The component of a vector vecr along X-axis will have maximum value if (a) vecr is along positive Y-axis (b) v

NCERT Exemplar · Q4

\vec{r}

(a)

\vec{r}

is along positive

Y

-axis

(b)

\vec{r}

is along positive

X

-axis

(c)

\vec{r}

makes an angle of

45^\circ

with the

X

-axis

(d)

\vec{r}

is along negative

Y

-axis

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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 …

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The component of a vector along an axis is maximized when the vector points directly along that axis. The answer is (B).

When we project a vector onto an axis, we are asking: how much of this vector "lives" in that direction? Mathematically, if $\vec{r}$ makes an angle $\theta$ with the $X$ -axis and has magnitude $r$ , its $X$ -component is $r_x = r \cos\theta$ .

The cosine function tells us everything. It reaches its maximum value of $1$ when $\theta = 0°$ , meaning the vector points directly along the positive $X$ -axis. At any other angle, $\cos\theta < 1$ , so the component shrinks.

Let me walk through each option to see this clearly:

Option (A): $\vec{r}$ along positive $Y$ -axis

Here $\theta = 90°$ , so $r_x = r\cos(90°) = r \cdot 0 = 0$ . The component vanishes entirely because the vector is perpendicular to the $X$ -axis.
Option (B): $\vec{r}$ along positive $X$ -axis

Now $\theta = 0°$ , giving $r_x = r\cos(0°) = r \cdot 1 = r$ . The entire magnitude of the vector contributes to the $X$ -component. This is the maximum possible value.
Option (C): $\vec{r}$ at $45°$ to the $X$ -axis

We get $r_x = r\cos(45°) = r \cdot \frac{1}{\sqrt{2}} = \frac{r}{\sqrt{2}} \approx 0.707r$ . This is less than the full magnitude $r$ .
Option (D): $\vec{r}$ along negative $Y$ -axis

This means $\theta = 270°$ (or $-90°$ ), so $r_x = r\cos(270°) = r \cdot 0 = 0$ . Again, perpendicular means zero component. …

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Concept: Optimizing a Trigonometric Function Instead of Comparing Four Cases

Method: Calculus Extremum Test on $f(\theta) = r\cos\theta$ (Never Evaluate the Four Given Options)

Rather than substituting each of the four listed directions into $r_x = r\cos\theta$ and comparing the four resulting numbers, this method treats $r_x$ as a function of the free angle $\theta$ and finds where that function is maximized using ordinary single-variable calculus -- the four options are then read off as special cases of a single continuous curve, not four separate calculations.

Steps

Write the $X$ -component as an explicit function of the angle. For a vector of fixed magnitude $r$ making angle $\theta$ with the $X$ -axis:

f(\theta) = r_x(\theta) = r\cos\theta, \qquad \theta \in [0^\circ, 360^\circ)

Differentiate and locate critical points.

f'(\theta) = -r\sin\theta

Setting $f'(\theta)=0$ (with $r\neq0$ ) gives $\sin\theta=0$ , i.e. $\theta = 0^\circ$ or $\theta=180^\circ$ -- these are the only two candidates for an extremum on the whole circle, without ever mentioning $90^\circ$ or $45^\circ$ .

Classify each critical point with the second-derivative test.

f''(\theta) = -r\cos\theta

At $\theta=0^\circ$ : $f''(0^\circ) = -r < 0$ (since $r>0$ ) $\Rightarrow$ local maximum.
At $\theta=180^\circ$ : $f''(180^\circ) = +r > 0$ $\Rightarrow$ local minimum.

Evaluate the function at the maximum.

f(0^\circ) = r\cos0^\circ = r

This is the single largest value $r_x$ can ever take, for any direction of $\vec r$ -- it is a global maximum since it's the only local max and $f$ is bounded by $\pm r$ . …

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

COMEDK 2026Set 2026-M1 markMCQ
$90^{\circ}$
›Reveal solutionSolution
Using the parallelogram law with the condition that the resultant is perpendicular to the smaller force gives $A^2-B^2=R^2$ . With $A+B=50$ and $R=20$ , we get $A-B=8$ , so $A=29\ \mathrm{N}$ and $B=21\ \mathrm{N}$ . The correct option is (C).

Concept and Intuition

When two forces act at a point, their resultant's magnitude and direction are given by the parallelogram law. Here, we know the resultant is perpendicular to the smaller force. That means the smaller force has no component along the resultant's direction — it is entirely "sideways" to the resultant. This geometric condition gives a clean relationship between the two forces and the resultant, which, together with the sum of their magnitudes, lets us solve for each.

Step-by-step solution

Set up variables and given conditions Let the two forces be $A$ and $B$ , with $A > B$ (so $B$ is the smaller force). Given:

$A + B = 50 \quad \text{(1)}$
Resultant magnitude $R = 20$ N.

The resultant is at $90^\circ$ to the smaller force $B$ .

Apply the parallelogram law For two forces $A$ and $B$ with an angle $\theta$ between them, the resultant magnitude is:

$R^2 = A^2 + B^2 + 2AB\cos\theta$
Also, the angle $\alpha$ that the resultant makes with force $B$ satisfies:
$\tan\alpha = \frac{A\sin\theta}{B + A\cos\theta}$
Here, $\alpha = 90^\circ$ (resultant perpendicular to $B$ ), so $\tan 90^\circ$ is undefined, meaning the denominator must be zero:
$B + A\cos\theta = 0 \quad \Rightarrow \quad \cos\theta = -\frac{B}{A}$

Substitute $\cos\theta$ into the resultant equation From $R^2 = A^2 + B^2 + 2AB\cos\theta$ , replace $\cos\theta$ :

$20^2 = A^2 + B^2 + 2AB\left(-\frac{B}{A}\right)$
Simplify:
$400 = A^2 + B^2 - 2B^2 = A^2 - B^2$
So:
$A^2 - B^2 = 400 \quad \text{(2)}$

Solve the system of equations From (1): $A = 50 - B$ . Substitute into (2):

$(50 - B)^2 - B^2 = 400$
Expand:
$2500 - 100B + B^2 - B^2 = 400$ $2500 - 100B = 400$ $100B = 2100 \quad \Rightarrow \quad B = 21$
Then $A = 50 - 21 = 29$ .

