If A, B, C, D are the points with position vectors $\hat{i}+\hat{j}-\hat{k}$, $2\hat{i}-\hat{j}+3\hat{k}$, $2\hat{i}-3\hat{k}$, $3\hat{i}-2\hat{j}+\hat{k}$, respectively, find the projection of $\vec{AB}$ along $\vec{CD}$.

If A, B, C, D are the points with position vectors hati+hatj-hatk, 2hati-hatj+3hatk, 2hati-3hatk, 3hati-2hatj+

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\hat{i}+\hat{j}-\hat{k}

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Concept understanding — Vector Projection

Vector Projection

Picture a stick leaning in sunlight with the sun directly overhead: the shadow it casts on the ground is the projection of the stick onto the ground. The stick is your vector, the ground is the direction you project onto, and the shadow tells you how much of the stick lies along that direction.

That is the whole idea: projection answers "how much of this vector points in that particular direction?"

The Geometry

Take two vectors $\vec{a}$ and $\vec{b}$ . The projection of $\vec{a}$ onto $\vec{b}$ is a new vector that

lies along the line of $\vec{b}$ (parallel to $\vec{b}$ ), and
has length equal to how much of $\vec{a}$ points along $\vec{b}$ .

Note

The scalar projection is a number; the vector projection is a vector — same information, but the vector version also carries direction.

The Formula

For $\vec{b} \neq \vec{0}$ ,

\text{proj}_{\vec{b}}\,\vec{a} = \frac{\vec{a}\cdot\vec{b}}{\lVert\vec{b}\rVert^{2}}\,\vec{b}, \qquad \text{comp}_{\vec{b}}\,\vec{a} = \frac{\vec{a}\cdot\vec{b}}{\lVert\vec{b}\rVert}.

Why it works: $\vec{a}\cdot\vec{b}$ measures how much $\vec{a}$ "agrees" with $\vec{b}$ (positive if aligned, negative if opposed, zero if perpendicular). Dividing by $\lVert\vec{b}\rVert^{2}$ turns that into the signed length of the shadow relative to $\vec{b}$ , and multiplying by $\vec{b}$ places that length along $\vec{b}$ .

A Quick Example

Let $\vec{a}=(3,4)$ and $\vec{b}=(1,1)$ (the line $y=x$ ):

$\vec{a}\cdot\vec{b}=3+4=7$ , and $\lVert\vec{b}\rVert^{2}=2$
$\text{proj}_{\vec{b}}\,\vec{a}=\frac{7}{2}(1,1)=(3.5,\,3.5)$

The shadow sits exactly on the line $y=x$ . …

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$\vec{AB}=\vec{CD}=\hat{i}-2\hat{j}+4\hat{k}$ , so the projection of $\vec{AB}$ along $\vec{CD}$ is $\dfrac{21}{\sqrt{21}}=\sqrt{21}$ .

What is being asked

The scalar projection of $\vec{AB}$ along $\vec{CD}$ measures how much of $\vec{AB}$ lies in the direction of $\vec{CD}$ . It is

\text{projection}=\frac{\vec{AB}\cdot\vec{CD}}{|\vec{CD}|}.

Step 1: build the two vectors

\vec{AB}=\vec{B}-\vec{A}=(2-1)\hat{i}+(-1-1)\hat{j}+(3-(-1))\hat{k}=\hat{i}-2\hat{j}+4\hat{k}

\vec{CD}=\vec{D}-\vec{C}=(3-2)\hat{i}+(-2-0)\hat{j}+(1-(-3))\hat{k}=\hat{i}-2\hat{j}+4\hat{k}

(They happen to be the same vector.)

Step 2: dot product and magnitude

$\vec{AB}\cdot\vec{CD}=(1)(1)+(-2)(-2)+(4)(4)=1+4+16=21$ …

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

KEAM 2024Set eng-2024-06074 marksMCQ
$A(0,3,-3)$
›Reveal solutionSolution

$\overrightarrow{AB}=B-A=(1,-2,4)$ , $\overrightarrow{AC}=C-A=(2,-3,6)$ . Dot product $=2+6+24=32$ ; $|\overrightarrow{AC}|=\sqrt{4+9+36}=7$ . Projection $=\dfrac{32}{7}$ .

Compute the vectors:
$\overrightarrow{AB}=(1-0,\,1-3,\,1-(-3))=(1,-2,4),$ $\overrightarrow{AC}=(2-0,\,0-3,\,3-(-3))=(2,-3,6).$
Dot product: $\overrightarrow{AB}\cdot\overrightarrow{AC}=(1)(2)+(-2)(-3)+(4)(6)=2+6+24=32$ . …
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KEAM 2023Set eng-2023-P2-B24 marksMCQ
$\vec{a} = 2\hat{i} - 3\hat{j} + 4\hat{k}$
›Reveal solutionSolution
The projection is $\dfrac43$ .

Concept and Intuition

The scalar projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ .

Step-by-Step Solution

Dot product: $\vec a\cdot\vec b=(2)(1)+(-3)(2)+(4)(2)=2-6+8=4$ .

$|\vec b|=\sqrt{1+4+4}=3$ .

Projection $=\dfrac{4}{3}$ .

