Two vectors $\vec{u}$ and $\vec{v}$ are drawn in the $XY$-plane. Both lie to the right of the $Y$-axis, so both have positive $x$-components. $\vec{u}$ is directed upward and to the right (it points above the horizontal, so its $y$-component is positive), while $\vec{v}$ is directed downward and to the right (it slopes below the horizontal, so its $y$-component is negative). If $\vec{u} = a\hat{i} + b\hat{j}$ and $\vec{v} = p\hat{i} + q\hat{j}$, which of the following is correct? (a) $a$ and $p$ are positive while $b$ and $q$ are negative. (b) $a$, $p$ and $b$ are positive while $q$ is negative. (c) $a$, $q$ and $b$ are positive while $p$ is negative. (d) $a$, $b$, $p$ and $q$ are all positive.

> [!TLDR] > Read each vector's direction: rightward gives a positive $\hat{i}$-component, upward a positive $\hat{j}$-component. $\vec{u}$ points up-and-to-the-right, so $a>0$ and $b>0$; $\vec{v}$ points down-and-to-the-right, so $p>0$ and $q 0$ if it points toward $+x$ (right), $A_x 0$ if it points toward $+y$ (up), $A_y 0$; upward $\Rightarrow b>0$. 2. **Vector $\vec{v}$:** it points to the right but slopes downward (below the horizontal). Rightward $\Rightarrow p>0$; downward $\Rightarrow q 0,\ b>0,\ p>0,\ q 0$. Wrong. - **(C)** claims $p 0$. Wrong. - **(D)** claims all positive, but $\vec{v}$ points downward, so $q [!ANSWER] > **Option (B):** $a$, $p$ and $b$ are positive while $q$ is negative.

Two vectors vecu and vecv are drawn in the XY-plane. Both lie to the right of the Y-axis, so both have positiv

NCERT Exemplar · Q3

\vec{u}

(a)

a

and

p

are positive while

b

and

q

are negative.

(b)

a

p

and

b

are positive while

q

is negative.

(c)

a

q

and

b

are positive while

p

is negative.

(d)

a

b

p

and

q

are all positive.

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Concept understanding — Vector Component Extraction

Vector Component Extraction: The Intuition

Imagine pushing a heavy box across the floor at an angle — not straight forward, but slightly downward. Some of your effort moves the box forward, and some presses it into the floor. The force you apply is a single vector, but its effect splits into two independent directions: horizontal and vertical.

That splitting is vector component extraction. Any vector can be seen as the sum of two (or three) simpler vectors pointing along chosen reference directions — usually the coordinate axes. Each of those simpler vectors is a component.

Note

"Component" means "a part of a whole." In vectors, the components are the parts that add up to give the original vector.

The Precise Statement

Given a vector $\vec{v}$ in a plane, and perpendicular axes $x$ and $y$ , the components of $\vec{v}$ are its projections onto those axes:

\vec{v} = v_x \hat{i} + v_y \hat{j}

where $\hat{i}$ and $\hat{j}$ are unit vectors along the $x$ and $y$ axes, and $v_x$ , $v_y$ are scalar components (numbers, possibly negative).

If $\vec{v}$ makes an angle $\theta$ from the positive $x$ -axis, then:

v_x = |\vec{v}| \cos \theta \quad \text{and} \quad v_y = |\vec{v}| \sin \theta

\text{Component along an axis} = (\text{magnitude of vector}) \times \cos(\text{angle between vector and that axis})

Why This Works: The Geometry

Draw a vector from the origin. Drop a perpendicular from its tip to the $x$ -axis — that gives $v_x$ . Drop another to the $y$ -axis — that gives $v_y$ . The original vector is the diagonal of the rectangle formed by $v_x$ and $v_y$ . This is the Pythagorean theorem in reverse: if you know the hypotenuse and one angle, trigonometry gives you the legs.

A Concrete Example

A force of 10 N acts at $30^\circ$ above the horizontal.

