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Concept 01 / 06Vector Algebra

Vectors: Basic Concepts, Addition, Position Vectors and the Section Formula

Vectors as directed quantities, types of vectors (zero, unit, equal, collinear, coplanar), the triangle and parallelogram laws, properties of addition and scalar multiplication, position vectors, the section formula and vector proofs in geometry.

Figure 1
A vector is a quantity with both magnitude and direction, represented by a directed line segment AB⃗\vec{AB} from its initial point AA to its terminal point BB. Its magnitude ∣AB⃗∣|\vec{AB}| is the length of the segment. Two vectors are equal if they have the same magnitude and direction, wherever they are drawn; vectors in this sense are 'free' and can be slid parallel to themselves. Types of vectors. The zero vector 0⃗\vec{0} has zero magnitude and no definite direction. A unit vector has magnitude 11; the unit vector along a⃗\vec{a} is a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}. Collinear (parallel) vectors lie along the same or parallel lines, regardless of sense; a⃗\vec{a} and b⃗\vec{b} are collinear iff b⃗=λa⃗\vec{b} = \lambda\vec{a} for some scalar λ\lambda. Coplanar vectors lie in (or parallel to) one plane. Any two vectors are coplanar; three vectors may or may not be. The negative −a⃗-\vec{a} has the same magnitude and the opposite direction. Addition. Triangle law: to add b⃗\vec{b} to a⃗\vec{a}, place the tail of b⃗\vec{b} at the head of a⃗\vec{a}; the sum goes from the tail of a⃗\vec{a} to the head of b⃗\vec{b}. Parallelogram law: place both at a common tail; the sum is the diagonal from that point, as in the figure. The other diagonal is b⃗−a⃗\vec{b} - \vec{a} (from the head of a⃗\vec{a} to the head of b⃗\vec{b}). Addition is commutative and associative, 0⃗\vec{0} is the identity, and a⃗+(−a⃗)=0⃗\vec{a} + (-\vec{a}) = \vec{0}. For a closed polygon traversed in order, the sum of the side vectors is 0⃗\vec{0}. Triangle inequality: ∣a⃗+b⃗∣≤∣a⃗∣+∣b⃗∣|\vec{a} + \vec{b}| \le |\vec{a}| + |\vec{b}|, with equality iff the vectors point the same way. Also ∣a⃗−b⃗∣≥∣∣a⃗∣−∣b⃗∣∣|\vec{a} - \vec{b}| \ge ||\vec{a}| - |\vec{b}||. Scalar multiplication. λa⃗\lambda\vec{a} has magnitude ∣λ∣∣a⃗∣|\lambda||\vec{a}| and the direction of a⃗\vec{a} if λ>0\lambda > 0, opposite if λ<0\lambda < 0. It distributes over vector and scalar addition. Position vectors. Fix an origin OO. The position vector of a point PP is OP⃗\vec{OP}, often written p⃗\vec{p}. Then for any two points AB⃗=b⃗−a⃗\vec{AB} = \vec{b} - \vec{a} ('head minus tail'). This single identity turns geometric statements into algebra. Section formula. The point dividing ABAB internally in the ratio m:nm : n has position vector mb⃗+na⃗m+n\frac{m\vec{b} + n\vec{a}}{m + n}; externally, mb⃗−na⃗m−n\frac{m\vec{b} - n\vec{a}}{m - n}. The midpoint is a⃗+b⃗2\frac{\vec{a} + \vec{b}}{2} and the centroid of a triangle is a⃗+b⃗+c⃗3\frac{\vec{a} + \vec{b} + \vec{c}}{3}. Collinearity of points. Three points A,B,CA, B, C are collinear iff AB⃗\vec{AB} and AC⃗\vec{AC} are parallel, iff there are scalars x,y,zx, y, z, not all zero, with xa⃗+yb⃗+zc⃗=0⃗x\vec{a} + y\vec{b} + z\vec{c} = \vec{0} and x+y+z=0x + y + z = 0. Similarly, four points are coplanar iff there is such a relation with four scalars summing to zero. Vector methods in geometry. Many