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.

KEAM 2026Set eng-2026-04174 marksMCQ
$\vec{r} = -(5 + 4s)\hat{i} + (7 - 2s)\hat{j} + (3 + 4s)\hat{k}$
›Reveal solutionSolution

The line's direction is the coefficient of the parameter $s$ ; normalize it.

From $\vec r=-(5+4s)\hat i+(7-2s)\hat j+(3+4s)\hat k$ , the coefficient of $s$ is $-4\hat i-2\hat j+4\hat k$ .

Magnitude $=\sqrt{16+4+16}=6$ . Unit vector $=\dfrac{-4\hat i-2\hat j+4\hat k}{6}=\dfrac{-2\hat i-\hat j+2\hat k}{3}$ .

✓Final answer
$The correct option is (A) .$
KEAM 2023Set eng-2023-P2-B24 marksMCQ
$\vec{a} = \hat{i} - \hat{j} + 2\hat{k}$
›Reveal solutionSolution
The unit vector along a is (1/sqrt6)(i - j + 2k); scaling by 5 gives (1/sqrt6)(5i - 5j + 10k).

Concept and Intuition

A vector of magnitude m in the direction of a is m * (a/|a|).

Step-by-Step Solution

|a| = sqrt(1^2 + (-1)^2 + 2^2) = sqrt(6).

Unit vector = (1/sqrt6)(i - j + 2k).

Times 5: (5/sqrt6)(i - j + 2k) = (1/sqrt6)(5i - 5j + 10k).

Common Mistakes

Dropping the 1/sqrt6 normalisation (option A) — that vector has magnitude 5 sqrt6, not 5.

✓Final answer
$The correct option is (C) — (1/sqrt6)(5i - 5j + 10k).$

ANSWER: C
KEAM 2023Set eng-2023-P2-B24 marksMCQ
$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$
›Reveal solutionSolution
a - b = 3j - k, with magnitude sqrt10, so the unit vector is (1/sqrt10)(3j - k).

Concept and Intuition

Subtract componentwise, then divide by the magnitude to get the unit vector.

Step-by-Step Solution

a - b = (i + j + 2k) - (i - 2j + 3k) = 0i + 3j - k.

|a - b| = sqrt(0 + 9 + 1) = sqrt10.

Unit vector = (1/sqrt10)(3j - k).

Common Mistakes

Sign errors in j and k components when subtracting.

✓Final answer
$The correct option is (B) — (1/sqrt10)(3j - k).$

ANSWER: B
KEAM 2022Set eng-2022-P2-B14 marksMCQ
$\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$
›Reveal solutionSolution
The unit vector is $\frac{1}{7}(3\hat i+6\hat j-2\hat k)$ .

Concept and Intuition

Add the vectors, then divide by the magnitude of the sum to normalize.

Step-by-Step Solution

$\vec a+\vec b=(2+1)\hat i+(3+3)\hat j+(-4+2)\hat k=3\hat i+6\hat j-2\hat k$ .

$|\vec a+\vec b|=\sqrt{3^2+6^2+(-2)^2}=\sqrt{9+36+4}=\sqrt{49}=7$ .

Unit vector $=\frac{1}{7}(3\hat i+6\hat j-2\hat k)$ .

Common Mistakes

Adding component errors (especially the $\hat k$ : $-4+2=-2$ ).

Not simplifying $\sqrt{49}=7$ and leaving an irrational denominator.

✓Final answer
$\frac{1}{7}(3\hat i+6\hat j-2\hat k)$

ANSWER: C

←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

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes

Concept and Intuition

Step-by-Step Solution

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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