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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Cartesian Vectors in 2D and 3D

Drawing arrows works for two vectors, but it gets slow, and in 3D it gets hard. Put the vector on a coordinate grid and it becomes a list of numbers. Every operation turns into arithmetic on those numbers.

How It Works

Components. A Cartesian vector is written v⃗=[a,b]\vec{v} = [a, b] in 2D or [a,b,c][a, b, c] in 3D: how far it moves along xx, along yy and along zz. In unit vector form, [a,b,c]=ai⃗+bj⃗+ck⃗[a, b, c] = a\vec{i} + b\vec{j} + c\vec{k}, where i⃗\vec{i}, j⃗\vec{j} and k⃗\vec{k} are length 11 along the axes.

The vector between two points. Tip minus tail: from A(x1,y1,z1)A(x_1, y_1, z_1) to B(x2,y2,z2)B(x_2, y_2, z_2), AB⃗=[x2−x1,  y2−y1,  z2−z1]\vec{AB} = [x_2 - x_1, \; y_2 - y_1, \; z_2 - z_1]. Order matters, and BA⃗=−AB⃗\vec{BA} = -\vec{AB}.

Operations. Add, subtract and multiply by a scalar one component at a time. [1,2,3]+[4,−1,0]=[5,1,3][1, 2, 3] + [4, -1, 0] = [5, 1, 3] and 3[1,2,3]=[3,6,9]3[1, 2, 3] = [3, 6, 9]. The tip to tail picture is still true; the arithmetic just does it for you.

Magnitude. Pythagoras in each dimension: ∣[a,b,c]∣=a2+b2+c2|[a, b, c]| = \sqrt{a^2 + b^2 + c^2}. The distance between two points is the magnitude of the vector between them.

Unit vectors. Divide a vector by its magnitude to get a vector of length 11 in the same direction: v⃗∣v⃗∣\frac{\vec{v}}{|\vec{v}|}.

Collinear vectors. Two vectors are parallel exactly when one is a scalar multiple of the other. Find kk from one pair of components, then check it works for all the others.

Equal vectors have equal components, which is how you find a missing point: in parallelogram ABCDABCD, AD⃗=BC⃗\vec{AD} = \vec{BC}.

5 Worked Examples

Easy first, test level last. Try each one on paper, then open the solution.

  1. Easy

    For u⃗=[3,−1,2]\vec{u} = [3, -1, 2] and v⃗=[−1,4,0]\vec{v} = [-1, 4, 0], find 2u⃗−3v⃗2\vec{u} - 3\vec{v}.
    Show solution
    1. 2u⃗=[6,−2,4]2\vec{u} = [6, -2, 4] and 3v⃗=[−3,12,0]3\vec{v} = [-3, 12, 0].
    2. Subtract component by component: [6+3,  −2−12,  4−0][6 + 3, \; -2 - 12, \; 4 - 0].

    Answer: [9,−14,4][9, -14, 4]

  2. Easy

    Find AB⃗\vec{AB} and ∣AB⃗∣|\vec{AB}| for A(2,−1,3)A(2, -1, 3) and B(5,3,3)B(5, 3, 3).
    Show solution
    1. Tip minus tail: AB⃗=[5−2,  3−(−1),  3−3]=[3,4,0]\vec{AB} = [5 - 2, \; 3 - (-1), \; 3 - 3] = [3, 4, 0].
    2. ∣AB⃗∣=9+16+0=5|\vec{AB}| = \sqrt{9 + 16 + 0} = 5.

    Answer: AB⃗=[3,4,0]\vec{AB} = [3, 4, 0] and ∣AB⃗∣=5|\vec{AB}| = 5

  3. Medium

    Find the unit vector in the direction of v⃗=[2,−3,6]\vec{v} = [2, -3, 6].
    Show solution
    1. ∣v⃗∣=4+9+36=49=7|\vec{v}| = \sqrt{4 + 9 + 36} = \sqrt{49} = 7.
    2. Divide each component by 77. Check: 4+9+3649=1\frac{4 + 9 + 36}{49} = 1.

    Answer: [27,−37,67]\left[\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}\right]

  4. Medium

    Find aa and bb so that [a,6,−4][a, 6, -4] and [2,b,8][2, b, 8] are collinear.
    Show solution
    1. Collinear means [a,6,−4]=k[2,b,8][a, 6, -4] = k[2, b, 8] for some kk.
    2. The zz components fix kk: −4=8k-4 = 8k, so k=−12k = -\frac{1}{2}.
    3. Then a=−12(2)=−1a = -\frac{1}{2}(2) = -1 and 6=−12b6 = -\frac{1}{2}b, so b=−12b = -12.

    Answer: a=−1a = -1 and b=−12b = -12

  5. Test level

    A(1,−2,3)A(1, -2, 3), B(4,0,1)B(4, 0, 1) and C(6,3,2)C(6, 3, 2) are three vertices of parallelogram ABCDABCD. Find DD and the lengths of both diagonals.
    Show solution
    1. In ABCDABCD, AD⃗=BC⃗=[2,3,1]\vec{AD} = \vec{BC} = [2, 3, 1], so D=(1+2,  −2+3,  3+1)=(3,1,4)D = (1 + 2, \; -2 + 3, \; 3 + 1) = (3, 1, 4).
    2. Check: the diagonals bisect each other. The midpoint of ACAC is (3.5,0.5,2.5)(3.5, 0.5, 2.5), and so is the midpoint of BDBD.
    3. AC⃗=[5,5,−1]\vec{AC} = [5, 5, -1], so ∣AC⃗∣=25+25+1=51|\vec{AC}| = \sqrt{25 + 25 + 1} = \sqrt{51}.
    4. BD⃗=[−1,1,3]\vec{BD} = [-1, 1, 3], so ∣BD⃗∣=1+1+9=11|\vec{BD}| = \sqrt{1 + 1 + 9} = \sqrt{11}.

    Answer: D(3,1,4)D(3, 1, 4); the diagonals are 51≈7.14\sqrt{51} \approx 7.14 and 11≈3.32\sqrt{11} \approx 3.32

Learn It on Martian Lab

Martian Lab teaches this in the Algebraic Vectors lesson of its MCV4U Cartesian Vectors unit. Watch it, then take the Geometric Vectors practice tests to see if it stuck.

Watch the Algebraic Vectors lesson free

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Free Geometric Vectors practice test, with solutions

More Geometric Vectors topics: Adding and Subtracting Vectors · Force and Velocity Vector Problems

More MCV4U topics: Evaluating Limits · Chain Rule · Optimization Problems · Related Rates · Curve Sketching Steps · Dot Product and Projections · Cross Product, Area and Torque · Equations of Planes · Intersection of Three Planes · Continuity · Derivative from First Principles · Power, Product and Quotient Rules · Velocity and Acceleration · Increasing, Decreasing and Critical Points · Concavity and the Second Derivative Test · Derivatives of e^x and a^x · Derivatives of Sin, Cos and Tan · Equations of Lines in 3D · Intersection of Lines and Planes · Distance from a Point to a Line or Plane

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