classical theorems become one-line calculations. The line joining the midpoints of two sides of a triangle is parallel to the third side and half as long: a⃗+c⃗2−a⃗+b⃗2=c⃗−b⃗2\frac{\vec{a} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2} = \frac{\vec{c} - \vec{b}}{2}. The diagonals of a parallelogram bisect each other: both midpoints equal a⃗+c⃗2\frac{\vec{a} + \vec{c}}{2} when b⃗+d⃗=a⃗+c⃗\vec{b} + \vec{d} = \vec{a} + \vec{c}. Key Takeaways: - a^=a⃗∣a⃗∣\hat{a} = \frac{\vec{a}}{|\vec{a}|}; a⃗∥b⃗  ⟺  b⃗=λa⃗\vec{a} \parallel \vec{b} \iff \vec{b} = \lambda\vec{a}. - Parallelogram diagonals: a⃗+b⃗\vec{a} + \vec{b} and b⃗−a⃗\vec{b} - \vec{a}. - AB⃗=b⃗−a⃗\vec{AB} = \vec{b} - \vec{a} (head minus tail). - Section formula: mb⃗+na⃗m+n\frac{m\vec{b} + n\vec{a}}{m + n}; centroid a⃗+b⃗+c⃗3\frac{\vec{a} + \vec{b} + \vec{c}}{3}. - Collinear points: xa⃗+yb⃗+zc⃗=0⃗x\vec{a} + y\vec{b} + z\vec{c} = \vec{0} with x+y+z=0x + y + z = 0, not all zero. - Sum of vectors around a closed polygon is 0⃗\vec{0}. - Pitfall: writing AB⃗=a⃗−b⃗\vec{AB} = \vec{a} - \vec{b} (tail minus head). Brain-Teaser 1: ABCDEFABCDEF is a regular hexagon with centre OO. Show that AB⃗+AC⃗+AD⃗+AE⃗+AF⃗=3AD⃗\vec{AB} + \vec{AC} + \vec{AD} + \vec{AE} + \vec{AF} = 3\vec{AD}. Solution: Pair the vectors symmetric about the diagonal ADAD. In a regular hexagon, ABDEABDE and ACDFACDF are rectangles, and this makes AB⃗+AE⃗\vec{AB} + \vec{AE} and AC⃗+AF⃗\vec{AC} + \vec{AF} each equal to AD⃗\vec{AD}. Indeed AE⃗=BD⃗\vec{AE} = \vec{BD} (opposite sides of the rectangle ABDEABDE), so AB⃗+AE⃗=AB⃗+BD⃗=AD⃗\vec{AB} + \vec{AE} = \vec{AB} + \vec{BD} = \vec{AD}; similarly AF⃗=CD⃗\vec{AF} = \vec{CD}, so AC⃗+AF⃗=AD⃗\vec{AC} + \vec{AF} = \vec{AD}. Adding the remaining AD⃗\vec{AD} gives 3AD⃗3\vec{AD}. Symmetric pairing plus the triangle law does the whole job without coordinates. Brain-Teaser 2: Show that the points with position vectors a⃗−2b⃗+3c⃗\vec{a} - 2\vec{b} + 3\vec{c}, 2a⃗+3b⃗−4c⃗2\vec{a} + 3\vec{b} - 4\vec{c} and −7b⃗+10c⃗-7\vec{b} + 10\vec{c} are collinear, where a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} are non-coplanar. Solution: PQ⃗=(2a⃗+3b⃗−4c⃗)−(a⃗−2b⃗+3c⃗)=a⃗+5b⃗−7c⃗\vec{PQ} = (2\vec{a} + 3\vec{b} - 4\vec{c}) - (\vec{a} - 2\vec{b} + 3\vec{c}) = \vec{a} + 5\vec{b} - 7\vec{c}. PR⃗=(−7b⃗+10c⃗)−(a⃗−2b⃗+3c⃗)=−a⃗−5b⃗+7c⃗=−PQ⃗\vec{PR} = (-7\vec{b} + 10\vec{c}) - (\vec{a} - 2\vec{b} + 3\vec{c}) = -\vec{a} - 5\vec{b} + 7\vec{c} = -\vec{PQ}. Since PR⃗=−PQ⃗\vec{PR} = -\vec{PQ}, the points are collinear, with PP the midpoint of QRQR. Non-coplanarity of a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} is what lets us compare coefficients safely. Brain-Teaser 3: Prove that the medians of a triangle are concurrent, and find the point. Solution: Let the vertices be a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}. The median from AA goes to b⃗+c⃗2\frac{\vec{b} + \vec{c}}{2}. The point dividing it in the ratio 2:12 : 1 is 2⋅b⃗+c⃗2+1⋅a⃗3=a⃗+b⃗+c⃗3\frac{2 \cdot \frac{\vec{b} + \vec{c}}{2} + 1 \cdot \vec{a}}{3} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}. This expression is symmetric in a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}, so the same point lies on all three medians: they are concurrent at the centroid a⃗+b⃗+c⃗3\frac{\vec{a} + \vec{b} + \vec{c}}{3}. Symmetry of the formula replaces the need to intersect lines.