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.

Andhra Pradesh BieapMCQ· 1mImportance★★★★★est

50% · 34/68 Questions

✓ Free question

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.

AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
$\bar{A} = -\hat{i} + 2\hat{j} + 3\hat{k}$
›Reveal solutionSolution
This is a direct vector-addition problem: knowing the resultant of $\vec A,\vec B,\vec C$ is $2\hat k$ , we solve for $\vec C$ by subtraction. The answer is $\vec C=3\hat i-\hat j+3\hat k$ .

Concept and Intuition

If three vectors sum to a known resultant, the unknown one is just the resultant minus the sum of the known ones — vector subtraction is done component-by-component, independently along $\hat i$ , $\hat j$ , $\hat k$ .

Step-by-Step Solution

Given $\vec A = -\hat i+2\hat j+3\hat k$ and $\vec B=-2\hat i-\hat j-4\hat k$ .

Add them: $\vec A+\vec B = (-1-2)\hat i+(2-1)\hat j+(3-4)\hat k = -3\hat i+\hat j-\hat k$ .

The resultant $\vec A+\vec B+\vec C$ is a vector along $+\hat z$ with magnitude 2, i.e. $\vec A+\vec B+\vec C=2\hat k$ .

Solve for $\vec C$ : $\vec C = 2\hat k - (\vec A+\vec B) = 2\hat k - (-3\hat i+\hat j-\hat k) = 3\hat i-\hat j+(2+1)\hat k = 3\hat i-\hat j+3\hat k$ .

Common Mistakes

Sign slip while distributing the minus sign over $(\vec A+\vec B)$ 's components.

Forgetting that the resultant vector is $2\hat k$ (not just "magnitude 2" applied incorrectly to $\vec C$ alone before subtraction).

✓Final answer
$3\hat{i} - \hat{j} + 3\hat{k}$

ANSWER: A
AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
$\bar{a}$
›Reveal solutionSolution
The vector equals the sum of its parallel and perpendicular components — add them and square the magnitude.

Concept and Intuition

Any vector decomposes uniquely into a component along a given direction plus a component perpendicular to it; the two given pieces ARE that decomposition, so $\bar a$ is simply their vector sum, no projection formula needed.

Step-by-Step Solution

$\bar a = \dfrac{7}{25}(4\bar i+5\bar j+3\bar k) + \dfrac1{25}(47\bar i-10\bar j-46\bar k)$ .

$\bar i$ -component: $\dfrac{7\times4+47}{25}=\dfrac{28+47}{25}=\dfrac{75}{25}=3$ .

$\bar j$ -component: $\dfrac{7\times5-10}{25}=\dfrac{35-10}{25}=\dfrac{25}{25}=1$ .

$\bar k$ -component: $\dfrac{7\times3-46}{25}=\dfrac{21-46}{25}=\dfrac{-25}{25}=-1$ .

So $\bar a=3\bar i+\bar j-\bar k$ .

$|\bar a|^2=3^2+1^2+(-1)^2=9+1+1=11$ .

Common Mistakes

Trying to re-derive $\bar a$ via a projection formula instead of recognizing the two given vectors already sum to $\bar a$ .

Arithmetic slip combining the $\dfrac{7}{25}$ and $\dfrac1{25}$ fractions.

✓Final answer
$11$

ANSWER: C
AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
$^{-1}$
›Reveal solutionSolution
The key idea is that the string’s length is constant, so the component of block A’s velocity along the string must equal block B’s upward speed. Using the 60° angle, we find block B’s speed is $10 \cos 60^\circ = 5\ \text{m/s}$ . The correct option is (B).

The problem involves a classic constrained motion setup: two blocks connected by a string that passes over a fixed pulley. The string’s length doesn’t change, so the rate at which the string shortens on one side must equal the rate at which it lengthens on the other. Here, block A moves horizontally, pulling the string along the 60° direction; block B moves vertically. The trick is to relate A’s horizontal speed to the speed of the string segment that actually moves B.

Identify the constraint

The string is inextensible. Therefore, the speed of block B (which moves straight up or down) equals the speed at which the string is being pulled along its own direction from block A’s side. That is, the component of A’s velocity parallel to the string is what matters.

