A vector $\vec P$ has magnitude 5 units and points in a direction making an angle of $37^\circ$ with the x-axis. A second vector $\vec Q$ has rectangular components $Q_x = 4$ units and $Q_y = 3$ units. Are $\vec P$ and $\vec Q$ equal vectors? Justify your answer using both magnitude and direction.

A vector vec P has magnitude 5 units and points in a direction making an angle of 37^° with the x-axis. A seco

Example · Example 1

\vec P

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Concept understanding — Vector Equality

Vector Equality

When are two vectors the same vector? A vector carries only two pieces of information — magnitude (length) and direction. It does not carry a fixed starting point. So two vectors are equal when they agree on both magnitude and direction, no matter where each one happens to be drawn.

Precise definition

Two vectors $\vec{a}$ and $\vec{b}$ are equal, written $\vec{a}=\vec{b}$ , if and only if

they have the same magnitude, $|\vec{a}|=|\vec{b}|$ , and
they have the same direction.

Important

Position is irrelevant. An arrow drawn at the top-left of the page and an identical-length, identical-direction arrow at the bottom-right are the same vector. This is why such vectors are called "free" vectors — you may slide them anywhere without changing them.

In component form

Equality becomes a simple check on coordinates. If

\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},\qquad \vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k},

then

\vec{a}=\vec{b}\iff a_1=b_1,\; a_2=b_2,\; a_3=b_3.

Equal vectors have equal corresponding components — a single vector equation is really three scalar equations at once.

Don't confuse it with neighbouring ideas

Watch out

Equal vectors are not the same as coinitial vectors (which merely share a starting point) or collinear vectors (which are parallel but may differ in length or point oppositely). Two vectors of equal length pointing in opposite directions are not equal — they are negatives of each other, $\vec{a}=-\vec{b}$ .

Why it matters

Because equal vectors can be moved freely, we are allowed to shift vectors to a common tail before adding them, to compare forces acting at different points, and to represent every point by a position vector from the origin. Vector equality is what makes the whole "slide it wherever you like" freedom of vector algebra legitimate.

Vector equality is a foundational definition in the NCERT Class 12 Vector Algebra chapter, frequently tested through short conceptual CBSE board questions that distinguish it from coinitial or collinear vectors. Students revising "vector algebra class 12 important questions" for boards or JEE Main should treat this component-matching test as a quick sanity check before any vector proof.

$\vec Q$ has components $Q_x = 4$ , $Q_y = 3$ , so its magnitude is $Q = \sqrt{4^2+3^2} = \sqrt{25} = 5$ units, matching the magnitude of $\vec P$ . Its direction is $\theta = \tan^{-1}(3/4)$ , which — using the standard 3-4-5 triangle approximation used throughout this course — is $37^\circ$ with the x-axis, matching the direction of $\vec P$ . Since both the magnitude (5 units) and the direction ( $37^\circ$ ) match exactly, $\vec P$ and $\vec Q$ are equal vectors, even though they were specified in different ways (one by magnitude-and-angle, the other by components). [!ANSWER] Yes — $\vec P$ and $\vec Q$ are equal vectors (both have magnitude 5 units and direction $37^\circ$ with the x-axis).

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

CBSE 2026Set ANNUAL1 markMCQ
$Q.Assertion (A): For vector \vec{a} = \hat{i} + 2\hat{j} and vector \vec{b} = 2\hat{i} + \hat{j}, |\vec{a}| = |\vec{b}|. Reason (R): Vector \vec{a} and vector \vec{b} are equal vectors.$
(a)
(i) Both A and R are correct and R is the correct explanation of A.
(b)
(ii) Both A and R are correct but R is not the correct explanation of A.
(c)
(iii) A is correct but R is incorrect.
(d)
(iv) Both A and R are incorrect.
›Reveal solutionSolution
A is true ( $|\vec a|=|\vec b|=\sqrt5$ ) but R is false (the vectors are not equal) — option (iii).

Concept. The magnitude of $x\hat i+y\hat j$ is $\sqrt{x^2+y^2}$ . Two vectors are equal only if all their corresponding components are equal — equal magnitude alone is not enough.

Steps.

$\vec a=\hat i+2\hat j\ \Rightarrow\ |\vec a|=\sqrt{1^2+2^2}=\sqrt5$ .

$\vec b=2\hat i+\hat j\ \Rightarrow\ |\vec b|=\sqrt{2^2+1^2}=\sqrt5$ .

So $|\vec a|=|\vec b|$ : Assertion (A) is correct.

Components: $\vec a$ has $(1,2)$ while $\vec b$ has $(2,1)$ ; these differ, so $\vec a\neq\vec b$ : Reason (R) is incorrect.

✓Final answer
$A is correct but R is incorrect — option (iii) .$
CBSE 2025Set 65/1/11 markMCQ
$\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$
›Reveal solutionSolution

Two vectors are perpendicular when their dot product is zero. Computing $\vec{a} \cdot \vec{b} = 0$ confirms orthogonality, so the answer is (B).

When you're given two vectors and asked about their relationship, you need to check three fundamental properties: parallelism (same or opposite direction), perpendicularity (right angles), and magnitude (length). Each has a clear algebraic test.

