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.

GUJCET 2020Set 071 markMCQ
$2x - 3y + 4z - 6 = 0$
›Reveal solutionSolution

The foot of the perpendicular from the origin lies along the normal $\vec n=(2,-3,4)$ .

Concept: The foot is $(2t,-3t,4t)$ for some $t$ , and it lies on the plane $2x-3y+4z-6=0$ :

$2(2t)-3(-3t)+4(4t)-6=0\ \Rightarrow\ (4+9+16)t=6\ \Rightarrow\ 29t=6\ \Rightarrow\ t=\frac{6}{29}.$ …
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GUJCET 2021Set 151 markMCQ
$2x - 3y + 4z - 12 = 0$
›Reveal solutionSolution

The foot of the perpendicular from the origin is along the plane's normal.

Concept. For plane $ax+by+cz+d=0$ , foot from origin $=\dfrac{-d}{a^2+b^2+c^2}(a,b,c)$ . …
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GSEB Higher Secondary Certificate (HSC) Examination 2026Set ANNUAL1 markMCQ
$\hat i+3\hat j+7\hat k$
(a) $60/114$
(b) $60/\sqrt{114}$
(c) $66/114$
(d) $66/\sqrt{114}$
›Reveal solutionSolution

Projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ .

$\vec a\cdot\vec b=(1)(7)+(3)(-1)+(7)(8)=7-3+56=60$ . $|\vec b|=\sqrt{49+1+64}=\sqrt{114}$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2025Set ANNUAL1 markMCQ
$\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$
(a) $\dfrac{3\sqrt{6}}{5}$
(b) $\dfrac{5\sqrt{2}}{3}$
(c) $\dfrac{5\sqrt{3}}{2}$
(d) $\dfrac{5\sqrt{6}}{3}$
›Reveal solutionSolution

Projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a\cdot\vec b}{|\vec b|}$ .

$\vec a\cdot\vec b=2(1)+3(2)+2(1)=10$ . $|\vec b|=\sqrt{1+4+1}=\sqrt6$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2024Set ANNUAL1 markMCQ
$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$
(a) $\frac{10}{\sqrt{6}}$
(b) $\frac{\sqrt{10}}{6}$
(c) $\frac{\sqrt{10}}{17}$
(d) $\frac{10}{\sqrt{17}}$
›Reveal solutionSolution

Projection of $\vec a$ on $\vec b$ is $\frac{\vec a\cdot\vec b}{|\vec b|}$ .

$\vec a\cdot\vec b=1(2)+2(3)+1(2)=10$ . $|\vec b|=\sqrt{4+9+4}=\sqrt{17}$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2019Set ANNUAL1 markMCQ
$(-1, 2, -1)$
(a) $-\frac{1}{\sqrt6}$
(b) $\frac{1}{\sqrt6}$
(c) $1$
(d) $-1$
›Reveal solutionSolution

The (scalar) projection of a vector on $\hat i$ is simply its $x$ -component.

…

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$A(1, 2, -3)$
(a) $8x + 8y + 14z - 47 = 0$
(b) $8x - 8y - 14z + 47 = 0$
(c) $8x - 8y - 14z - 47 = 0$
(d) $8x + 8y + 14z + 47 = 0$
›Reveal solutionSolution

The mirror plane is the perpendicular bisector of the segment joining a point and its image.

Here $A(1,2,-3)$ and image $B(-3,6,4)$ .

Normal direction $=B-A=(-4,4,7)$ .

Midpoint $M=\left(\frac{1-3}{2},\frac{2+6}{2},\frac{-3+4}{2}\right)=\left(-1,4,\tfrac12\right)$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$\bar{x} = (1, 2, 1)$
(a) $\frac{5}{\sqrt{14}}$
(b) $-\frac{5}{14}$
(c) $\frac{5}{14}$
(d) $-\frac{5}{\sqrt{14}}$
›Reveal solutionSolution

Component (scalar projection) of $\bar x$ along $\bar y$ is $\frac{\bar x\cdot\bar y}{|\bar y|}$ .

$\bar x\cdot\bar y=(1)(2)+(2)(-3)+(1)(-1)=2-6-1=-5$ .

$|\bar y|=\sqrt{4+9+1}=\sqrt{14}$ .

…

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GSEB Higher Secondary Certificate (HSC) Examination 2018Set ANNUAL1 markMCQ
$(1, 2, -3)$
(a) $x + 2y - 3z = 0$
(b) $\frac{x}{1} + \frac{y}{2} - \frac{z}{3} = 1$
(c) $x + 2y - 3z = -6$
(d) $x + 2y - 3z = 14$
›Reveal solutionSolution

The foot of the perpendicular from the origin is both the normal direction and a point on the plane.

The foot $N(1,2,-3)$ gives the normal $(1,2,-3)$ . The plane $x+2y-3z=d$ passes through $N$ : …

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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

What is being asked

Step 1: build the two vectors

Step 2: dot product and magnitude

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