Resolve A’s velocity along the string

Block A moves horizontally to the right at $10\ \text{m/s}$ . The string makes a $60^\circ$ angle with the horizontal (measured at A). The component of A’s velocity along the string is:

$v_{\text{string}} = v_A \cos 60^\circ = 10 \times \frac{1}{2} = 5\ \text{m/s}.$
This is the speed at which the string is being pulled from A’s side.

Relate to block B’s speed Since the string passes over a fixed pulley, the other side of the string (attached to B) moves at the same speed as the string itself. Thus block B’s speed is exactly that component:

$v_B = 5\ \text{m/s}.$
The direction: as A moves right, it pulls the string, so B moves upward.

Watch out

A common mistake is to think that B’s speed equals A’s full speed (10 m/s) or to use the vertical component ( $10 \sin 60^\circ \approx 8.66\ \text{m/s}$ ). But only the component along the string transfers motion because the string can only transmit force and motion along its own length.

Tip

If the string were horizontal at A, then $v_B = v_A$ . If it were vertical, $v_B = 0$ (since A’s horizontal motion wouldn’t pull the string). The cosine factor smoothly interpolates between these extremes.

✓Final answer
$The correct option is (B). ANSWER: B$
AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
$\bar{p}$
›Reveal solutionSolution
A right-triangle (Pythagorean) resolution of a 2D vector into its rectangular components; $x=24$ units.

Concept and Intuition

Any 2D vector's magnitude and its two perpendicular (x and y) components form a right triangle, so $|\vec p|^2 = p_x^2+p_y^2$ . Knowing the magnitude and one component lets you solve for the other.

Step-by-Step Solution

Given $|\vec p| = 25$ , $p_y = 7$ .

$p_x^2 = |\vec p|^2 - p_y^2 = 625 - 49 = 576$ .

$p_x = \sqrt{576} = 24$ units.

Common Mistakes

Adding the components instead of using Pythagoras ( $25-7=18$ , a distractor option).

Forgetting to square before subtracting.

✓Final answer
$The correct option is (A) — 24 units.$

ANSWER: A
AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
$\alpha$
›Reveal solutionSolution
This tests the fundamental identity of direction cosines, $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ . Rearranging gives $\sin^2\alpha+\sin^2\beta=1+\cos^2\gamma$ , option (C).

Concept and Intuition

For any vector in 3D space, the direction cosines with respect to the three coordinate axes always satisfy $l^2+m^2+n^2=1$ where $l=\cos\alpha$ , $m=\cos\beta$ , $n=\cos\gamma$ . This single identity is the key to relating any combination of these angles' sines and cosines.

Step-by-Step Solution

Direction cosine identity: $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ .

Use $\sin^2\theta=1-\cos^2\theta$ for each of $\alpha,\beta$ :

$\sin^2\alpha+\sin^2\beta=(1-\cos^2\alpha)+(1-\cos^2\beta)=2-(\cos^2\alpha+\cos^2\beta).$

From the identity, $\cos^2\alpha+\cos^2\beta=1-\cos^2\gamma$ .

Substitute: $\sin^2\alpha+\sin^2\beta=2-(1-\cos^2\gamma)=1+\cos^2\gamma$ .

Common Mistakes

Forgetting the direction-cosine identity entirely and trying to treat $\alpha,\beta,\gamma$ as independent angles (they are not — they're constrained by the vector's single direction in space).

Sign slip turning $1+\cos^2\gamma$ into $1-\cos^2\gamma$ (which would equal $\sin^2\gamma$ , option (A) — a close but wrong distractor).

✓Final answer
$1 + \cos^2\gamma$

ANSWER: C
AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
$y = 2^{x+2}$
›Reveal solutionSolution
The dot-product conditions simply pin down the $x$ -coordinates of $P$ and $Q$ ; plugging into the curve equation gives their $y$ -coordinates, and a direct vector combination finishes the problem.