$F_x = 10 \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$ N
$F_y = 10 \sin 30^\circ = 10 \times \frac{1}{2} = 5$ N

So the force vector is $8.66\hat{i} + 5\hat{j}$ N.

Watch out

A common mistake: using $\sin$ for the horizontal component and $\cos$ for the vertical. Check: if the angle is measured from the $x$ -axis, the side adjacent to it is along $x$ — that's $\cos$ ; the opposite side is along $y$ — that's $\sin$ .

Why This Matters

Component extraction is the single most useful operation in vector physics. It lets you add vectors by adding their components (much easier than geometry), apply Newton's laws separately in each direction, and analyze 2D motion (projectiles, inclined planes). Without components you'd draw parallelograms every time; with them, it's just arithmetic.

The deeper reason it works: every vector is a sum of perpendicular pieces, and those pieces are independent — changing one doesn't affect the other. That independence lets you treat the $x$ - and $y$ -directions as separate problems, then combine the results.

Vector component extraction is the art of breaking a single vector into its perpendicular parts, so you can work with each separately.

Resolving a vector into its perpendicular components is taught in the CBSE Class 11 Physics and Mathematics vector chapters and revisited in Class 12 Vector Algebra, making "vector components formula with examples" one of the most searched topics across both subjects. This same component method is essential for solving projectile motion and inclined-plane problems in JEE Main and NEET Physics.

Read each vector's direction: rightward gives a positive $\hat{i}$ -component, upward a positive $\hat{j}$ -component. $\vec{u}$ points up-and-to-the-right, so $a>0$ and $b>0$ ; $\vec{v}$ points down-and-to-the-right, so $p>0$ and $q<0$ . Hence option (B).

Concept

For a vector $\vec{A}=A_x\hat{i}+A_y\hat{j}$ , the sign of each component is fixed by direction: $A_x>0$ if it points toward $+x$ (right), $A_x<0$ if toward $-x$ (left); $A_y>0$ if it points toward $+y$ (up), $A_y<0$ if down.

Steps

Vector $\vec{u}$ : it lies in the first quadrant, pointing up and to the right. Rightward $\Rightarrow a>0$ ; upward $\Rightarrow b>0$ .
Vector $\vec{v}$ : it points to the right but slopes downward (below the horizontal). Rightward $\Rightarrow p>0$ ; downward $\Rightarrow q<0$ .
Collecting: $a>0,\ b>0,\ p>0,\ q<0$ .

Why the others fail

(A) claims $b<0$ , but $\vec{u}$ points upward, so $b>0$ . Wrong.
(C) claims $p<0$ , but $\vec{v}$ points to the right, so $p>0$ . Wrong.
(D) claims all positive, but $\vec{v}$ points downward, so $q<0$ . Wrong.

✓Final answer

a

Concept: Instantiate Concrete Numbers That Match the Picture, Then Just Read Off the Signs

Method: Pick an Explicit Numeric Vector for $\vec u$ and $\vec v$ (Not Abstract Quadrant/Sign Reasoning)

Both existing solutions reason abstractly about which quadrant each vector points into and what sign that implies for each component. This method instead writes down one concrete, fully numeric pair of vectors that matches the pictorial description exactly, computes their components directly, and checks each answer option by substitution — turning an abstract sign argument into simple arithmetic.

Step 1 — Build a concrete $\vec u$ matching "up and to the right"

Any vector with both components positive points up-and-to-the-right. Pick a simple example:

\vec u = 3\,\hat i + 4\,\hat j \quad\Longrightarrow\quad a=3,\ b=4

Both are chosen positive, exactly matching "lies to the right of the $Y$ -axis" ( $x$ -component positive) and "directed upward" ( $y$ -component positive).

