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MHT-CET Maths · Vectors

Magnitude, Components, and Unit Vectors

How to write a vector in i, j, k 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 18 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 mostly MODERATE, and it's the same expansion every time.

Concept 1 of 6: Vectors and component form

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.
Practise 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

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.
Practise 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

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}.
Practise 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

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}|.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q104Moderate

Example 4 · Vectors · Magnitude, Components, and Unit Vectors

∣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

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

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}|.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q107Hard

Example 5 · Vectors · Magnitude, Components, and Unit Vectors

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

"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

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}.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 9th May Shift 2 · Q134Moderate

Example 6 · Vectors · Magnitude, Components, and Unit Vectors

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

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)

Test yourself on Vectors

20 past MHT-CET questions from this chapter, timed at 36 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Vectors chapter testDrill all 18 questions

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