Concept and Intuition

$\vec{OP}\cdot\bar i$ is just the $x$ -component of the position vector $\vec{OP}$ (the dot product with the unit vector $\bar i$ picks out the $x$ -coordinate). So these conditions directly specify where on the curve $P$ and $Q$ sit.

Step-by-Step Solution

$\vec{OP}\cdot\bar i=-1 \Rightarrow$ $P$ has $x=-1$ . Since $P$ lies on $y=2^{x+2}$ : $y_P = 2^{-1+2}=2^1=2$ . So $P=(-1,2)$ .

$\vec{OQ}\cdot\bar i=2 \Rightarrow$ $Q$ has $x=2$ . Then $y_Q=2^{2+2}=2^4=16$ . So $Q=(2,16)$ .

$\vec{OQ}=2\bar i+16\bar j$ , $\vec{OP}=-\bar i+2\bar j$ .

$\vec{OQ}-4\vec{OP} = (2\bar i+16\bar j) - 4(-\bar i+2\bar j) = (2+4)\bar i+(16-8)\bar j = 6\bar i+8\bar j$ .

Common Mistakes

Forgetting to substitute the found $x$ -values back into the curve equation to get $y_P, y_Q$ (i.e., stopping after finding just the $x$ -coordinates).

Sign errors when distributing the $-4$ across $\vec{OP}$ 's components.

✓Final answer
$6\bar i+8\bar j$

ANSWER: C
AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$\vec{P} = 3\hat{i} + 4\hat{j}$
›Reveal solutionSolution
This tests finding the scalar component of a vector along a given direction using the dot product with the unit vector of that direction.

Concept and Intuition

The component of $\vec P$ along a direction is the projection of $\vec P$ onto the unit vector of that direction: $\vec P\cdot\hat n$ , where $\hat n$ is obtained by normalizing the given direction vector.

Step-by-Step Solution

Direction vector: $\hat i+2\hat j$ , magnitude $=\sqrt{1^2+2^2}=\sqrt5$ .

Unit vector: $\hat n=\dfrac{\hat i+2\hat j}{\sqrt5}$ .

$\vec P=3\hat i+4\hat j$ .

Component $=\vec P\cdot\hat n=\dfrac{3(1)+4(2)}{\sqrt5}=\dfrac{3+8}{\sqrt5}=\dfrac{11}{\sqrt5}$ .

Common Mistakes

Forgetting to normalize the direction vector before taking the dot product.

Arithmetic slip: $3(1)+4(2)=11$ , not $8$ or $5$ .

✓Final answer
$\dfrac{11}{\sqrt5}$

ANSWER: B
AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
$\vec{A}$
›Reveal solutionSolution
This is the standard problem of finding the component of $(\vec A-\vec B)$ along $(\vec A+\vec B)$ for two mutually perpendicular vectors — a scalar cannot have a directional component, so the intended quantity must be a vector combination, and only this reading matches a listed option exactly.

Concept and Intuition

The component of any vector $\vec V$ along a direction $\hat n$ is $\vec V\cdot\hat n$ . Here the natural vector to project is $(\vec A-\vec B)$ along $(\vec A+\vec B)$ . Since $\vec A\perp\vec B$ , we know $\vec A\cdot\vec B=0$ , which simplifies both the dot product and the magnitude of $\vec A+\vec B$ nicely (Pythagoras-like).

Step-by-Step Solution

Unit vector along $\vec A+\vec B$ : $\hat n=\dfrac{\vec A+\vec B}{|\vec A+\vec B|}$ .

Since $\vec A\perp\vec B$ , $\vec A\cdot\vec B=0$ , so $|\vec A+\vec B|^2=|A|^2+|B|^2+2\vec A\cdot\vec B=|A|^2+|B|^2$ , giving $|\vec A+\vec B|=\sqrt{|A|^2+|B|^2}$ .

Component of $(\vec A-\vec B)$ along $\hat n$ : $(\vec A-\vec B)\cdot\hat n=\dfrac{(\vec A-\vec B)\cdot(\vec A+\vec B)}{|\vec A+\vec B|}$ .

