MHT-CET Maths · Vectors

Magnitude, Components, and Unit Vectors

How to write a vector in î ĵ k̂ component form, measure its length, build a unit vector pointing in any direction, and — the MHT-CET workhorse — find the magnitude of a sum of vectors from given angles or perpendicularity conditions.

All Vectors notesMHT-CET Maths strategy

Why this matters

This is the on-ramp to the whole Vectors chapter: every later technique (dot product, cross product, scalar triple product) starts by writing vectors in component form and reading off a magnitude. Across the 9 PYQs here, ONE shape dominates — finding the length of a combination like a + b + c by expanding |a + b + c|² = Σ|·|² + 2Σ(a·b) and using the angle or perpendicularity data to evaluate the dot-product cross terms. Master that single identity (plus the unit-vector-along-a-diagonal construction) and you have the entire subtopic; the difficulty mix is MODERATE-to-HARD, but it's the same expansion every time.

Concept 1 of 6

Vectors and component form

Intuition

Every vector in 3-D space can be built from three standard perpendicular unit vectors i^,j^,k^\hat{i}, \hat{j}, \hat{k} (pointing along the x,y,zx, y, z axes). Writing a vector as a1i^+a2j^+a3k^a_1\hat{i} + a_2\hat{j} + a_3\hat{k} is just saying "go a1a_1 steps along xx, a2a_2 along yy, a3a_3 along zz."

Definition

A vector a⃗\vec{a} is written in component form as a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, where:

  • a1,a2,a3a_1, a_2, a_3 are its components (also called scalar components) along the axes,
  • i^,j^,k^\hat{i}, \hat{j}, \hat{k} are the standard basis — mutually perpendicular unit vectors of length 1.

A null (zero) vector 0⃗\vec{0} has all components zero and no direction; a unit vector has magnitude 1; collinear (parallel) vectors are scalar multiples of each other.

Component form

a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}
  • a1,a2,a3a_1, a_2, a_3components along i^,j^,k^\hat{i}, \hat{j}, \hat{k}
  • i^,j^,k^\hat{i}, \hat{j}, \hat{k}standard perpendicular unit vectors

Diagram · component form (drag to rotate)

xyzv

Step along x (2.4 î), then y ( 1.6 ĵ), then z ( 2.0 k̂) to reach the tip: v = 2.4 î + 1.6 ĵ + 2.0 k̂. Any vector is the sum of its axis components, and |v| = √(x² + y² + z²).

Worked example

Write the vector from the origin to the point P(−1,5,2)P(-1, 5, 2) in component form, and state its components.
Practice this concept4 quick reps

A missing axis means a zero component, not a 2-D vector

3i^−4k^3\hat{i} - 4\hat{k} lives in 3-D with a2=0a_2 = 0. When you square components for a magnitude, the missing term contributes 02=00^2 = 0 — don't drop the slot or miscount the axes.

Concept 2 of 6

Magnitude of a vector and distance between two points

Intuition

The magnitude of a vector is its length — extended Pythagoras on its components. The displacement from AA to BB is AB→=b⃗−a⃗\overrightarrow{AB} = \vec{b} - \vec{a} (head minus tail), and the distance ABAB is the magnitude of that displacement.

Definition

For a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}, the magnitude is ∣a⃗∣=a12+a22+a32|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}. For points A,BA, B with position vectors a⃗,b⃗\vec{a}, \vec{b}: the displacement is AB→=b⃗−a⃗\overrightarrow{AB} = \vec{b} - \vec{a} and the distance is AB=∣b⃗−a⃗∣AB = |\vec{b} - \vec{a}|.

Magnitude and distance

∣a⃗∣=a12+a22+a32AB=∣b⃗−a⃗∣|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \qquad AB = |\vec{b} - \vec{a}|
  • a1,a2,a3a_1, a_2, a_3components of a⃗\vec{a}
  • a⃗,b⃗\vec{a}, \vec{b}position vectors of AA and BB

Diagram · magnitude = √(x² + y²)

x = 4y = 3|v| = 5

The components x and y are the legs of a right triangle; the vector is the hypotenuse, so |v| = √(x² + y²) = √(16 + 9) = 5. In 3-D the same idea adds a third leg: |v| = √(x² + y² + z²).

