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.

JKBOSE Class 12 Annual Regular Examination 2025Set SZ2 marks
$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$
›Reveal solutionSolution

Divide the vector by its magnitude; the unit vector is $\dfrac{1}{\sqrt6}(\hat{i}+\hat{j}+2\hat{k})$ .

For $\vec{a} = \hat{i}+\hat{j}+2\hat{k}$ , the magnitude is:

$|\vec{a}| = \sqrt{1^2+1^2+2^2} = \sqrt{1+1+4} = \sqrt{6}$

The unit vector in the direction of $\vec{a}$ is:

$\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt6} = \frac{1}{\sqrt6}\hat{i} + \frac{1}{\sqrt6}\hat{j} + \frac{2}{\sqrt6}\hat{k}$

✓Final answer
$\hat{a} = \dfrac{1}{\sqrt6}\hat{i}+\dfrac{1}{\sqrt6}\hat{j}+\dfrac{2}{\sqrt6}\hat{k}$
JKBOSE Class 12 Annual Regular Examination 2023Set ANNUAL2 marks
$5\hat{i} - \hat{j} + 2\hat{k}$
›Reveal solutionSolution

Find the unit vector along $\vec a = 5\hat i - \hat j + 2\hat k$ , then scale it to magnitude $8$ .

Let $\vec a = 5\hat i - \hat j + 2\hat k$ .

Step 1 — magnitude of $\vec a$ .

$|\vec a| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25+1+4} = \sqrt{30}$

Step 2 — unit vector along $\vec a$ .

$\hat a = \frac{\vec a}{|\vec a|} = \frac{5\hat i - \hat j + 2\hat k}{\sqrt{30}}$

Step 3 — scale to magnitude 8. The vector of magnitude $8$ in the direction of $\vec a$ is $8\hat a$ :

$8\hat a = \frac{8}{\sqrt{30}}\left(5\hat i - \hat j + 2\hat k\right) = \frac{40}{\sqrt{30}}\hat i - \frac{8}{\sqrt{30}}\hat j + \frac{16}{\sqrt{30}}\hat k$

Rationalising (multiplying numerator and denominator by $\sqrt{30}$ , and using $\sqrt{30}/30 = 1/\sqrt{30}$ , i.e. $40/\sqrt{30} = 40\sqrt{30}/30 = 4\sqrt{30}/3$ , similarly for the others):

$8\hat a = \frac{4\sqrt{30}}{3}\hat i - \frac{4\sqrt{30}}{15}\hat j + \frac{8\sqrt{30}}{15}\hat k$

✓Final answer
$\dfrac{8}{\sqrt{30}}\left(5\hat i - \hat j + 2\hat k\right) = \dfrac{4\sqrt{30}}{3}\hat i - \dfrac{4\sqrt{30}}{15}\hat j + \dfrac{8\sqrt{30}}{15}\hat k$
JKBOSE Class 12 Annual Regular Examination 2019Set WZ2 marks
$\overrightarrow{PQ}$
›Reveal solutionSolution

Find $\vec{PQ}=Q-P$ , then divide by its magnitude.

$P(1,2,3)$ , $Q(4,5,6)$

$\vec{PQ} = (4-1,5-2,6-3) = (3,3,3)$

$|\vec{PQ}| = \sqrt{9+9+9}=\sqrt{27}=3\sqrt3$

Unit vector $=\dfrac{(3,3,3)}{3\sqrt3} = \dfrac{1}{\sqrt3}(1,1,1) = \dfrac{1}{\sqrt3}(\hat i+\hat j+\hat k)$

✓Final answer
$\overrightarrow{PQ}$
JKBOSE Class 12 Annual Regular Examination 2018Set WZ2 marks
$\vec{a} = 2\hat{i}+3\hat{j}+\hat{k}$
›Reveal solutionSolution

Divide the vector by its own magnitude.

Given $\vec a=2\hat i+3\hat j+\hat k$ .

$|\vec a|=\sqrt{2^2+3^2+1^2}=\sqrt{4+9+1}=\sqrt{14}$

$\hat a=\dfrac{\vec a}{|\vec a|}=\dfrac{2\hat i+3\hat j+\hat k}{\sqrt{14}}$

✓Final answer
$\hat a = \dfrac{2}{\sqrt{14}}\hat i+\dfrac{3}{\sqrt{14}}\hat j+\dfrac{1}{\sqrt{14}}\hat k$

←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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