Expand numerator: $(\vec A-\vec B)\cdot(\vec A+\vec B)=|A|^2-\vec A\cdot\vec B+\vec B\cdot\vec A-|B|^2=|A|^2-|B|^2$ (using $\vec A\cdot\vec B=0$ ).

So the component $=\dfrac{|A|^2-|B|^2}{\sqrt{|A|^2+|B|^2}}$ .

Common Mistakes

Treating $\vec A\cdot\vec B$ (a scalar, and zero here since perpendicular) as if it were a vector that could have a "component" — this makes the literal question ill-posed, so the intended vector combination $(\vec A-\vec B)$ must be recognized.

Forgetting that perpendicularity simplifies $|\vec A+\vec B|$ to $\sqrt{|A|^2+|B|^2}$ (no cross term).

✓Final answer
$\dfrac{|\vec{A}|^2 - |\vec{B}|^2}{\sqrt{|\vec{A}|^2 + |\vec{B}|^2}}$

ANSWER: C
AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
$\vec{P}$
›Reveal solutionSolution
Setting the resultant's x-component to zero and its y-component to $2P$ , and using $|\vec Q|=10$ , gives $P=2\sqrt5$ m.

Concept and Intuition

Vector addition in components: if the sum of two vectors points purely along one axis, the components along the other axis must cancel exactly. This gives one equation; combined with the given magnitude of $\vec Q$ , we get a second equation — enough to solve for $P$ .

Step-by-Step Solution

Let $\vec P = (P, 0)$ since it's along the x-axis, and $\vec Q=(Q_x,Q_y)$ with $|\vec Q|=10\Rightarrow Q_x^2+Q_y^2=100$ .

Resultant $\vec R=\vec P+\vec Q=(P+Q_x,\ Q_y)$ . Since $\vec R$ is along the y-axis, its x-component is zero: $P+Q_x=0\Rightarrow Q_x=-P$ .

$\vec R$ 's magnitude is given as $2P$ (twice that of $\vec P$ ), and since $\vec R$ is purely along y: $|R|=|Q_y|=2P$ .

Substitute $Q_x=-P$ and $Q_y=\pm2P$ into $Q_x^2+Q_y^2=100$ : $P^2+4P^2=100\Rightarrow5P^2=100\Rightarrow P^2=20\Rightarrow P=\sqrt{20}=2\sqrt5$ m.

Common Mistakes

Forgetting that $Q_x$ must be exactly $-P$ (not just "some component") for the resultant to be purely vertical.

Arithmetic slip simplifying $\sqrt{20}$ — it's $2\sqrt5$ , not $\sqrt{10}$ or $5\sqrt2$ .

✓Final answer
$2\sqrt{5}$

ANSWER: D
AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
$40\ N$
›Reveal solutionSolution
Since the two rectangular components combine via Pythagoras to give the resultant, the missing component is $\sqrt{40^2-(20\sqrt3)^2}=20\,N$ .

Concept and Intuition

Rectangular components of a force are mutually perpendicular, so the magnitude of the resultant is the hypotenuse of a right triangle whose legs are the two components: $F_{resultant}^2=F_x^2+F_y^2$ .

Step-by-Step Solution

Let the resultant force be $F=40\,N$ , one component $F_1=20\sqrt3\,N$ , and the other component $F_2$ unknown.

By Pythagoras: $F^2=F_1^2+F_2^2 \Rightarrow F_2^2 = F^2-F_1^2$ .

Compute $F_1^2=(20\sqrt3)^2=400\times3=1200$ .

Compute $F^2=40^2=1600$ .

$F_2^2=1600-1200=400\Rightarrow F_2=20\,N$ .

Common Mistakes

Adding the components directly instead of using the Pythagorean (vector) relation.

Squaring $20\sqrt3$ incorrectly (forgetting the factor of $3$ from $(\sqrt3)^2$ ).

✓Final answer
$The correct option is (B) — 20 N.$

ANSWER: B

←Previous questionQ2 — Which one of the following statements is true? (a) A scalar quantity i…→Next questionQ4 — The component of a vector vecr along X-axis will have maximum value if…

Concept

Steps

Why the others fail

🎓Unlock everything free for 14 days

Unlock everything free for 14 days