Worked example

Points AA and BB have position vectors a⃗=i^+2j^+2k^\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k} and b⃗=5i^+5j^+14k^\vec{b} = 5\hat{i} + 5\hat{j} + 14\hat{k}. Find the distance ABAB.
Practice this conceptself-check · 4 quick reps

AB→=b⃗−a⃗\overrightarrow{AB} = \vec{b} - \vec{a} — head minus tail

Reversing it gives BA→\overrightarrow{BA}. The distance (magnitude) is the same either way, but the direction flips — and direction matters the moment the result feeds a dot product or an angle.

Magnitude needs every component squared

For 3i^+4j^3\hat{i} + 4\hat{j}, the answer is 9+16=5\sqrt{9 + 16} = 5, not 3+4=73 + 4 = 7. Add the SQUARES, then take ONE square root at the end.

Concept 3 of 6

Unit vector along a given direction

Intuition

To keep a vector's direction but throw away its length, divide it by its own magnitude — the result is a unit vector (length 1). To go the other way, multiply a unit vector by whatever magnitude you want.

Definition

For any non-zero a⃗\vec{a}, the unit vector along it is a^=a⃗∣a⃗∣\hat{a} = \dfrac{\vec{a}}{|\vec{a}|}. A vector of magnitude rr in the same direction as a⃗\vec{a} is r a^=r∣a⃗∣ a⃗r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}. Every unit vector's components square-sum to 1.

Unit vector

a^=a⃗∣a⃗∣r a^=r∣a⃗∣ a⃗\hat{a} = \dfrac{\vec{a}}{|\vec{a}|} \qquad r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}
  • a^\hat{a}unit vector along a⃗\vec{a}
  • ∣a⃗∣|\vec{a}|magnitude of a⃗\vec{a}
  • rrdesired magnitude of the scaled vector

Visualization · slide k, scale the vector

vk·v
k·v = (4.0, 2.0)|k·v| = |k|·|v| = 4.47

Multiplying by k scales the length by |k| and keeps the same line. k > 1 stretches, 0 < k < 1 shrinks, k < 0 flips to the opposite direction, and k = 0 collapses it to the zero vector.

Worked example

Find a vector of magnitude 10 in the direction of a⃗=4i^−3j^\vec{a} = 4\hat{i} - 3\hat{j}.
Practice this conceptself-check · 4 quick reps

Divide by the magnitude, don't subtract it

The unit vector is a⃗/∣a⃗∣\vec{a}/|\vec{a}| — scale every component by the same 1/∣a⃗∣1/|\vec{a}|. A common slip is normalising only one component or dividing by the wrong length.

Concept 4 of 6

Magnitude of a sum from angles or perpendicularity

Intuition

To find the length of a combination like a⃗+b⃗+c⃗\vec{a} + \vec{b} + \vec{c}, never compute components — instead SQUARE the magnitude. The square expands into the sum of each vector's squared length plus twice every pairwise dot product. Here we treat a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta as a recalled Std-11 fact (the dot product gets its full treatment on a later page); all you need is that perpendicular vectors have dot product 0, and that a given angle fixes each cross term.

Definition

Expand the square of the magnitude:

∣a⃗+b⃗+c⃗∣2=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2+2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})
Each cross term is a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta. Two ways the cross terms get pinned down:

  • Given an angle θ\theta between each pair → each dot product is ∣⋅∣∣⋅∣cos⁡θ|\cdot||\cdot|\cos\theta.
  • Perpendicularity conditions like a⃗⊥(b⃗+c⃗)\vec{a}\perp(\vec{b}+\vec{c}), b⃗⊥(c⃗+a⃗)\vec{b}\perp(\vec{c}+\vec{a}), c⃗⊥(a⃗+b⃗)\vec{c}\perp(\vec{a}+\vec{b}) → adding the three gives 2(a⃗⋅b⃗+b⃗⋅c⃗+c⃗⋅a⃗)=02(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0, so ALL cross terms vanish together and ∣a⃗+b⃗+c⃗∣2=∣a⃗∣2+∣b⃗∣2+∣c⃗∣2|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2.

