If $\vec{a}$ is any non-zero vector, then $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$ equals ________.

If veca is any non-zero vector, then (veca·hati)hati+(veca·hatj)hatj+(veca·hatk)hatk equals ____.

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\vec{a}

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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 expression is the component form of $\vec{a}$ itself. Since $\hat{i},\hat{j},\hat{k}$ are orthonormal basis vectors, the sum of the projections onto each axis reconstructs the original vector. The answer is $\vec{a}$ .

The key here is to recognise what the expression $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$ actually means.

When you take the dot product $\vec{a}\cdot\hat{i}$ , you get the scalar component of $\vec{a}$ along the $x$ -axis. Multiplying that scalar by the unit vector $\hat{i}$ gives you the vector component of $\vec{a}$ in the $x$ -direction. The same holds for $\hat{j}$ and $\hat{k}$ .

So the sum is simply the decomposition of $\vec{a}$ into its Cartesian components, added back together. That sum must equal $\vec{a}$ itself.

Let’s verify it step by step.

Write $\vec{a}$ in component form.

Any vector in 3D can be written as $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ , where $a_1, a_2, a_3$ are real numbers.
Compute each dot product.
- $\vec{a}\cdot\hat{i} = (a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot\hat{i} = a_1$ (since $\hat{i}\cdot\hat{i}=1$ , $\hat{j}\cdot\hat{i}=0$ , $\hat{k}\cdot\hat{i}=0$ ).
- Similarly, $\vec{a}\cdot\hat{j} = a_2$ and $\vec{a}\cdot\hat{k} = a_3$ .
Multiply each scalar by its unit vector.
- $(\vec{a}\cdot\hat{i})\hat{i} = a_1\hat{i}$
- $(\vec{a}\cdot\hat{j})\hat{j} = a_2\hat{j}$
- $(\vec{a}\cdot\hat{k})\hat{k} = a_3\hat{k}$
Add them up.

$a_1\hat{i} + a_2\hat{j} + a_3\hat{k} = \vec{a}$ …

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

TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
$\vec{A}$
›Reveal solutionSolution
The component of one vector along another is given by the dot product divided by the magnitude of the reference vector. Setting the given condition equal leads to a simple ratio of magnitudes: $\frac{|\vec{A}|}{|\vec{B}|} = \sqrt{2}$ , which corresponds to option (C) $2:1$ after squaring.

The key idea here is understanding what "component of a vector along another vector" means. The component of $\vec{A}$ along $\vec{B}$ is the scalar projection: how much of $\vec{A}$ points in the direction of $\vec{B}$ . This is given by $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$ . Similarly, the component of $\vec{B}$ along $\vec{A}$ is $\frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$ . The problem tells us one is twice the other, which sets up a simple equation relating the magnitudes.

Write the given condition mathematically. The component of $\vec{A}$ along $\vec{B}$ is $\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|}$ . The component of $\vec{B}$ along $\vec{A}$ is $\frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}$ . The problem states:

$\frac{\vec{A} \cdot \vec{B}}{|\vec{B}|} = 2 \cdot \frac{\vec{A} \cdot \vec{B}}{|\vec{A}|}.$

Simplify the equation. Assuming $\vec{A}$ and $\vec{B}$ are not perpendicular (otherwise the dot product is zero and the condition is trivial or undefined), we can divide both sides by $\vec{A} \cdot \vec{B}$ (which is nonzero). This gives:

$\frac{1}{|\vec{B}|} = \frac{2}{|\vec{A}|}.$

Solve for the ratio of magnitudes. Cross-multiplying:
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TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
$\vec{a} = 4\vec{i} + 5\vec{j} - 3\vec{k}$
›Reveal solutionSolution
The component of $\vec{b}$ parallel to $\vec{a}$ is the projection of $\vec{b}$ onto $\vec{a}$ . Its magnitude is $\frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}|} = 2\sqrt{2}$ , so the correct option is (A).

Concept & Intuition

When we ask for the “component of $\vec{b}$ parallel to $\vec{a}$ ,” we mean: how much of $\vec{b}$ points in the same direction as $\vec{a}$ ? This is exactly the scalar projection of $\vec{b}$ onto $\vec{a}$ , given by the formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ . The sign tells us if it’s in the same or opposite direction; the magnitude is the absolute value. So we just compute the dot product and the length of $\vec{a}$ , then divide.

Step-by-step solution

Write the vectors in component form

$\vec{a} = (4, 5, -3)$ and $\vec{b} = (6, -2, -2)$ .

Compute the dot product $\vec{a} \cdot \vec{b}$

$\vec{a} \cdot \vec{b} = 4 \cdot 6 + 5 \cdot (-2) + (-3) \cdot (-2) = 24 - 10 + 6 = 20.$

Compute the magnitude of $\vec{a}$

$|\vec{a}| = \sqrt{4^2 + 5^2 + (-3)^2} = \sqrt{16 + 25 + 9} = \sqrt{50} = 5\sqrt{2}.$

Find the scalar projection (component parallel to $\vec{a}$ ) The component of $\vec{b}$ parallel to $\vec{a}$ is
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Q.If $\vec{a}$ is any non-zero vector, then $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$ equals ________.

Concept understanding — Vector Component Extraction

Vector Component Extraction: The Intuition

The Precise Statement

Why This Works: The Geometry

A Concrete Example

Why This Matters …

Method: Reconstructing a Vector from Its Projections onto $\hat{i},\hat{j},\hat{k}$

Steps

Common Mistakes

Q.If a⃗\vec{a}a is any non-zero vector, then (a⃗⋅i^)i^+(a⃗⋅j^)j^+(a⃗⋅k^)k^(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}(a⋅i^)i^+(a⋅j^​)j^​+(a⋅k^)k^ equals ________.

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Q.If $\vec{a}$ is any non-zero vector, then $(\vec{a}\cdot\hat{i})\hat{i}+(\vec{a}\cdot\hat{j})\hat{j}+(\vec{a}\cdot\hat{k})\hat{k}$ equals ________.

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