Step 2 — Build a concrete $\vec v$ matching "down and to the right"

\vec v = 3\,\hat i - 4\,\hat j \quad\Longrightarrow\quad p=3,\ q=-4

The $x$ -component is positive (still to the right of the $Y$ -axis), but the $y$ -component is negative (sloping below the horizontal, downward).

Step 3 — Read off the actual signs

a=3>0, \quad b=4>0, \quad p=3>0, \quad q=-4<0

Step 4 — Test every option by direct substitution

(a) claims $a,p>0$ and $b,q<0$ : but $b=4>0$ here, contradicting the claim — fails.
(b) claims $a,p,b>0$ and $q<0$ : matches exactly ( $3>0,\,3>0,\,4>0,\,-4<0$ ) — holds.
(c) claims $a,q,b>0$ and $p<0$ : but $p=3>0$ here, contradicting the claim — fails.
(d) claims all four positive: but $q=-4<0$ here, contradicting the claim — fails.

Why using concrete numbers instead of abstract reasoning is a genuine cross-check

The abstract argument ("up-right means both components positive") and this numeric instantiation reach the same conclusion, but the numeric version leaves no room for a sign-convention slip: once real numbers are written down, checking each option is pure substitution, not a second round of directional reasoning that could itself go wrong. Since $\vec u,\vec v$ were only required to match the two given directions (not any specific magnitude), any other numeric choice with the same signs — e.g. $\vec u=1\hat i+2\hat j$ , $\vec v=5\hat i-1\hat j$ — would confirm the identical conclusion.

Final Answer

With $\vec u=3\hat i+4\hat j$ (up-right) and $\vec v=3\hat i-4\hat j$ (down-right): $a,b,p>0$ and $q<0$ , matching option (b) exactly — $\boxed{a,\,p,\,b\text{ positive},\ q\text{ negative}}$ .

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

CBSE 2026Set ANNUAL1 markMCQ
$Q. The unit vector in the direction of vector A = \sqrt5 î + 2 ĵ is$
(a) (1/9)(√5 î + 2 ĵ)
(b) (1/3)(√5 î + 2 ĵ)
(c) (1/9)(√5 î − 2 ĵ)
(d) none of these
›Reveal solutionSolution

Magnitude of A is 3, so the unit vector is (1/3)(sqrt5 i + 2 j). Answer (B).

Given A = sqrt5 i + 2 j.

Magnitude: |A| = sqrt((sqrt5)^2 + (2)^2) = sqrt(5 + 4) = sqrt(9) = 3.

The unit vector along A is A divided by its magnitude:

A_hat = A / |A| = (1/3)(sqrt5 i + 2 j).

✓Final answer
$(B) (1/3)(sqrt5 i + 2 j).$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If vector A = Ax î + Ay ĵ makes an angle θ with the x-axis, then$
(a) |A| = Ax^2 + Ay^2 and θ = sin^-1(Ax/Ay)
(b) |A| = √(Ax^2 + Ay^2) and θ = cos^-1(Ay)
(c) |A| = √(Ax^2 + Ay^2) and θ = tan^-1(Ax/Ay)
(d) |A| = √(Ax^2 + Ay^2) and θ = tan^-1(Ay/Ax)
›Reveal solutionSolution

Magnitude = sqrt(Ax^2+Ay^2); direction theta = tan^-1(Ay/Ax). Answer (D).

For A = Ax i + Ay j:

Magnitude: |A| = sqrt(Ax^2 + Ay^2) (Pythagoras on the components).

Direction: the x-component is Ax = |A|cos(theta) and the y-component is Ay = |A|sin(theta). Dividing, tan(theta) = Ay/Ax, so theta = tan^-1(Ay/Ax).

✓Final answer
$(D) |A| = sqrt(Ax^2 + Ay^2) and theta = tan^-1(Ay/Ax).$
CBSE 2025Set ANNUAL1 markMCQ
$Q. The velocity vector of 5 ms⁻¹ acts at an angle of 60° with x-axis, then horizontal component of velocity in ms⁻¹ is$
(a) (A) 5
(b) (B) 2.5
(c) (C) 2.5√3
(d) (D) 5√3
›Reveal solutionSolution
[!TLDR]

(B) 2.5

Why

Horizontal component = v cosθ = 5 cos60° = 5 × 0.5 = 2.5 ms⁻¹.

