Scalar product

Two vectors point in different directions and have different lengths — how do you boil that relationship down to a single number? The scalar product (or dot product) does exactly this: it multiplies two vectors together and returns a plain real number that captures both how long they are and how aligned they are. It is the workhorse of the ISC Class 12 Vector Algebra unit, sitting alongside the vector (cross) product as one of the two named ways to multiply vectors. This article builds the definition from the geometry of projection, shows how it collapses into a simple sum of products once vectors are written in components, and uses it for the three jobs examiners actually ask for: finding the angle between two vectors, testing perpendicularity, and computing a projection. Two fully worked numeric problems anchor every formula.

The core insight

The scalar product a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ=a1b1+a2b2+a3b3\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta = a_1b_1+a_2b_2+a_3b_3 turns two vectors into one number that is zero exactly when the vectors are perpendicular.

The core idea

Definition. For nonzero vectors a⃗\vec a and b⃗\vec b with angle θ\theta between them (0≤θ≤π0 \le \theta \le \pi), the scalar product is

a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ,\vec a \cdot \vec b = |\vec a||\vec b|\cos\theta,

where: ∣a⃗∣|\vec a| and ∣b⃗∣|\vec b| are the magnitudes (lengths) of the two vectors and θ\theta is the angle measured between them when their tails are placed together. The result is a scalar, not a vector — hence the name — and it can be positive, negative, or zero depending on θ\theta.

Why it is a "projection" measurement. Drop a perpendicular from the tip of b⃗\vec b onto the line of a⃗\vec a. The foot of that perpendicular marks off a length on a⃗\vec a's line called the projection of b⃗\vec b on a⃗\vec a, equal to ∣b⃗∣cos⁡θ|\vec b|\cos\theta. Multiplying that projected length by ∣a⃗∣|\vec a| gives exactly the dot product, so

proja⃗b⃗=a⃗⋅b⃗∣a⃗∣.\text{proj}_{\vec a}\vec b = \frac{\vec a \cdot \vec b}{|\vec a|}.

This is why the dot product is large and positive when two vectors point roughly the same way, zero when they are perpendicular (there is nothing to project), and negative when they point roughly opposite ways.

Figure
Two vectors from a common origin with the projection of one onto the other marked as a dashed drop
Dot product as length times projected length vector a vector b theta projection of b on a, length b cos theta

Key properties. The dot product is commutative, a⃗⋅b⃗=b⃗⋅a⃗\vec a \cdot \vec b = \vec b \cdot \vec a, and distributive over addition, a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec a\cdot(\vec b+\vec c) = \vec a\cdot\vec b + \vec a\cdot\vec c. For a scalar kk, (ka⃗)⋅b⃗=k(a⃗⋅b⃗)(k\vec a)\cdot\vec b = k(\vec a\cdot\vec b). Taking θ=0\theta = 0 in the definition gives a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2, so a vector dotted with itself is simply the square of its own length. And because cos⁡θ=0\cos\theta = 0 only at θ=π/2\theta = \pi/2, nonzero vectors satisfy a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b — the standard perpendicularity test. For the standard basis vectors, i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1 and i^⋅j^=j^⋅k^=k^⋅i^=0\hat i\cdot\hat j=\hat j\cdot\hat k=\hat k\cdot\hat i=0, since each pair is mutually perpendicular unit vectors.

Component formula and the angle between two vectors

If a⃗=a1i^+a2j^+a3k^\vec a = a_1\hat i+a_2\hat j+a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b = b_1\hat i+b_2\hat j+b_3\hat k, then

a⃗⋅b⃗=a1b1+a2b2+a3b3,\vec a\cdot\vec b = a_1b_1+a_2b_2+a_3b_3,

where: a1,a2,a3a_1, a_2, a_3 and b1,b2,b3b_1, b_2, b_3 are the respective components along i^,j^,k^\hat i, \hat j, \hat k. Combining this with the definition gives the standard way to find the angle between two vectors:

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣.\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}.

