

• Add two vectors graphically using both the triangle law and the parallelogram law, and explain why they give the same resultant.
• Compute a resultant's magnitude and direction using $R=\sqrt{A^2+B^2+2AB\cos\theta}$ and $\tan\alpha = B\sin\theta/(A+B\cos\theta)$ for any given angle $\theta$.
• Construct $-\vec B$ (same length, reversed direction) and use it to perform vector subtraction as $\vec A+(-\vec B)$.
• State and justify, from the diagrams, the commutative and associative properties of vector addition.
• Recognize and instantly evaluate the special cases: parallel ($\theta=0^\circ$, $R=A+B$), antiparallel ($\theta=180^\circ$, $R=|A-B|$), and perpendicular ($\theta=90^\circ$, $R=\sqrt{A^2+B^2}$).
• Identify the common triangle/interior-angle sign trap ($\theta$ versus $180^\circ-\theta$) before substituting into the cosine formula.

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