Magnitude of a sum (squared)

∣a⃗+b⃗+c⃗∣2=∑∣a⃗∣2+2∑(a⃗⋅b⃗),a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ|\vec{a} + \vec{b} + \vec{c}|^2 = \sum|\vec{a}|^2 + 2\sum(\vec{a}\cdot\vec{b}), \qquad \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta
  • ∑∣a⃗∣2\sum|\vec{a}|^2∣a⃗∣2+∣b⃗∣2+∣c⃗∣2|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2
  • ∑(a⃗⋅b⃗)\sum(\vec{a}\cdot\vec{b})all three pairwise dot products
  • θ\thetaangle between the pair of vectors in a dot product

Visualization · add two vectors tip-to-tail

aba+b
a + b = (5, 4)|a + b| = 6.40|a| + |b| = 7.29

The dashed amber arrow is b moved to the head of a — the resultant (indigo) runs from the shared tail to that head, which is also the diagonal of the parallelogram. Notice |a + b| equals |a| + |b| only when a and b point the same way.

Worked example

If ∣a⃗∣=1|\vec{a}| = 1, ∣b⃗∣=2|\vec{b}| = 2, ∣c⃗∣=2|\vec{c}| = 2 and each angle between pairs is 60∘60^\circ, find ∣a⃗+b⃗+c⃗∣|\vec{a} + \vec{b} + \vec{c}|.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4VectorsMODERATE
∣a⃗∣=2,∣b⃗∣=3,∣c⃗∣=5|\vec{a}| = 2, |\vec{b}| = 3, |\vec{c}| = 5 and each angle between pairs is 60°60°, then the value of ∣a⃗+b⃗+c⃗∣|\vec{a} + \vec{b} + \vec{c}| is

[Q104 · 14th May Shift 2 · 2024]

Add the cross terms with the factor of 2

The expansion is ∑∣⋅∣2+2∑(a⃗⋅b⃗)\sum|\cdot|^2 + 2\sum(\vec{a}\cdot\vec{b}). Forgetting the 22 halves every cross term — a frequent wrong answer when the angle version has non-zero dot products.

Perpendicularity conditions cancel ALL cross terms at once

The three conditions a⃗⊥(b⃗+c⃗)\vec{a}\perp(\vec{b}+\vec{c}) etc. don't say each individual dot product is zero — they say their SUM is zero. That's all you need: the whole 2∑(a⃗⋅b⃗)2\sum(\vec{a}\cdot\vec{b}) term drops, leaving just ∑∣⋅∣2\sqrt{\sum|\cdot|^2}.

Take the square root at the very end

You compute ∣a⃗+b⃗+c⃗∣2|\vec{a}+\vec{b}+\vec{c}|^2 first. For magnitudes 1,8,41, 8, 4 with zero cross terms it's 8181, and the answer is 81=9\sqrt{81} = 9 — NOT 8181. The distractor that leaves the squared value un-rooted is the classic trap.
Drill 3 more on magnitude of a sum from angles or perpendicularity

Concept 5 of 6

Magnitude of a sum with projection-equality and a perpendicular pair

Intuition

A second flavour of the same expansion. Instead of stating angles directly, the question gives an indirect handle: "the projection of b⃗\vec{b} along a⃗\vec{a} equals the projection of c⃗\vec{c} along a⃗\vec{a}" forces (b⃗−c⃗)⋅a⃗=0(\vec{b}-\vec{c})\cdot\vec{a} = 0, and "b⃗⊥c⃗\vec{b}\perp\vec{c}" forces b⃗⋅c⃗=0\vec{b}\cdot\vec{c} = 0. Feed both into the squared-magnitude expansion and the cross terms collapse.

Definition

Two facts unlock these:

  • Equal projections along a⃗\vec{a}: projection of b⃗\vec{b} on a⃗\vec{a} is a⃗⋅b⃗∣a⃗∣\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|}; equating it to c⃗\vec{c}'s projection gives a⃗⋅b⃗=a⃗⋅c⃗\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c}, i.e. (b⃗−c⃗)⋅a⃗=0(\vec{b}-\vec{c})\cdot\vec{a} = 0.
  • Perpendicular pair: b⃗⊥c⃗⇒b⃗⋅c⃗=0\vec{b}\perp\vec{c}\Rightarrow\vec{b}\cdot\vec{c} = 0.

