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Grade 12 · MCV4U · Calculus & Vectors

MCV4U: Dot Product and Projections

The dot product takes two vectors and gives back a single number. That number tells you how much the vectors point the same way, which is how you find angles, test for perpendicular vectors and build projections.

How It Works

There are two ways to compute it, and they always agree.

From lengths and the angle: u⃗⋅v⃗=∣u⃗∣∣v⃗∣cos⁡θ\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta, where θ\theta is the angle between the vectors placed tail to tail.

From components: multiply matching components and add. For u⃗=[u1,u2,u3]\vec{u} = [u_1, u_2, u_3] and v⃗=[v1,v2,v3]\vec{v} = [v_1, v_2, v_3], the dot product is u1v1+u2v2+u3v3u_1v_1 + u_2v_2 + u_3v_3.

The answer is a scalar, never a vector. A vector dotted with itself gives its length squared, u⃗⋅u⃗=∣u⃗∣2\vec{u} \cdot \vec{u} = |\vec{u}|^2. Order does not matter, and it distributes over addition like ordinary multiplication.

The sign tells you the angle. Positive means the angle is acute. Zero means the vectors are perpendicular. Negative means the angle is obtuse. The zero case is the one you will use most: to show two vectors are perpendicular, or to find the unknown that makes them perpendicular, set the dot product to zero.

Finding an angle combines both formulas. Work out the dot product from the components, find both lengths, then solve cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}.

Projections answer the question: how much of u⃗\vec{u} lies along v⃗\vec{v}? Think of the shadow u⃗\vec{u} casts on the line of v⃗\vec{v}. The scalar projection is that shadow's signed length, u⃗⋅v⃗∣v⃗∣\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}. The vector projection points along v⃗\vec{v} with that length, u⃗⋅v⃗v⃗⋅v⃗v⃗\frac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}}\vec{v}.

Order matters for projections: projecting u⃗\vec{u} onto v⃗\vec{v} divides by the length of v⃗\vec{v}, the vector you project onto. Swap them and the answer changes.

Where it shows up next. Work is a dot product: force dotted with displacement, in joules, so only the part of the force along the motion counts. And in Lines and Planes, a zero dot product between a normal and a direction is how you tell that a line runs parallel to a plane.

5 Worked Examples

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

  1. Easy

    ∣u⃗∣=4|\vec{u}| = 4, ∣v⃗∣=6|\vec{v}| = 6 and the angle between them is 120∘120^\circ. Find u⃗⋅v⃗\vec{u} \cdot \vec{v}.
    Show solution
    1. u⃗⋅v⃗=∣u⃗∣∣v⃗∣cos⁡θ=(4)(6)cos⁡120∘=24(−12)=−12\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta = (4)(6)\cos 120^\circ = 24\left(-\frac{1}{2}\right) = -12.

    Answer: −12-12. Negative, as it should be for an obtuse angle.

  2. Easy

    Are u⃗=[2,−1,3]\vec{u} = [2, -1, 3] and v⃗=[4,5,−1]\vec{v} = [4, 5, -1] perpendicular?
    Show solution
    1. u⃗⋅v⃗=2(4)+(−1)(5)+3(−1)=8−5−3=0\vec{u} \cdot \vec{v} = 2(4) + (-1)(5) + 3(-1) = 8 - 5 - 3 = 0.

    Answer: Yes. The dot product is 00, so the vectors are perpendicular.

  3. Medium

    Find the angle between a⃗=[1,2,2]\vec{a} = [1, 2, 2] and b⃗=[3,0,4]\vec{b} = [3, 0, 4].
    Show solution
    1. a⃗⋅b⃗=3+0+8=11\vec{a} \cdot \vec{b} = 3 + 0 + 8 = 11.
    2. ∣a⃗∣=1+4+4=3|\vec{a}| = \sqrt{1 + 4 + 4} = 3 and ∣b⃗∣=9+0+16=5|\vec{b}| = \sqrt{9 + 0 + 16} = 5.
    3. cos⁡θ=11(3)(5)=1115\cos\theta = \frac{11}{(3)(5)} = \frac{11}{15}, so θ=cos⁡−1(1115)\theta = \cos^{-1}\left(\frac{11}{15}\right).

    Answer: θ≈42.8∘\theta \approx 42.8^\circ

  4. Medium

    Find kk so that [k,3,−2][k, 3, -2] and [2,k,6][2, k, 6] are perpendicular.
    Show solution
    1. Perpendicular means the dot product is zero: 2k+3k−12=02k + 3k - 12 = 0.
    2. 5k=125k = 12, so k=125k = \frac{12}{5}.
    3. Check: 2(125)+3(125)−12=605−12=02\left(\frac{12}{5}\right) + 3\left(\frac{12}{5}\right) - 12 = \frac{60}{5} - 12 = 0.

    Answer: k=125=2.4k = \frac{12}{5} = 2.4

  5. Test level

    For u⃗=[5,1,3]\vec{u} = [5, 1, 3] and v⃗=[2,−1,2]\vec{v} = [2, -1, 2], find the scalar and vector projections of u⃗\vec{u} onto v⃗\vec{v}. Then write u⃗\vec{u} as a vector along v⃗\vec{v} plus a vector perpendicular to v⃗\vec{v}.
    Show solution
    1. u⃗⋅v⃗=10−1+6=15\vec{u} \cdot \vec{v} = 10 - 1 + 6 = 15 and ∣v⃗∣=4+1+4=3|\vec{v}| = \sqrt{4 + 1 + 4} = 3.
    2. Scalar projection: 153=5\frac{15}{3} = 5.
    3. Vector projection: 159[2,−1,2]=[103,−53,103]\frac{15}{9}[2, -1, 2] = \left[\frac{10}{3}, -\frac{5}{3}, \frac{10}{3}\right].
    4. The rest of u⃗\vec{u}: [5,1,3]−[103,−53,103]=[53,83,−13][5, 1, 3] - \left[\frac{10}{3}, -\frac{5}{3}, \frac{10}{3}\right] = \left[\frac{5}{3}, \frac{8}{3}, -\frac{1}{3}\right].
    5. Check it is perpendicular to v⃗\vec{v}: 103−83−23=0\frac{10}{3} - \frac{8}{3} - \frac{2}{3} = 0.

    Answer: Scalar projection 55; vector projection [103,−53,103]\left[\frac{10}{3}, -\frac{5}{3}, \frac{10}{3}\right]; u⃗=[103,−53,103]+[53,83,−13]\vec{u} = \left[\frac{10}{3}, -\frac{5}{3}, \frac{10}{3}\right] + \left[\frac{5}{3}, \frac{8}{3}, -\frac{1}{3}\right].

Learn It on Martian Lab

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

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