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.

AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\bar a=2\bar i-3\bar j+5\bar k$
›Reveal solutionSolution
Compute $\bar a-\bar b$ , normalize it, then scale to magnitude 28; the answer is $12\bar i-24\bar j+8\bar k$ .

Concept and Intuition

Any vector of a required magnitude $m$ in the direction of a vector $\bar v$ is $m\cdot\dfrac{\bar v}{|\bar v|}$ — scale the unit vector along $\bar v$ by $m$ .

Step-by-Step Solution

$\bar a-\bar b = (2-(-1))\bar i+(-3-3)\bar j+(5-3)\bar k = 3\bar i-6\bar j+2\bar k$ .

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

Unit vector along $\bar a-\bar b$ : $\dfrac{1}{7}(3\bar i-6\bar j+2\bar k)$ .

Required vector of magnitude 28: $28\times\dfrac{1}{7}(3\bar i-6\bar j+2\bar k)=4(3\bar i-6\bar j+2\bar k)=12\bar i-24\bar j+8\bar k$ .

Common Mistakes

Computing $\bar b-\bar a$ instead of $\bar a-\bar b$ , which flips every sign.

Forgetting to divide by the magnitude before rescaling.

✓Final answer
$12\bar i-24\bar j+8\bar k$

ANSWER: B
AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
$2\bar{i}+4\bar{j}-3\bar{k}$
›Reveal solutionSolution
Tests the coplanarity condition (scalar triple product = 0) followed by unit-vector normalization; the answer is (D).

Concept and Intuition

Three vectors $\bar u,\bar v,\bar w$ are coplanar exactly when their scalar triple product $[\bar u\ \bar v\ \bar w] = \bar u\cdot(\bar v\times\bar w)$ vanishes — equivalently, the determinant formed from their rectangular components is zero. This is because a nonzero triple product measures the (signed) volume of the parallelepiped spanned by the three vectors; coplanar vectors span zero volume.

Step-by-Step Solution

Write the determinant condition for $\bar u=2\bar i+4\bar j-3\bar k$ , $\bar v=-\bar i+2\bar j+3\bar k$ , $\bar w=p\bar i-2\bar j+\bar k$ :

$\begin{vmatrix}2&4&-3\\-1&2&3\\p&-2&1\end{vmatrix}=0$

Expand along the first row: $2\big(2\cdot1-3\cdot(-2)\big)-4\big((-1)\cdot1-3p\big)+(-3)\big((-1)(-2)-2p\big)$ $=2(8)-4(-1-3p)-3(2-2p)=16+4+12p-6+6p=14+18p$ .

Set $14+18p=0\Rightarrow p=-\dfrac{14}{18}=-\dfrac{7}{9}$ .

Compute the target vector: $9p\bar i-4\bar j+4\bar k = 9\left(-\dfrac{7}{9}\right)\bar i-4\bar j+4\bar k=-7\bar i-4\bar j+4\bar k$ .

Its magnitude: $\sqrt{(-7)^2+(-4)^2+4^2}=\sqrt{49+16+16}=\sqrt{81}=9$ .

Unit vector $=\dfrac{1}{9}(-7\bar i-4\bar j+4\bar k)$ .

Common Mistakes

Forgetting to substitute $p$ back into the target vector expression $9p\bar i-4\bar j+4\bar k$ (using the original $p\bar i-2\bar j+\bar k$ instead).

Sign errors in the cofactor expansion of the $3\times3$ determinant.

✓Final answer
$\dfrac{1}{9}(-7\bar{i}-4\bar{j}+4\bar{k})$

ANSWER: D

←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

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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