Vector Operations>Law of parallelogram of vectors

"If two vectors of the same type start from the same point and form two adjacent sides of a parallelogram, then their sum (resultant) is represented by the diagonal of the parallelogram that starts from the same point."

Parallelogram law of vector addition

Consider the parallelogram with vectors \[\vec P\] and \[\vec Q\] making an angle θ between them. To find |\[\vec R\]|, we use the following steps:

1. Drop a perpendicular from C to the extended line OA, meeting it at point D

2. In the right triangle ODC, apply the Pythagorean theorem: OC² = OD² + DC².

3. Since OD = OA + AD, and from geometry:
   AD = AC cos θ = Q cos θ
   DC = AC sin θ = Q sin θ
   OA = P

4. Substituting these values:
   OC² = (P + Q cos⁡θ)² + (Q sin⁡θ)²  

5. Expanding and simplifying:
   R² = P² + Q² + 2PQ cos ⁡θ

R = \[\sqrt{P^{2}+Q^{2}+2PQ\cos\theta}\]   --- (Equation 1)

Finding the Direction of \[\vec R\]:

To find the angle α that \[\vec R\] makes with \[\vec P\]:

From the right triangle ODC:
\[\tan\alpha=\frac{DC}{OD}=\frac{Q\sin\theta}{P+Q\cos\theta}\]

\[\alpha=\tan^{-1}\left(\frac{Q\sin\theta}{P+Q\cos\theta}\right)\]    --- (Equation 2)

Question: A boat travels at 20 km/h northward in still water. A current flows at 5 km/h eastward. Find the boat's actual velocity.

Solution:

• \[\vec P\] = 20 km/h (North)
• \[\vec Q\] = 5 km/h (East)
• Angle between them θ = 90°

Magnitude:

R = \[\sqrt{20^2+5^2+2(20)(5)\cos90°}\]

R = \[\sqrt{400+25+0}=\sqrt{425}\] = 20.61 km/h

Direction:

\[\alpha=\tan^{-1}\left(\frac{5\sin90°}{20+5\cos90°}\right)=\tan^{-1}\left(\frac{5}{20}\right)=14°04^{\prime}\]

Answer: The boat travels at 20.61 km/h in a direction 14°04' east of north.