If a⃗ is a non-zero vector of magnitude 'a' and λ – Class 12 Mathematics | ConceptFirst

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Q. If a⃗ is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λa⃗ is a unit vector if -

(i) λ = 1

(ii) λ = −1

(iii) a = 1/|λ|

(iv) a = |λ|

Mizoram MbseMizoram Board of School Education HSSLC 2025MCQ· 1mImportance★★★★★

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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 …

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A unit vector has magnitude 1; set $|\lambda\vec a|=1$ and solve for $a=|\vec a|$ .

$\vec a$ has magnitude $a=|\vec a|$ . Then $|\lambda\vec a| = |\lambda|\,|\vec a| = |\lambda|\,a$ .

…

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

Showing the 12 most recent of 40 on this concept.

CBSE 2026Set V11 markMCQ
$\vec{a}$
(a) $\lambda=1$
(b) $\lambda=-1$
(c) $a=|\lambda|$
(d) $a=\frac{1}{|\lambda|}$
›Reveal solutionSolution

Requiring $|\lambda\vec a|=1$ gives $|\lambda|a=1$ , i.e. $a=\dfrac{1}{|\lambda|}$ ; answer (d).

The magnitude of $\lambda\vec a$ is

$|\lambda\vec a|=|\lambda|\,|\vec a|=|\lambda|\,a.$ …
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CBSE 2026Set ANNUAL1 markMCQ
$\vec{a} = \hat{i}+\hat{j}+2\hat{k}$
(a) $\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{5}}$
(b) $\frac{\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$
(c) $\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$
(d) $\frac{\hat{i}+\hat{j}+2\hat{k}}{\sqrt{6}}$
›Reveal solutionSolution

A unit vector along $\vec a$ is $\vec a/|\vec a|$ .

$|\vec a| = \sqrt{1^2+1^2+2^2}=\sqrt6$ .

Unit vector $=\dfrac{\hat i+\hat j+2\hat k}{\sqrt6}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\vec{a} = 2\hat{i} - 7\hat{j} - 3\hat{k}$
(a) $\dfrac{2\hat{i}-7\hat{j}-3\hat{k}}{\sqrt{62}}$
(b) $2\hat{i}-7\hat{j}-3\hat{k}$
(c) $\dfrac{1}{\sqrt{62}}$
(d) None of these
›Reveal solutionSolution

A unit vector in the direction of $\vec a$ is $\hat a = \dfrac{\vec a}{|\vec a|}$ .

$|\vec a| = \sqrt{2^2+(-7)^2+(-3)^2} = \sqrt{4+49+9} = \sqrt{62}$ .

…

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CBSE 2026Set ANNUAL1 mark
$\vec{a} = \hat{i}+\hat{j}+2\hat{k}$
›Reveal solutionSolution

Compute the magnitude of $\vec a$ , then divide the vector by its magnitude to get the unit vector.

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

Step 1: Find the magnitude.

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

Step 2: Divide by the magnitude.

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

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

…

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CBSE 2025Set ANNUAL1 markMCQ
$\vec a=\hat i+7\hat j+4\hat k$
(i) $\dfrac{1}{7}(2\hat i+6\hat j-3\hat k)$
(ii) $\dfrac{1}{7}(-2\hat i+6\hat j+3\hat k)$
(iii) $\dfrac{1}{7}(-2\hat i-6\hat j+3\hat k)$
(iv) $\dfrac{1}{7}(2\hat i-6\hat j+3\hat k)$
›Reveal solutionSolution

Find $\vec a - \vec b$ , its magnitude, then divide by the magnitude.

$\vec a - \vec b = (1-3)\hat i + (7-1)\hat j + (4-1)\hat k = -2\hat i + 6\hat j + 3\hat k$

Magnitude:

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

Unit vector: …

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CBSE 2025Set ANNUAL1 markMCQ
$\vec{a}$
(a) 0
(b) -1
(c) 1
(d) 3
(a) $\lambda=1$
(b) $\lambda=-1$
(c) $a=|\lambda|$
(d) $a=\dfrac{1}{|\lambda|}$
›Reveal solutionSolution

A vector is a unit vector when its magnitude equals $1$ ; use $|\lambda\vec a|=|\lambda||\vec a|$ .

