Find the unit vector in the direction of sum of vectors veca=2hati-hatj+hatk and vecb=2hatj+hatk.

NCERT Exemplar · Q1

\vec{a}=2\hat{i}-\hat{j}+\hat{k}

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Concept understanding — Unit Vector Scaling

Unit Vector Scaling: From Intuition to Precision

Imagine you're drawing an arrow on graph paper. It has a direction and a length. Now suppose you want to keep the direction exactly the same, but make the arrow exactly one unit long. That's the core idea of unit vector scaling: take any vector and shrink or stretch it so its length becomes 1, without changing where it points.

The Intuition First

Think of a vector as a "directed step." A step of 3 metres north-east is a vector of length 3 in the north-east direction. To get a unit vector in the same direction, you'd take a step of exactly 1 metre north-east — scaling the original down by a factor of 3.

The key insight: direction is independent of length. A vector pointing north-east at length 5 and one at length 1 share the same direction. Unit vector scaling isolates that direction by forcing the length to be exactly 1.

The Precise Statement

\hat{\mathbf{v}} = \frac{\mathbf{v}}{\|\mathbf{v}\|}

Here $\mathbf{v}$ is any non-zero vector, $\|\mathbf{v}\|$ is its magnitude, and $\hat{\mathbf{v}}$ ("v-hat") is the unit vector in the same direction. The operation: divide each component by the vector's length.

Example in 2D

Take $\mathbf{v} = (3, 4)$ . Its length is:

\|\mathbf{v}\| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5

The unit vector is $\hat{\mathbf{v}} = \left( \frac{3}{5}, \frac{4}{5} \right)$ .

Check: $\sqrt{(3/5)^2 + (4/5)^2} = \sqrt{25/25} = 1$ . Direction unchanged — the ratio $3:4$ is preserved.

Example in 3D

For $\mathbf{v} = (2, -1, 2)$ :

\|\mathbf{v}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3

\hat{\mathbf{v}} = \left( \frac{2}{3}, -\frac{1}{3}, \frac{2}{3} \right)

Why This Matters

Unit vectors are the building blocks of direction. In physics they represent pure directions for forces, velocities, or fields; in computer graphics, camera orientations and light directions. In mathematics they simplify dot products and projections — the dot product of a unit vector with another vector directly gives the component of that vector along the unit vector's direction.

Watch out

You cannot scale the zero vector to a unit vector — division by zero is undefined. The zero vector has no direction to preserve.

The One-Line Summary

Unit vector scaling takes any non-zero vector and divides it by its own length, producing a vector of length 1 that points exactly where the original pointed.

Normalising a vector into a unit vector is a routine computation throughout the NCERT Class 12 Vector Algebra chapter and appears constantly in CBSE board numericals and JEE Main problems. "Unit vector formula class 12 with examples" is a common search among students building up to direction-cosine and dot-product questions.

$\vec{a}+\vec{b}=2\hat{i}+\hat{j}+2\hat{k}$ has magnitude $3$ , so the required unit vector is $\dfrac13(2\hat{i}+\hat{j}+2\hat{k})$ .

The idea

A unit vector points the same way as a given vector but has length $1$ . To build one you divide the vector by its own magnitude. Here the given vector is the sum $\vec{a}+\vec{b}$ , so first add, then normalise.

Add the vectors

Write $\vec{b}$ with its zero $\hat{i}$ -component: $\vec{b}=0\hat{i}+2\hat{j}+\hat{k}$ .

\vec{a}+\vec{b}=(2+0)\hat{i}+(-1+2)\hat{j}+(1+1)\hat{k}=2\hat{i}+\hat{j}+2\hat{k}

Magnitude of the sum

|\vec{a}+\vec{b}|=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=\sqrt{9}=3

Normalise

\hat{u}=\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{2\hat{i}+\hat{j}+2\hat{k}}{3}=\frac23\hat{i}+\frac13\hat{j}+\frac23\hat{k}

Check: $\left(\tfrac23\right)^2+\left(\tfrac13\right)^2+\left(\tfrac23\right)^2=\dfrac{4+1+4}{9}=1$ , confirming it is a unit vector.

✓Final answer

\dfrac{2}{3}\hat{i}+\dfrac{1}{3}\hat{j}+\dfrac{2}{3}\hat{k}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2023Set A-21 markMCQ
$\hat{i}$
›Reveal solutionSolution

The component of one vector in the direction of another is the scalar projection $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$ — divide by the magnitude of the direction vector, not of $\vec{a}$ .