Common Mistakes …
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KEAM 2026Set eng-2026-04214 marksMCQ
$\vec{a}=\hat{i}+\hat{j}+\hat{k}$
›Reveal solutionSolution

Projection of $\vec a$ on $\vec b$ equals $\frac{\vec a\cdot\vec b}{|\vec b|}=\frac{1}{\sqrt3}$ .

With $\vec a=\hat i+\hat j+\hat k$ , $\vec b=\hat i-\hat j+\hat k$ :

$\vec a\cdot\vec b=1-1+1=1,\qquad |\vec b|=\sqrt{1+1+1}=\sqrt3.$ …
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KEAM 2025Set eng-2025-04294 marksMCQ
$\vec{a} = 3\hat{i} - \hat{j} - 2\hat{k}$
›Reveal solutionSolution

$\vec a\cdot\vec b=3-2+6=7$ and $|\vec b|=\sqrt{14}$ ; the scalar projection is $\frac{7}{\sqrt{14}}=\frac{\sqrt{14}}{2}$ . …

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KEAM 2025Set eng-2025-04254 marksMCQ
$\vec{a} = \alpha\hat{i} - 3\hat{j} - 2\hat{k}$
›Reveal solutionSolution

The scalar projection of $\vec a$ on $\vec c$ is $\tfrac{\vec a\cdot\vec c}{|\vec c|}$ ; set it to $2$ and solve for $\alpha$ .

$\vec a = \alpha\hat i - 3\hat j - 2\hat k$ , $\vec c = \hat i - 2\hat j + 2\hat k$ .
$\vec a \cdot \vec c = \alpha(1) + (-3)(-2) + (-2)(2) = \alpha + 6 - 4 = \alpha + 2.$ $|\vec c| = \sqrt{1 + 4 + 4} = 3.$
Projection: …
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KEAM 2025Set eng-2025-04234 marksMCQ
$|\vec{a}| = 12$
›Reveal solutionSolution
The angle is $\dfrac{\pi}{6}$ .

Concept and Intuition

The scalar projection of $\vec a$ on $\vec b$ equals $|\vec a|\cos\theta$ . Set it equal to the given value and solve for $\theta$ .

Step-by-Step Solution

Projection $=|\vec a|\cos\theta=12\cos\theta=6\sqrt3$ .

$\cos\theta=\dfrac{6\sqrt3}{12}=\dfrac{\sqrt3}{2}$ .

$\theta=\dfrac{\pi}{6}$ .

Common Mistakes …
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KEAM 2026Set eng-2026-04184 marksMCQ
$\vec{b}$
›Reveal solutionSolution

The scalar projection of $\vec b$ on $\vec a$ is $|\vec b|\cos\theta$ ; solving $|\vec b|\cdot\tfrac12=12$ gives $|\vec b|=24$ .

The projection of $\vec b$ on $\vec a$ equals $|\vec b|\cos\theta$ where $\theta=60^\circ$ : …
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KEAM 2024Set eng-2024-06054 marksMCQ
$\vec{a}$
›Reveal solutionSolution

Expanding the squared magnitude with $\vec a\cdot\vec b=\cos\theta$ and unit vectors gives $\sin^2\theta$ .

With $|\vec a|=|\vec b|=1$ and $\vec a\cdot\vec b=\cos\theta$ : …
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KEAM 2025Set eng-2025-04284 marksMCQ
$5, 6, 8$
›Reveal solutionSolution

The length is the root of the sum of squares of the axis projections: $\sqrt{125}=5\sqrt5$ .

If the projections of a line segment on the coordinate axes are $5,6,8$ , then the length of the segment is …
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KEAM 2024Set eng-2024-06064 marksMCQ
$\vec{a},\vec{b},\vec{c}$
›Reveal solutionSolution

Project onto $\vec a$ : $\vec a\cdot(\vec b+\vec c)=\vec a\cdot\vec b+\vec a\cdot\vec c$ .

Divide the dot-product identity by $|\vec a|$ :
$|\vec b+\vec c|\cos45^\circ=|\vec b|\cos30^\circ+|\vec c|\cos60^\circ.$
Compute the right side:

$\sqrt6\cdot\frac{\sqrt3}{2}+2\sqrt2\cdot\frac12=\frac{\sqrt{18}}{2}+\sqrt2=\frac{3\sqrt2}{2}+\sqrt2=\frac{5\sqrt2}{2}.$ …
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Q.If A, B, C, D are the points with position vectors $\hat{i}+\hat{j}-\hat{k}$ , $2\hat{i}-\hat{j}+3\hat{k}$ , $2\hat{i}-3\hat{k}$ , $3\hat{i}-2\hat{j}+\hat{k}$ , respectively, find the projection of $\vec{AB}$ along $\vec{CD}$ .

Concept understanding — Vector Projection

Vector Projection

The Geometry

The Formula

A Quick Example

What is being asked

Step 1: build the two vectors

Step 2: dot product and magnitude

Method: Scalar projection of one vector along another

Steps

Common Mistakes

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

Concept and Intuition

Step-by-Step Solution

Common Mistakes …

What is being asked

Step 1: build the two vectors

Step 2: dot product and magnitude

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