Given $|\vec a|=a\ (\ne0)$ and scalar $\lambda\ (\ne0)$ , the magnitude of $\lambda\vec a$ is

$|\lambda\vec a|=|\lambda|\,|\vec a|=|\lambda|\,a.$

For $\lambda\vec a$ to be a unit vector we need $|\lambda\vec a|=1$ :

$|\lambda|\,a=1\ \Rightarrow\ a=\frac{1}{|\lambda|}.$

Checking the options: (a) $\lambda=1$ and (b) $\lambda=-1$ force a specific $\lambda$ but ignore $a$ ; (c) $a=|\lambda|$ is the reciprocal of the correct relation. Only (d) $a=\frac{1}{|\lambda|}$ works.

…

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CBSE 2025Set ANNUAL1 markMCQ
$\vec a=\hat i+\hat j+2\hat k$
(a) $\dfrac{1}{2}\hat i+\dfrac{1}{2}\hat j+\hat k$
(b) $\dfrac{1}{\sqrt 3}\hat i+\dfrac{1}{\sqrt 3}\hat j+\dfrac{2}{\sqrt 3}\hat k$
(c) $\dfrac{1}{\sqrt 5}\hat i+\dfrac{1}{\sqrt 5}\hat j+\dfrac{2}{\sqrt 5}\hat k$
(d) $\dfrac{1}{\sqrt 6}\hat i+\dfrac{1}{\sqrt 6}\hat j+\dfrac{2}{\sqrt 6}\hat k$
›Reveal solutionSolution

Unit vector $=\vec a/|\vec a|$ ; here $|\vec a|=\sqrt6$ .

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

$|\vec a|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt6$

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

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CBSE 2025Set ANNUAL1 markMCQ
$Q. If a⃗ is a non-zero vector of magnitude 'a' and λ is a non-zero scalar, then λa⃗ is a unit vector if -$
(i) λ = 1
(ii) λ = −1
(iii) a = 1/|λ|
(iv) a = |λ|
›Reveal solutionSolution

A unit vector has magnitude 1; set $|\lambda\vec a|=1$ and solve for $a=|\vec a|$ .

$\vec a$ has magnitude $a=|\vec a|$ . Then $|\lambda\vec a| = |\lambda|\,|\vec a| = |\lambda|\,a$ .

…

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CBSE 2024Set A11 markMCQ
$\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$
(a) $\dfrac{\hat{i} - \hat{j} - 2\hat{k}}{6}$
(b) $\dfrac{\hat{i} + \hat{j} + 2\hat{k}}{\sqrt{6}}$
(c) $\dfrac{\hat{i} - \hat{j} + 2\hat{k}}{6}$
(d) $\dfrac{\hat{i} + \hat{j} - 2\hat{k}}{\sqrt{6}}$
›Reveal solutionSolution

Divide $\vec a$ by its magnitude $\sqrt6$ to get the unit vector, so (b). …

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CBSE 2024Set ANNUAL1 mark
$x(\hat{i}+\hat{j}+\hat{k})$
›Reveal solutionSolution

Set the magnitude of $x(\hat i+\hat j+\hat k)$ equal to 1 and solve for x.

The vector is $x\hat i + x\hat j + x\hat k$ . Its magnitude is:

$\left|x(\hat i+\hat j+\hat k)\right| = \sqrt{x^2+x^2+x^2} = |x|\sqrt{3}$

For this to be a unit vector: …

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CBSE 2024Set D1 markMCQ
$\vec{a} = \vec{i} + \vec{j} + 2\vec{k}$
(a) $\frac{\vec{i} + \vec{j} + \vec{k}}{\sqrt{6}}$
(b) $\frac{\vec{i} + \vec{j} + 2\vec{k}}{\sqrt{6}}$
(c) $\frac{\vec{i} + \vec{j} + 2\vec{k}}{6}$
(d) $\frac{\vec{i} + \vec{j} + \vec{k}}{6}$
›Reveal solutionSolution

Unit vector $= \vec{a}/|\vec{a}|$ .

$|\vec{a}| = |\vec{i} + \vec{j} + 2\vec{k}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}.$ Hence …

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CBSE 2024Set A1 mark
$x\cdot(\hat{i}+\hat{j}+\hat{k})$
›Reveal solutionSolution

If $x(\hat i+\hat j+\hat k)$ is a unit vector then $x=\pm\dfrac{1}{\sqrt3}$ .

The magnitude of $\hat i+\hat j+\hat k$ is $\sqrt{1^2+1^2+1^2}=\sqrt3$ . For $x(\hat i+\hat j+\hat k)$ to be a unit vector we need $|x|\sqrt3=1$ , …

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