Compare the components for the following vector eq – Class 11 Physics

IV. Exercises · Q13

Q. Compare the components for the following vector equations.

(a)

T\hat{j} - mg\hat{j} = ma\hat{j}

(b)

\vec{T} + \vec{F} = \vec{A} + \vec{B}

(c)

\vec{T} - \vec{F} = \vec{A} - \vec{B}

(d)

T\hat{j} - mg\hat{j} = -ma\hat{j}

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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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Step 1. (a) $T\hat j-mg\hat j=ma\hat j$ : both sides are along $\hat j$ only, so comparing the single (y) component directly gives $T-mg=ma$ (this is the tension equation for a lift accelerating upward, e.g. Newton's second law applied to a mass in an accelerating lift).

Step 2. (b) $\vec T+\vec F=\vec A+\vec B$ : resolving all four vectors into $x$ , $y$ , $z$ components and equating each axis separately gives three scalar equations: $T_x+F_x=A_x+B_x$ , $T_y+F_y=A_y+B_y$ , $T_z+F_z=A_z+B_z$ .

Step 3. (c) $\vec T-\vec F=\vec A-\vec B$ : similarly, $T_x-F_x=A_x-B_x$ , $T_y-F_y=A_y-B_y$ , $T_z-F_z=A_z-B_z$ . …

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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 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.

…

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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).

…

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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

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

…

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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|. …

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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

…

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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 $(-5,7)$ .

…

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

…

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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 …

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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 compon …

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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

…

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Q.Compare the components for the following vector equations.

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