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+

NCERT Exemplar · Q12

\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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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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