Check the "smaller force" condition We assumed $B$ is the smaller force, but here $B = 21$ and $A = 29$ — that's fine, $B < A$ . However, does the resultant being perpendicular to the smaller force hold? Yes, because we used that condition. But wait — let's verify the resultant magnitude:

$A^2 - B^2 = 29^2 - 21^2 = 841 - 441 = 400 \quad \checkmark$
So $R = \sqrt{400} = 20$ N. This matches option (C): 29 N, 21 N. …
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KCET 2024Set D-21 markMCQ
$-$
›Reveal solutionSolution

$\vec\omega$ lies along the rotation axis (the vertical axle, $\pm Z$ ); apply the right-hand rule to the sense drawn in the figure — the near edge moves $-Y$ , which forces $\vec\omega$ along $-Z$ .

Step 1 — The concept: angular velocity is an axial vector.

Unlike a linear velocity, $\vec\omega$ does not point along the direction anything is moving. It points along the axis of rotation, and its sense is given by the right-hand rule: curl the fingers of the right hand in the sense of the rotation, and the extended thumb gives the direction of $\vec\omega$ . Its magnitude is $\omega = d\theta/dt$ .

Step 2 — Fix the axis.

The fan hangs from a vertical down-rod and spins about that rod. The dashed axle is vertical, and in the given frame $Z$ points straight up. So $\vec\omega$ must be along $+Z$ or $-Z$ — the horizontal directions $X$ and $Y$ are impossible, immediately eliminating options (A) and (B). The only question left is the sign.

Step 3 — Read the sense of rotation from the figure.

The frame drawn is right-handed: $Z$ up, $Y$ to the right, $X$ out of the page towards the viewer. The rotation arrow is an ellipse round the axle whose near (front, $+X$ side) arc runs from right to LEFT, while the far arc runs left to right.

So a blade tip that is momentarily nearest the viewer (position $\vec r$ along $+\hat X$ ) has velocity $\vec v$ pointing in the $-\hat Y$ direction.

Step 4 — Apply $\vec v = \vec\omega \times \vec r$ to pin the sign.

Write $\vec\omega = \omega_z \hat Z$ and take the blade tip at $\vec r = r\hat X$ :
$\vec v = \omega_z \hat Z \times r\hat X = \omega_z r\,(\hat Z \times \hat X) = \omega_z r\,\hat Y .$
But the figure says this point moves along $-\hat Y$ , i.e. $\vec v = -|v|\hat Y$ . Matching:

$\omega_z r = -|v| \;\Longrightarrow\; \omega_z < 0 .$ …
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COMEDK 2023Set 2023-M1 markMCQ
$\vec{p}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$
›Reveal solutionSolution

The area of a parallelogram with adjacent sides $\vec p,\vec q$ is $|\vec p\times\vec q|=\sqrt{684}$ sq units.

With $\vec p=5\hat i-4\hat j+3\hat k$ and $\vec q=3\hat i+2\hat j-\hat k$ :
$\vec p\times\vec q=\begin{vmatrix}\hat i&\hat j&\hat k\\5&-4&3\\3&2&-1\end{vmatrix}$
$\hat i:\;(-4)(-1)-(3)(2)=4-6=-2$

$\hat j:\;-[(5)(-1)-(3)(3)]=-[-5-9]=14$

$\hat k:\;(5)(2)-(-4)(3)=10+12=22$ …
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COMEDK 2022Set 20221 markMCQ
$^\circ$
›Reveal solutionSolution

The force to which the resultant is perpendicular is therefore 10/sqrt(3) kg-wt (the other force is Q = 20/sqrt(3)).

Concept: resolve the two forces; the resultant is perpendicular to one of them, so the component of the resultant along that force is zero.

Let P lie along the x-axis and Q make 120 degrees with it.

Q components: (-Q/2, +Q sqrt(3)/2)

Resultant components: (P - Q/2, Q sqrt(3)/2)

Resultant perpendicular to P => x-component = 0:

P - Q/2 = 0 -> Q = 2P

Magnitude of the resultant: …
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COMEDK 2021Set 20211 markMCQ
$\widehat i - 3\widehat j + 2\widehat k$
›Reveal solutionSolution

Check: (4i + 3j - 5k) + (-3i - 3j + 5k) = i, which is indeed the unit vector along X.

Concept: Vector addition; a unit vector along the X-axis is i.

First add the two given vectors:

(i - 3j + 2k) + (3i + 6j - 7k) = (1+3)i + (-3+6)j + (2-7)k = 4i + 3j - 5k

Let the required vector be v. Then

(4i + 3j - 5k) + v = i

v = i - (4i + 3j - 5k) = (1 - 4)i + (0 - 3)j + (0 + 5)k

v = -3i - 3j + 5k …
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COMEDK 2021Set 2021-B1 markMCQ
$\vec{A} = \hat{\imath} - 3\hat{\jmath} + 2\hat{k}$
›Reveal solutionSolution

$\vec B = \hat\imath - \vec A = 3\hat\jmath - 2\hat k$ .

A unit vector along the X-axis is $\hat\imath = (1,0,0)$ . We require
$\vec A + \vec B = \hat\imath.$
Therefore …
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