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

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\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.

GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$\vec a = \hat i - 2\hat j + 3\hat k$
(a) $\dfrac{1}{\sqrt{14}}(5\hat i-10\hat j+15\hat k)$
(b) $-\dfrac{1}{\sqrt{14}}(5\hat i-10\hat j+15\hat k)$
(c) $\dfrac{1}{\sqrt{14}}(\hat i-2\hat j+3\hat k)$
(d) $-\dfrac{1}{\sqrt{14}}(\hat i-2\hat j+3\hat k)$
›Reveal solutionSolution

Find the unit vector along $\vec a$ and scale it to magnitude 5.

$|\vec a|=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}$ . Unit vector $=\dfrac{1}{\sqrt{14}}(\hat i-2\hat j+3\hat k)$ .

Magnitude-5 vector $=5\cdot\dfrac{1}{\sqrt{14}}(\hat i-2\hat j+3\hat k)=\dfrac{1}{\sqrt{14}}(5\hat i-10\hat j+15\hat k)$ .

✓Final answer
$\dfrac{1}{\sqrt{14}}(5\hat i-10\hat j+15\hat k)$
GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$\vec{a}$
(a) $a = \dfrac{1}{|\lambda|}$
(b) $a = |\lambda|$
(c) $\lambda = -1$
(d) $\lambda = 1$
›Reveal solutionSolution

A unit vector has magnitude 1, so $|\lambda\vec a|=|\lambda|\,a=1$ .

$|\lambda\vec a|=|\lambda||\vec a|=|\lambda|a=1 \Rightarrow a=\dfrac{1}{|\lambda|}$ .

✓Final answer
$a=\dfrac{1}{|\lambda|}$
GSEB Higher Secondary Certificate (HSC) Examination 2022Set ANNUAL1 markMCQ
$5\hat{i} - \hat{j} + 2\hat{k}$
(a) $\frac{40}{\sqrt{30}}\hat{i} - \frac{8}{\sqrt{30}}\hat{j} + \frac{16}{\sqrt{30}}\hat{k}$
(b) $40\hat{i} - 8\hat{j} + 16\hat{k}$
(c) $\frac{4}{3}\hat{i} - \frac{8}{30}\hat{j} + \frac{16}{30}\hat{k}$
(d) None
›Reveal solutionSolution

Scale the unit vector $\hat v = \vec v/|\vec v|$ by the required magnitude 8.

$\vec v = 5\hat i - \hat j + 2\hat k$ , $|\vec v| = \sqrt{25+1+4} = \sqrt{30}.$

Required vector $= 8\,\hat v = \dfrac{8}{\sqrt{30}}(5\hat i - \hat j + 2\hat k) = \dfrac{40}{\sqrt{30}}\hat i - \dfrac{8}{\sqrt{30}}\hat j + \dfrac{16}{\sqrt{30}}\hat k.$

✓Final answer
$\tfrac{40}{\sqrt{30}}\hat{i}-\tfrac{8}{\sqrt{30}}\hat{j}+\tfrac{16}{\sqrt{30}}\hat{k}$
GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$\bar{x} = (2, 3, \sqrt3)$
(a) $\left(\frac{1}{2}, \frac{3}{2}, \frac{\sqrt3}{4}\right)$
(b) $\left(\frac{1}{4}, \frac{3}{4}, \frac{\sqrt3}{4}\right)$
(c) $\left(\frac{1}{2}, \frac{3}{4}, \frac{\sqrt3}{4}\right)$
(d) $\left(\frac{1}{4}, \frac{3}{2}, \frac{\sqrt3}{2}\right)$
›Reveal solutionSolution

A unit vector along $\bar x$ is $\bar x/|\bar x|$ .

$|\bar x| = \sqrt{2^2+3^2+(\sqrt3)^2} = \sqrt{4+9+3} = \sqrt{16}=4$ .

Unit vector $= \frac{1}{4}(2,3,\sqrt3) = \left(\frac{1}{2},\frac{3}{4},\frac{\sqrt3}{4}\right)$ .

✓Final answer
$\left(\frac12,\frac34,\frac{\sqrt3}{4}\right)$
GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$3\sqrt{21}$
(a) $(-12, -3, 6)$
(b) $\frac{1}{\sqrt{21}}(12, 3, -6)$
(c) $(12, 3, -6)$
(d) $\frac{1}{\sqrt{21}}(4, 1, -2)$
›Reveal solutionSolution

Multiply the unit vector in the given direction by the required magnitude.

$|(4,1,-2)|=\sqrt{16+1+4}=\sqrt{21}$ , so the unit vector is $\dfrac{1}{\sqrt{21}}(4,1,-2)$ .

Required vector $=3\sqrt{21}\cdot\dfrac{1}{\sqrt{21}}(4,1,-2)=3(4,1,-2)=(12,3,-6)$ .

✓Final answer
$(12,3,-6)$

←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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