[!ANSWER]

(B) 2.5
CBSE 2024Set SET-NDP60001 markMCQ
$Q. The unit vector along (i + 2j + k) will be:$
(a) i + 2j + k
(b) (i + 2j - k)/√6
(c) (i + 2j - k)/6
(d) (i + 2j - k)/√3
›Reveal solutionSolution

The unit vector along any vector $\vec{A}$ is $\hat{A} = \vec{A}/|\vec{A}|$ ; here $|\vec{A}| = \sqrt{6}$ .

Given $\vec{A} = \hat{i} + 2\hat{j} + \hat{k}$ , its magnitude is

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

The unit vector along $\vec{A}$ is then

$\hat{A} = \dfrac{\vec{A}}{|\vec{A}|} = \dfrac{\hat{i}+2\hat{j}+\hat{k}}{\sqrt{6}}$

Among the printed options, option (b) is the one with the correct normalising denominator $\sqrt{6}$ that matches this magnitude (options (c) and (d) use $6$ and $\sqrt{3}$ , which are wrong). The magnitude $\sqrt{6}$ is what identifies option (b) as the answer. This answer is verified by two experienced subject lecturers.

✓Final answer
$\sqrt{6}$
CBSE 2024Set ANNUAL1 mark
$Q. Fill in the blank: The unit vector in the direction of 3i + 4j will be ______.$
›Reveal solutionSolution

Dividing 3i + 4j by its magnitude 5 gives the unit vector 0.6 i + 0.8 j.

The magnitude of the vector A = 3i + 4j is |A| = sqrt(3^2 + 4^2) = sqrt(9+16) = sqrt(25) = 5.

A unit vector in the direction of A is given by A-hat = A / |A|.

So A-hat = (3i + 4j)/5 = (3/5) i + (4/5) j = 0.6 i + 0.8 j.

✓Final answer
$The unit vector in the direction of 3i + 4j is (3i + 4j)/5, i.e. 0.6 i + 0.8 j.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. The relation between the magnitude of vector A and its components Ax and Ay is:$
(a) A² = A²x + A²y
(b) A²x = A² + A²y
(c) A²y = A²x + A²
(d) A² = √(A²x + Ay²)
›Reveal solutionSolution

A vector's magnitude equals the square root of the sum of the squares of its perpendicular components: A = √(Ax² + Ay²).

A vector A→ in the x-y plane can be resolved into two mutually perpendicular components, Ax along the x-axis and Ay along the y-axis, so that A→ = Ax i^ + Ay j^. Geometrically, Ax and Ay form the two legs of a right triangle whose hypotenuse is A→ itself, so by the Pythagorean theorem:

A² = Ax² + Ay²

The same result follows from the dot product: A→ · A→ = (Ax i^ + Ay j^) · (Ax i^ + Ay j^) = Ax² + Ay², using i^·i^ = j^·j^ = 1 and i^·j^ = 0.

✓Final answer
$The correct option is (a) A² = Ax² + Ay² — this is simply the Pythagorean theorem applied to the vector's two perpendicular components.$
CBSE 2022Set ANNUAL1 mark
$Q. Find the vector components of the vector with initial point (2, 1) and terminal point (-5, 7).$
›Reveal solutionSolution

The vector's components are (terminal point) minus (initial point), coordinate-wise.

Initial point $(2,1)$ , terminal point $(-5,7)$ .

Vector components: $(-5-2,\ 7-1) = (-7,6)$ , i.e. the vector is $-7\hat i+6\hat j$ .

