Why doesn't 25 km/h plus 10 km/h make 35 km/h?
Velocities are vectors, and vectors don't add like plain numbers. In NCERT's Example 3.3 a boat heads north at 25 km/h while the water current pushes it at 10 km/h, 60° east of south. The two partly cancel. Add them properly, by components or by the law of cosines, and the resultant is about 22 km/h, roughly 23.4° east of north. You would get 35 km/h only if both pointed the same way.
Rulers are slow. Components are quick.
You can add vectors by drawing them head to tail and measuring the third side. NCERT doesn't call that wrong. It says the graphical method is "sometimes tedious" and has "limited accuracy". Misread your protractor by a degree and the answer drifts. Components fix that.
Write each vector with the unit vectors î and ĵ:
A = A_x î + A_y ĵ and B = B_x î + B_y ĵ
Vectors obey the commutative and associative laws, so you're allowed to regroup. Collect the î terms, then the ĵ terms:
R = (A_x + B_x) î + (A_y + B_y) ĵ
So R_x = A_x + B_x and R_y = A_y + B_y. That's Eq. 3.21. Each component of the resultant is the sum of the matching components, nothing more.
A quick example, and a warning
Take A = 3î + 4ĵ and B = 5î + 2ĵ. Adding the components gives R = 8î + 6ĵ, so |R| = √(64 + 36) = 10.
Compare the sizes. |A| = 5 and |B| = √29, about 5.39. Together that's about 10.39, not 10. Components add. Magnitudes don't.
In three dimensions you add a third line, R_z = A_z + B_z. With more vectors, or with subtraction, you work term by term. For T = a + b − c, T_x = a_x + b_x − c_x, and the same for y and z.
When you only know two sizes and an angle
Some problems give you |A|, |B| and the angle θ between them. That's Example 3.2.
Draw A as OP and B as OQ, tail to tail, and complete the parallelogram. The diagonal OS is R. From S, drop a perpendicular SN onto OP extended. Then ON = A + B cos θ and SN = B sin θ. Apply Pythagoras to triangle OSN, tidy up with cos²θ + sin²θ = 1, and you get:
R² = A² + B² + 2AB cos θ
That's the law of cosines. Try it at the edges. At θ = 0 both vectors line up and R = A + B. At 90° it shrinks to plain Pythagoras.
For direction, let α be the angle between R and A, and β the angle between R and B. Writing SN two ways gives R sin α = B sin θ. A second perpendicular finishes the job:
R / sin θ = A / sin β = B / sin α
That's the law of sines. Rearranged, sin α = (B/R) sin θ, or tan α = B sin θ / (A + B cos θ).
Back to the boat
This is the step to watch. The boat points north. The current points 60° east of south. Put the two arrows tail to tail and the angle between them is not 60°. It's 180° − 60° = 120°.
Since cos 120° = −½, the 2AB cos θ term turns negative:
R² = 625 + 100 − 250 = 475, so R ≈ 21.8 km/h, about 22 km/h.
For direction, call φ the angle between R and north. By the law of sines, sin φ = (10 × sin 120°) / 21.8 ≈ 0.397, so φ ≈ 23.4°. In the book's figure that's towards the east.
Check it with components. The current is about 8.66 km/h east and exactly 5 km/h south. Add the boat's 25 km/h north and you're left with about 8.66 east and 20 north. √(8.66² + 20²) ≈ 21.8 again.
Check yourself
P = 4î + ĵ and Q = 2î + 7ĵ, both in metres. What is |P + Q|? (Answer at the bottom.)
Questions students ask
How do you add two vectors using components?
Add the x-components to get R_x and the y-components to get R_y. Then |R| = √(R_x² + R_y²).
What is the law of cosines for vectors?
R² = A² + B² + 2AB cos θ, where θ is the angle between A and B placed tail to tail.
Why is the angle 120° and not 60° in the boat example?
The current points 60° east of south and the boat points north. Tail to tail, the angle between them is 180° − 60° = 120°.
Can the resultant be smaller than both vectors?
Yes. When two vectors point in nearly opposite directions and their sizes are close, cos θ is close to −1 and R can come out smaller than either.
Is the graphical method of adding vectors wrong?
No. NCERT calls it sometimes tedious, with limited accuracy. The analytical method avoids the drawing errors.
Answer: 10 m. P + Q = 6î + 8ĵ, and √(36 + 64) = 10.
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