Two vectors are parallel if one is a scalar multiple of the other: $\vec{a} = \lambda \vec{b}$ for some real $\lambda$ . Two vectors are perpendicular (orthogonal) if their dot product vanishes: $\vec{a} \cdot \vec{b} = 0$ . The magnitude of a vector $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$ is $|\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2}$ .

Let's systematically check each possibility.

Checking the options

1. Are the vectors parallel?

For $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ to be parallel, we'd need
$\frac{3}{1} = \frac{2}{-1} = \frac{-1}{1}$
The first ratio gives $3$ , the second gives $-2$ , and the third gives $-1$ . These are all different, so the vectors are not parallel. Option (A) is incorrect.

2. Are the vectors perpendicular?

Compute the dot product:
$\vec{a} \cdot \vec{b} = (3)(1) + (2)(-1) + (-1)(1) = 3 - 2 - 1 = 0$
The dot product is zero, which means the vectors are perpendicular. Option (B) looks promising.

3. Compare the magnitudes

Calculate $|\vec{a}|$ :
$|\vec{a}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$
Calculate $|\vec{b}|$ :
$|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3}$
Since $\sqrt{14} \approx 3.74$ and $\sqrt{3} \approx 1.73$ , we have $|\vec{a}| > |\vec{b}|$ . So option (C) is false and option (D) is false.

Tip

When checking perpendicularity, the dot product test is immediate and definitive. If $\vec{a} \cdot \vec{b} = 0$ , you're done — no need to compute angles or magnitudes unless the question specifically asks for them.

✓Final answer
$\vec{a} \perp \vec{b}$
CBSE 2024Set 65/1/11 markMCQ
$\begin{bmatrix} x+y & 2 \\ 5 & xy \end{bmatrix} = \begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
›Reveal solutionSolution
Two matrices are equal only when their corresponding entries are equal. Equating the entries gives $x+y=6$ and $xy=8$ , which are the sum and product of the roots of a quadratic. The required expression $\frac{24}{x}+\frac{24}{y}$ simplifies to $24\cdot\frac{x+y}{xy}=24\cdot\frac{6}{8}=18$ .

When two matrices are equal, every entry in the same position must match exactly. This is the fundamental idea of vector equality extended to matrices — think of it as comparing two tables of numbers cell by cell. No shortcuts, no tricks: if the top-left corner of the first matrix is $x+y$ , and the top-left corner of the second matrix is $6$ , then $x+y$ must equal $6$ . The same logic applies to every other position.

Here, both matrices are $2\times 2$ , so we get four equations. But notice that two of the entries are already identical ( $2$ and $5$ ), so they give no new information. The real constraints come from the other two positions.

Equate the (1,1) entries — the top-left corner:

$x + y = 6$

Equate the (2,2) entries — the bottom-right corner:

$xy = 8$

We now have two equations in two unknowns. You could solve for $x$ and $y$ individually (they are the roots of $t^2 - 6t + 8 = 0$ , so $t = 2$ or $4$ ), but the question doesn't ask for $x$ and $y$ separately — it asks for $\frac{24}{x} + \frac{24}{y}$ .

Tip

When you see a sum of reciprocals like $\frac{24}{x} + \frac{24}{y}$ , always try to rewrite it using $x+y$ and $xy$ :

$\frac{24}{x} + \frac{24}{y} = 24\left(\frac{1}{x} + \frac{1}{y}\right) = 24\left(\frac{x+y}{xy}\right)$

This avoids solving for $x$ and $y$ individually — a huge time-saver in exams.

Substitute the known values: $\frac{24}{x} + \frac{24}{y} = 24 \cdot \frac{x+y}{xy} = 24 \cdot \frac{6}{8} = 24 \cdot \frac{3}{4} = 18$

Watch out

A common mistake is to forget that $xy = 8$ and instead try to solve $x+y=6$ and $xy=8$ by guessing. Some students pick $x=2, y=4$ (correct) but then compute $\frac{24}{2}+\frac{24}{4}=12+6=18$ — which works here. But if the numbers were less friendly, guessing would fail. Always use the symmetric sum-and-product approach; it's foolproof.

✓Final answer
$\boxed{18}$
CBSE 2023Set ANNUAL1 markMCQ
$x\hat{i} + 2\hat{j} = 3\hat{i} - y\hat{j}$
(a) $3, -2$
(b) $2, 3$
(c) $-3, 2$
(d) none of these
›Reveal solutionSolution

Two vectors are equal exactly when their corresponding components are equal.

$x\hat i + 2\hat j = 3\hat i - y\hat j$ . Equate $\hat i$ -components: $x = 3$ . Equate $\hat j$ -components: $2 = -y \Rightarrow y = -2$ .

So $(x,y) = (3,-2)$ .

✓Final answer
$x=3,\ y=-2$
CBSE 2020Set ANNUAL1 mark
$Q. Define Negative of a vector.$
›Reveal solutionSolution

$-\vec{a}$ has the same magnitude as $\vec{a}$ but points in the opposite direction.

Concept. A vector has both magnitude and direction. Negating a vector keeps its length unchanged but flips its direction through $180^\circ$ .

Formal statement. The negative of a given vector $\vec{a}$ is defined as the vector whose magnitude is $|\vec{a}|$ and whose direction is exactly opposite to that of $\vec{a}$ . It is written $-\vec{a}$ , and satisfies $\vec{a}+(-\vec{a})=\vec{0}$ .

✓Final answer
$\vec{a}$

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