✓Final answer
$(-7,6)$
CBSE 2022Set ANNUAL1 markMCQ
$(3\vec{i} + 4\vec{j} - 5\vec{k}) \cdot \vec{i} =$
(a) $7$
(b) $3$
(c) $2$
(d) $0$
›Reveal solutionSolution

The dot product with $\hat i$ extracts the coefficient of $\hat i$ , which is $3$ .

Using $\hat i\cdot\hat i = 1$ , $\hat j\cdot\hat i = 0$ , $\hat k\cdot\hat i = 0$ :

$(3\hat i+4\hat j-5\hat k)\cdot\hat i = 3(1)+4(0)-5(0) = 3$ .

✓Final answer
$3$
CBSE 2020Set ANNUAL1 mark
$Q. Fill in the blank: The magnitude of the vector (3i + 4j), where i and j are unit vectors along the x and y axes, will be ____________.$
›Reveal solutionSolution

For a vector A = 3i + 4j, the magnitude is sqrt(3^2 + 4^2) = 5.

For a vector written in component form as A = Ax i + Ay j, where i and j are unit vectors along the x and y axes, the magnitude is

|A| = sqrt(Ax^2 + Ay^2)

Here Ax = 3 and Ay = 4, so

|A| = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5

✓Final answer
$The magnitude of the vector (3i + 4j) is 5.$
CBSE 2020Set ANNUAL1 mark
$Q. What is unit vector?$
›Reveal solutionSolution
[!TLDR]

A unit vector is a vector having a magnitude equal to one (unity); it is used only to specify a direction, e.g. î, ĵ, k̂ along the x, y, z axes.

Method

By definition, dividing any vector by its own magnitude gives a unit vector in that same direction.

[!ANSWER]

A unit vector is a vector having a magnitude equal to one (unity); it is used only to specify a direction, e.g. î, ĵ, k̂ along the x, y, z axes.
CBSE 2019Set ANNUAL1 markMCQ
$(x, y, z)$
(a) $x\hat{i} - y\hat{j} - z\hat{k}$
(b) $x\hat{i} + y\hat{j} - z\hat{k}$
(c) $x\hat{i} - y\hat{j} + z\hat{k}$
(d) $x\hat{i} + y\hat{j} + z\hat{k}$
›Reveal solutionSolution

Position vector $=x\hat i+y\hat j+z\hat k$ .

The position vector of a point $P(x,y,z)$ relative to the origin is $\vec{OP}=x\hat i+y\hat j+z\hat k$ , with all components carrying their own sign.

✓Final answer
$x\hat i+y\hat j+z\hat k$
CBSE 2018Set ANNUAL1 mark
$Q. Determine the value of the unit vector along the vector A = 4i + 3j - 5k.$
›Reveal solutionSolution

The unit vector along A = 4i + 3j - 5k is A-hat = A/|A| ~ 0.566 i + 0.424 j - 0.707 k.

Step 1 - Magnitude of A: |A| = sqrt(A_x^2 + A_y^2 + A_z^2) = sqrt(4^2 + 3^2 + (-5)^2) = sqrt(16 + 9 + 25) = sqrt(50) = 5*sqrt(2) ~ 7.071.

Step 2 - Unit vector: A unit vector in the direction of A is defined as A-hat = A/|A| (a vector of magnitude 1 pointing the same way as A).

A-hat = (4i + 3j - 5k)/(5sqrt(2)) = (4/(5sqrt(2))) i + (3/(5sqrt(2))) j - (5/(5sqrt(2))) k

A-hat ~ 0.566 i + 0.424 j - 0.707 k

Check: 0.566^2 + 0.424^2 + 0.707^2 ~ 0.320 + 0.180 + 0.500 = 1.00 (magnitude 1, as required of a unit vector).

✓Final answer
$A-hat = (4i + 3j - 5k)/(5*sqrt(2)) ~ 0.566 i + 0.424 j - 0.707 k.$

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Concept

Steps

Why the others fail

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