This one line does three jobs at once in ISC problems — angle-finding, perpendicularity-checking (set the numerator to zero), and, via the projection formula above, computing how much of one vector lies along another.

Derivation AFrom geometric definition to the component formula

The goal is to show, using nothing beyond the distributive property already stated and the dot products of the basis vectors i^,j^,k^\hat i, \hat j, \hat k, that the geometric definition a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta collapses into the purely algebraic sum a1b1+a2b2+a3b3a_1b_1+a_2b_2+a_3b_3 once both vectors are expanded in components — this is the bridge that makes every numeric ISC problem solvable by arithmetic instead of by measuring angles.

Step 1. Write both vectors in component form: a⃗=a1i^+a2j^+a3k^\vec a = a_1\hat i+a_2\hat j+a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b = b_1\hat i+b_2\hat j+b_3\hat k. Treat the dot as an ordinary product that distributes over the sums, exactly as multiplication distributes over addition of numbers.

Step 2. Expand the product term by term. Distributing a⃗⋅b⃗\vec a\cdot\vec b across all nine pairs gives a⃗⋅b⃗=a1b1(i^⋅i^)+a1b2(i^⋅j^)+a1b3(i^⋅k^)+a2b1(j^⋅i^)+a2b2(j^⋅j^)+a2b3(j^⋅k^)+a3b1(k^⋅i^)+a3b2(k^⋅j^)+a3b3(k^⋅k^)\vec a\cdot\vec b = a_1b_1(\hat i\cdot\hat i) + a_1b_2(\hat i\cdot\hat j) + a_1b_3(\hat i\cdot\hat k) + a_2b_1(\hat j\cdot\hat i) + a_2b_2(\hat j\cdot\hat j) + a_2b_3(\hat j\cdot\hat k) + a_3b_1(\hat k\cdot\hat i) + a_3b_2(\hat k\cdot\hat j) + a_3b_3(\hat k\cdot\hat k).

Step 3. Substitute the known basis products. Since i^,j^,k^\hat i, \hat j, \hat k are mutually perpendicular unit vectors, i^⋅i^=j^⋅j^=k^⋅k^=1\hat i\cdot\hat i=\hat j\cdot\hat j=\hat k\cdot\hat k=1 (each is ∣i^∣2cos⁡0=1|\hat i|^2\cos 0 = 1) while every mixed pair such as i^⋅j^\hat i\cdot\hat j equals ∣i^∣∣j^∣cos⁡90∘=0|\hat i||\hat j|\cos 90^\circ = 0. So six of the nine terms vanish and only the three "matching" terms survive.

Step 4. Read off the result. What remains is a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3, a purely algebraic recipe that always agrees with the geometric definition, since nothing but distributivity and the basis products (both already established) were used to reach it.

Worked example

Find the angle between a⃗=i^+2j^−2k^\vec a = \hat i + 2\hat j - 2\hat k and b⃗=3i^−6j^−2k^\vec b = 3\hat i - 6\hat j - 2\hat k. Then find λ\lambda for which p⃗=2i^+λj^+k^\vec p = 2\hat i + \lambda\hat j + \hat k is perpendicular to q⃗=i^−2j^+3k^\vec q = \hat i - 2\hat j + 3\hat k, and the projection of p⃗\vec p on q⃗\vec q at that value.

Step 1. Compute a⃗⋅b⃗\vec a\cdot\vec b using the component formula: a⃗⋅b⃗=(1)(3)+(2)(−6)+(−2)(−2)=3−12+4=−5\vec a\cdot\vec b = (1)(3)+(2)(-6)+(-2)(-2) = 3-12+4 = -5.

Step 2. Compute the magnitudes: ∣a⃗∣=12+22+(−2)2=9=3|\vec a| = \sqrt{1^2+2^2+(-2)^2} = \sqrt{9} = 3 and ∣b⃗∣=32+(−6)2+(−2)2=49=7|\vec b| = \sqrt{3^2+(-6)^2+(-2)^2} = \sqrt{49} = 7.