Then expand the target, grouping the difference: ∣a⃗+b⃗−c⃗∣2=∣a⃗∣2+∣b⃗−c⃗∣2+2a⃗⋅(b⃗−c⃗)|\vec{a}+\vec{b}-\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}-\vec{c}|^2 + 2\vec{a}\cdot(\vec{b}-\vec{c}). The last term is 0 (equal projections), and ∣b⃗−c⃗∣2=∣b⃗∣2+∣c⃗∣2|\vec{b}-\vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 (perpendicular pair).

Grouped expansion with a perpendicular pair

∣a⃗+b⃗−c⃗∣2=∣a⃗∣2+∣b⃗−c⃗∣2+2 a⃗⋅(b⃗−c⃗)|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b} - \vec{c}|^2 + 2\,\vec{a}\cdot(\vec{b} - \vec{c})
  • a⃗⋅(b⃗−c⃗)=0\vec{a}\cdot(\vec{b}-\vec{c}) = 0from equal projections of b⃗,c⃗\vec{b}, \vec{c} along a⃗\vec{a}
  • ∣b⃗−c⃗∣2=∣b⃗∣2+∣c⃗∣2|\vec{b}-\vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2from b⃗⊥c⃗\vec{b}\perp\vec{c}

Worked example

Let ∣a⃗∣=3|\vec{a}| = 3, ∣b⃗∣=6|\vec{b}| = 6, ∣c⃗∣=6|\vec{c}| = 6. If the projection of b⃗\vec{b} on a⃗\vec{a} equals the projection of c⃗\vec{c} on a⃗\vec{a}, and b⃗⊥c⃗\vec{b}\perp\vec{c}, find ∣a⃗+b⃗−c⃗∣|\vec{a} + \vec{b} - \vec{c}|.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 5VectorsHARD
Let the vectors a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} be such that ∣a⃗∣=2,∣b⃗∣=4|\vec{a}|=2, |\vec{b}|=4 and ∣c⃗∣=4|\vec{c}|=4. If the projection of b⃗\vec{b} on a⃗\vec{a} is equal to the projection of c⃗\vec{c} on a⃗\vec{a} and b⃗\vec{b} is perpendicular to c⃗\vec{c}, then the value of ∣a⃗+b⃗−c⃗∣|\vec{a}+\vec{b}-\vec{c}| is

[Q107 · 16th May Shift 1 · 2023]

"Equal projections along a⃗\vec{a}" means (b⃗−c⃗)⋅a⃗=0(\vec{b}-\vec{c})\cdot\vec{a} = 0, not b⃗=c⃗\vec{b} = \vec{c}

The projections being equal only forces a⃗⋅b⃗=a⃗⋅c⃗\vec{a}\cdot\vec{b} = \vec{a}\cdot\vec{c} — b⃗\vec{b} and c⃗\vec{c} can still differ wildly. Use it to kill exactly the a⃗\vec{a}-cross term, nothing more.

Watch the sign on the vector you subtract

In ∣a⃗+b⃗−c⃗∣|\vec{a}+\vec{b}-\vec{c}|, the c⃗\vec{c} term is subtracted, but ∣b⃗−c⃗∣2|\vec{b}-\vec{c}|^2 still ADDS the squared lengths when b⃗⊥c⃗\vec{b}\perp\vec{c} (the cross term −2b⃗⋅c⃗-2\vec{b}\cdot\vec{c} is zero). The minus sign only matters through the dot product, which vanishes here.
Drill 3 more on magnitude of a sum with projection-equality and a perpendicular pair

Concept 6 of 6

Unit vector parallel to a parallelogram's diagonal

Intuition

If two adjacent sides of a parallelogram are a⃗\vec{a} and b⃗\vec{b}, one diagonal runs from the shared vertex to the opposite corner — and by the triangle/parallelogram law that diagonal IS a⃗+b⃗\vec{a} + \vec{b}. To get a unit vector along it, add the sides, then divide by the resulting magnitude.

Definition

For a parallelogram with adjacent sides a⃗\vec{a} and b⃗\vec{b} from a common vertex:

  • The diagonal through that vertex is d⃗=a⃗+b⃗\vec{d} = \vec{a} + \vec{b}.
  • The other diagonal is a⃗−b⃗\vec{a} - \vec{b} (or b⃗−a⃗\vec{b} - \vec{a}).