Step 1 — Identify the vectors.

We want the component of $\vec{a}=\hat{i}$ in the direction of $\vec{b}=\hat{i}+\hat{j}+2\hat{k}$ .

Step 2 — Why this formula.

The component of $\vec{a}$ along $\vec{b}$ means "how much of $\vec{a}$ points the way $\vec{b}$ points". Since $\vec{a}\cdot\hat{b}=|\vec{a}|\cos\theta$ is exactly that shadow length, the scalar component is
$\text{comp}_{\vec{b}}\vec{a}=\vec{a}\cdot\hat{b}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}.$
Step 3 — Dot product.
$\vec{a}\cdot\vec{b}=(1)(1)+(0)(1)+(0)(2)=1$
Step 4 — Magnitude of $\vec{b}$ .
$|\vec{b}|=\sqrt{1^2+1^2+2^2}=\sqrt{6}$
Step 5 — Combine and rationalise.
$\text{comp}=\frac{1}{\sqrt{6}}=\frac{1}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}}=\frac{\sqrt{6}}{6}$
✓Final answer
$\dfrac{\sqrt{6}}{6}$

ANSWER: C
COMEDK 2023Set 2023-M1 markMCQ
$\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}$
›Reveal solutionSolution

$\mathbf a+\mathbf b-\mathbf c=-2\hat i+\hat j-2\hat k$ , magnitude $3$ ; the opposite-direction unit vector is $-\dfrac{(-2,1,-2)}{3}=\dfrac13(2\hat i-\hat j+2\hat k)$ .

$\mathbf a=2\hat i+\hat j-\hat k,\ \mathbf b=\hat i-\hat j,\ \mathbf c=5\hat i-\hat j+\hat k.$

$\mathbf a+\mathbf b-\mathbf c=(2+1-5)\hat i+(1-1+1)\hat j+(-1+0-1)\hat k=-2\hat i+\hat j-2\hat k.$

Magnitude $=\sqrt{4+1+4}=3.$

Unit vector in the same direction $=\dfrac13(-2\hat i+\hat j-2\hat k)$ ; in the opposite direction:
$-\dfrac13(-2\hat i+\hat j-2\hat k)=\dfrac13(2\hat i-\hat j+2\hat k).$
✓Final answer
$\frac{1}{3}(2\hat{\mathbf i}-\hat{\mathbf j}+2\hat{\mathbf k})$
COMEDK 2022Set 20221 markMCQ
$\mathbf{p}=\hat{i}+\hat{j}, \mathbf{q}=4 \hat{k}-\hat{j}$
›Reveal solutionSolution

Unit vector: (1/3)(i + 2j + 2k)

Concept: Unit vector = vector / its magnitude.

p = i + j , q = -j + 4k , r = i + k

3p = 3i + 3j

3p + q = 3i + (3 - 1) j + 4k = 3i + 2j + 4k

-2r = -2i - 2k

3p + q - 2r = (3 - 2) i + 2 j + (4 - 2) k = i + 2j + 2k

Magnitude:

|i + 2j + 2k| = sqrt(1 + 4 + 4) = sqrt(9) = 3

Unit vector:

(1/3)(i + 2j + 2k)

✓Final answer
$\frac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k})$

ANSWER: A

←Previous section · summarySummary →Next questionQ2 — If veca=hati+hatj+2hatk and vecb=2hati+hatj-2hatk, find the unit vecto…

Q.Find the unit vector in the direction of sum of vectors $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2\hat{j}+\hat{k}$ .

Concept understanding — Unit Vector Scaling

Unit Vector Scaling: From Intuition to Precision

The Intuition First

The Precise Statement

Example in 2D

Example in 3D

Why This Matters

The One-Line Summary

The idea

Add the vectors

Magnitude of the sum

Normalise

Method: Normalising a resultant into a unit vector

Steps

Common Mistakes

Q.Find the unit vector in the direction of sum of vectors a⃗=2i^−j^+k^\vec{a}=2\hat{i}-\hat{j}+\hat{k}a=2i^−j^​+k^ and b⃗=2j^+k^\vec{b}=2\hat{j}+\hat{k}b=2j^​+k^.

The idea

Add the vectors

Magnitude of the sum

Normalise

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Q.Find the unit vector in the direction of sum of vectors $\vec{a}=2\hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=2\hat{j}+\hat{k}$ .

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