Step 3. Apply the angle formula: cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−53×7=−521\cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \dfrac{-5}{3\times 7} = \dfrac{-5}{21}, so θ=cos⁡−1 ⁣(−521)\theta = \cos^{-1}\!\left(\dfrac{-5}{21}\right). The negative cosine tells us θ\theta is obtuse before any calculator is touched.

Step 4. For perpendicularity, set p⃗⋅q⃗=0\vec p\cdot\vec q = 0: (2)(1)+(λ)(−2)+(1)(3)=5−2λ=0(2)(1)+(\lambda)(-2)+(1)(3) = 5-2\lambda = 0, giving λ=52\lambda = \dfrac{5}{2}.

Step 5. With λ=52\lambda = \dfrac52, p⃗⋅q⃗=0\vec p\cdot\vec q = 0 by construction, so the projection of p⃗\vec p on q⃗\vec q is projq⃗p⃗=p⃗⋅q⃗∣q⃗∣=0∣q⃗∣=0\text{proj}_{\vec q}\vec p = \dfrac{\vec p\cdot\vec q}{|\vec q|} = \dfrac{0}{|\vec q|} = 0 — a perpendicular vector always has zero projection along the direction it is perpendicular to.

Answer: θ=cos⁡−1(−5/21)\theta = \cos^{-1}(-5/21) for the first pair; λ=5/2\lambda = 5/2 makes p⃗⊥q⃗\vec p \perp \vec q, and the projection of p⃗\vec p on q⃗\vec q is then 00.

Common traps

  • Forgetting a⃗⋅b⃗\vec a\cdot\vec b is a scalar: chaining it as if it were a vector (writing (a⃗⋅b⃗)⋅c⃗(\vec a\cdot\vec b)\cdot\vec c and expecting another dot product) is not a valid vector operation.
  • Reading a⃗⋅b⃗=0\vec a\cdot\vec b = 0 as "parallel" instead of "perpendicular" — zero dot product is exactly the perpendicularity condition.
  • Sign slips inside a1b1+a2b2+a3b3a_1b_1+a_2b_2+a_3b_3 when a component is negative — evaluate each product separately before summing.
  • Confusing the scalar projection a⃗⋅b⃗∣a⃗∣\frac{\vec a\cdot\vec b}{|\vec a|} (a number) with the projection vector, which additionally needs multiplying by the unit vector a^\hat a; ISC questions usually want the scalar unless stated otherwise.

Ask the tutor: want more practice finding an unknown vector component from a perpendicularity or angle condition — ask the AI tutor for a few extra scalar-product problems.

Why is the dot product called scalar even though it comes from two vectors?

Because the output is a single real number with no direction attached, unlike vector addition or the cross product, which both produce another vector. Geometrically it is ∣a⃗∣|\vec a| times a length (the projection of b⃗\vec b onto a⃗\vec a's direction), and a length times a length is just a number. That is also why you can never dot-multiply three vectors in a chain like a⃗⋅b⃗⋅c⃗\vec a\cdot\vec b\cdot\vec c without first deciding which pair to combine and what to do with the resulting scalar.

How do I remember which formula to use, the cos theta one or the component one?

Use the geometric definition ∣a⃗∣∣b⃗∣cos⁡θ|\vec a||\vec b|\cos\theta when the angle is already known or is what you are solving for. Use the component formula a1b1+a2b2+a3b3a_1b_1+a_2b_2+a_3b_3 whenever the vectors are given in i^,j^,k^\hat i,\hat j,\hat k form and you need a quick number. In practice most ISC problems give components, so you compute a⃗⋅b⃗\vec a\cdot\vec b from components first, then plug it into cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} if an angle is asked for.

If cos theta comes out negative, does that mean I made an arithmetic error?

Not necessarily — a negative dot product simply means the angle between the vectors is obtuse (between 90∘90^\circ and 180∘180^\circ), which is a perfectly valid geometric situation. In the worked example, cos⁡θ=−5/21\cos\theta = -5/21 is negative because the two given vectors point in broadly opposite general directions. Only double-check your arithmetic if the magnitude of the cosine value exceeds 11, since that is genuinely impossible and signals a computational mistake.

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