The unit vector parallel to the diagonal is d^=a⃗+b⃗∣a⃗+b⃗∣\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}.

Unit vector along a diagonal

d^=a⃗+b⃗∣a⃗+b⃗∣\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}
  • a⃗,b⃗\vec{a}, \vec{b}adjacent sides from the shared vertex
  • a⃗+b⃗\vec{a} + \vec{b}the diagonal through that vertex

Visualization · add two vectors tip-to-tail

aba+b
a + b = (5, 4)|a + b| = 6.40|a| + |b| = 7.29

The dashed amber arrow is b moved to the head of a — the resultant (indigo) runs from the shared tail to that head, which is also the diagonal of the parallelogram. Notice |a + b| equals |a| + |b| only when a and b point the same way.

Worked example

Two adjacent sides of a parallelogram are a⃗=i^+2j^+2k^\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k} and b⃗=3i^+2j^−6k^\vec{b} = 3\hat{i} + 2\hat{j} - 6\hat{k}. Find the unit vector parallel to the diagonal a⃗+b⃗\vec{a} + \vec{b}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 6VectorsMODERATE
Two adjacent sides of a parallelogram are 2i^−4j^+5k^2\hat{i}-4\hat{j}+5\hat{k} and i^−2j^−3k^\hat{i}-2\hat{j}-3\hat{k}, then the unit vector parallel to its diagonal is

[Q134 · 9th May Shift 2 · 2024]

Diagonal a⃗+b⃗\vec{a}+\vec{b}, not a⃗−b⃗\vec{a}-\vec{b}

A parallelogram has two diagonals: a⃗+b⃗\vec{a} + \vec{b} (through the shared vertex) and a⃗−b⃗\vec{a} - \vec{b} (the other one). The question usually wants the SUM; reading it as the difference gives a different unit vector entirely.

Normalise the diagonal's own magnitude, not a stray 77\sqrt{77}

Compute ∣a⃗+b⃗∣|\vec{a}+\vec{b}| AFTER adding — for 3i^−6j^+2k^3\hat{i}-6\hat{j}+2\hat{k} that is 9+36+4=7\sqrt{9+36+4} = 7. Distractors often divide by an unrelated magnitude (like 77\sqrt{77} from 49+28\sqrt{49 + 28}); always re-square the summed components.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

  • Vectors and component form

    Component form

    a⃗=a1i^+a2j^+a3k^\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}
  • Magnitude of a vector and distance between two points

    Magnitude and distance

    ∣a⃗∣=a12+a22+a32AB=∣b⃗−a⃗∣|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \qquad AB = |\vec{b} - \vec{a}|
  • Unit vector along a given direction

    Unit vector

    a^=a⃗∣a⃗∣r a^=r∣a⃗∣ a⃗\hat{a} = \dfrac{\vec{a}}{|\vec{a}|} \qquad r\,\hat{a} = \dfrac{r}{|\vec{a}|}\,\vec{a}
  • Magnitude of a sum from angles or perpendicularity

    Magnitude of a sum (squared)

    ∣a⃗+b⃗+c⃗∣2=∑∣a⃗∣2+2∑(a⃗⋅b⃗),a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ|\vec{a} + \vec{b} + \vec{c}|^2 = \sum|\vec{a}|^2 + 2\sum(\vec{a}\cdot\vec{b}), \qquad \vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta
  • Magnitude of a sum with projection-equality and a perpendicular pair

    Grouped expansion with a perpendicular pair

    ∣a⃗+b⃗−c⃗∣2=∣a⃗∣2+∣b⃗−c⃗∣2+2 a⃗⋅(b⃗−c⃗)|\vec{a} + \vec{b} - \vec{c}|^2 = |\vec{a}|^2 + |\vec{b} - \vec{c}|^2 + 2\,\vec{a}\cdot(\vec{b} - \vec{c})
  • Unit vector parallel to a parallelogram's diagonal

    Unit vector along a diagonal

    d^=a⃗+b⃗∣a⃗+b⃗∣\hat{d} = \dfrac{\vec{a} + \vec{b}}{|\vec{a} + \vec{b}|}

Watch out for (11)

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