Find all vectors of magnitude $3\sqrt{3}$ which are collinear with the vector $\hat{i} + \hat{j} + \hat{k}$.

Find all vectors of magnitude 3√(3) which are collinear with the vector hati + hatj + hatk.

Question

3\sqrt{3}

CBSECBSE Class XII Board 2023Subjective· 2mImportance★★★★★

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Collinear vectors are scalar multiples of the given vector. The unit vector along $\hat{i}+\hat{j}+\hat{k}$ is $\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$ , so vectors of magnitude $3\sqrt{3}$ are $\pm 3(\hat{i}+\hat{j}+\hat{k})$ .

Concept and Intuition

Two vectors are collinear if they lie along the same line — meaning one is a scalar multiple of the other. If a vector $\vec{a}$ is collinear with $\vec{b}$ , then $\vec{a} = \lambda \vec{b}$ for some real number $\lambda$ .

The magnitude of $\vec{a}$ is $|\vec{a}| = |\lambda| \, |\vec{b}|$ . So if we know the required magnitude and the magnitude of $\vec{b}$ , we can solve for $|\lambda|$ . The sign of $\lambda$ determines whether the vectors point in the same direction ( $\lambda > 0$ ) or opposite directions ( $\lambda < 0$ ).

Here, the given vector is $\vec{b} = \hat{i} + \hat{j} + \hat{k}$ . Its magnitude is $\sqrt{1^2+1^2+1^2} = \sqrt{3}$ . We want vectors of magnitude $3\sqrt{3}$ that are collinear with $\vec{b}$ .

Step-by-step solution

1. Find the magnitude of the given vector

|\hat{i} + \hat{j} + \hat{k}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}

2. Let the required vector be $\vec{v} = \lambda (\hat{i} + \hat{j} + \hat{k})$

Since $\vec{v}$ is collinear with $\hat{i}+\hat{j}+\hat{k}$ , it must be a scalar multiple.

3. Use the magnitude condition

|\vec{v}| = |\lambda| \cdot \sqrt{3} = 3\sqrt{3}

Therefore:

|\lambda| = \frac{3\sqrt{3}}{\sqrt{3}} = 3

So $\lambda = 3$ or $\lambda = -3$ .

Watch out

A common mistake is to forget the negative sign. Collinear does not mean "same direction" — it only means parallel. The vector can point opposite to the given vector and still be collinear.

4. Write the two vectors

For $\lambda = 3$ :

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

CBSE 2026Set 65/3/11 markMCQ
$\vec{a}$
›Reveal solutionSolution
The assertion is true because one vector is a scalar multiple of the other, making them collinear. The reason is false because the dot product of $\vec{a}$ and $-2\vec{a}$ is $-2|\vec{a}|^2$ , not zero. So (A) is true, (R) is false.

Let’s unpack this carefully. The question tests two separate ideas: what it means for vectors to be collinear, and what the dot product of a vector with its scalar multiple actually gives. Many students mix these up, so let’s build the intuition first.

Collinear vectors are vectors that lie along the same line or parallel lines. The key property: two non-zero vectors are collinear if and only if one is a scalar multiple of the other. That is, $\vec{b} = \lambda \vec{a}$ for some real number $\lambda$ . No angle condition or dot product zero is required — that’s for perpendicular vectors.

Now look at the assertion: $\vec{a}$ and $(-2\vec{a})$ are clearly related by $\lambda = -2$ . Since $-2$ is a real number, they are indeed collinear. The negative sign just means they point in opposite directions, but they’re still on the same line.

The reason claims $\vec{a} \cdot (-2\vec{a}) = 0$ . Let’s compute that:

$\vec{a} \cdot (-2\vec{a}) = -2 (\vec{a} \cdot \vec{a}) = -2 |\vec{a}|^2$ .

Since $\vec{a} \neq \vec{0}$ , $|\vec{a}|^2 > 0$ , so this product is negative, not zero. The reason is completely wrong.

Watch out

A common mistake is to think that collinear vectors have a dot product of zero. That’s actually the condition for perpendicular (orthogonal) vectors. Collinearity is about scalar multiples, not dot products.

So the assertion is true, the reason is false. In exam language, this means option (C) — Assertion true, Reason false.

Tip

Quick check: If two vectors are collinear, their dot product equals $\pm |\vec{a}||\vec{b}|$ , not zero. Zero dot product only happens when they are perpendicular (or one is zero).

Let’s walk through the reasoning step by step.

Check the assertion: Are $\vec{a}$ and $-2\vec{a}$ collinear? …
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CBSE 2026Set ANNUAL1 mark
$Q. Write True or False: Two collinear vectors are always equal in magnitude.$
›Reveal solutionSolution

False - collinear means parallel, not equal in length.

Two vectors are collinear if they are parallel (same or opposite direction), regardless of magnitude. For example $\hat i$ and $3\hat i$ are colli …

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CBSE 2022Set M1 mark
$Q. Define Collinear Vectors.$
›Reveal solutionSolution

Collinear vectors are vectors parallel to one and the same line.

Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions. Equivalently, $\vec{a}$ and $\vec{b}$ are collinear if $\vec{a}=\lambda\vec{b}$ for …
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Q.Find all vectors of magnitude $3\sqrt{3}$ which are collinear with the vector $\hat{i} + \hat{j} + \hat{k}$ .

Concept understanding — Collinear Vectors Properties

Collinear Vectors and Their Properties

The key property: one is a scalar multiple of the other

In component form

Other useful properties

Concept and Intuition

Step-by-step solution

Q.Find all vectors of magnitude 333\sqrt{3}33​ which are collinear with the vector i^+j^+k^\hat{i} + \hat{j} + \hat{k}i^+j^​+k^.

Concept and Intuition

Step-by-step solution

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Q.Find all vectors of magnitude $3\sqrt{3}$ which are collinear with the vector $\hat{i} + \hat{j} + \hat{k}$ .

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