ctor $\vec{p}$ whose magnitude is $|\vec{a}||\cos\theta|$, with the same direction as $l$ when $\cos\theta > 0$ and opposite when $\cos\theta < 0$. Its magnitude $|\vec{p}|$ is called the **projection** of $\vec{a}$ on $l$. > [!NOTE] > The textbook uses "projection" for the scalar (signed) magnitude, and "projection vector" for the full vector carrying both magnitude and direction. ## Observations and Key Results ### Observation 1: Projection Using a Unit Vector If $\hat{p}$ is the unit vector along a line $l$, the projection of $\vec{a}$ on $l$ is: $$ \vec{a} \cdot \hat{p} $$ because $\vec{a} \cdot \hat{p} = |\vec{a}||\hat{p}|\cos\theta = |\vec{a}|\cos\theta$, the signed magnitude of the projection. ### Observation 2: Projection of One Vector on Another The projection of $\vec{a}$ on another vector $\vec{b}$ is: $$ \vec{a} \cdot \hat{b} \quad \text{or} \quad \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} $$ where $\hat{b} = \frac{\vec{b}}{|\vec{b}|}$ is the unit vector in the direction of $\vec{b}$. > [!FORMULA] > **Projection of $\vec{a}$ on $\vec{b}$** > $$ \text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} $$ ### Observation 3: Special Angles - If $\theta = 0$, then $\cos\theta = 1$, so the projection vector of $\vec{a}$ is $\vec{a}$ itself. - If $\theta = \pi$, then $\cos\theta = -1$, so the projection vector of $\vec{a}$ is $-\vec{a}$ (opposite direction). ### Observation 4: Perpendicular Case If $\theta = \frac{\pi}{2}$ or $\theta = \frac{3\pi}{2}$, then $\cos\theta = 0$, so the projection vector of $\vec{a}$ is the **zero vector**. ## Direction Cosines and Projections on Axes If $\alpha$, $\beta$, $\gamma$ are the direction angles of $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, its direction cosines are: …184:T97b,The figure is a set of four panels, labelled (i) through (iv), each showing the same basic geometry: a horizontal directed line $l$ (with an arrowhead indicating its positive direction), a vector $\overrightarrow{AB}$ starting from a point $A$ on $l$, and a dashed perpendicular dropped from $B$ to meet $l$ at a foot $C$. The segment $AC$ along $l$ is the projection vector $\overrightarrow{p}$. What changes from panel to panel is the angle $\theta$ that $\overrightarrow{AB}$ makes with $l$, measured anticlockwise from $l$ to the vector. The four panels cover the four quadrants of the angle: - **(i)** $0^\circ < \theta < 90^\circ$ — the projection $\overrightarrow{p}$ points to the right (same direction as $l$). - **(ii)** $90^\circ < \theta < 180^\circ$ — the projection $\overrightarrow{p}$ points to the left (opposite to $l$). - **(iii)** $180^\circ < \theta < 270^\circ$ — again $\overrightarrow{p}$ points left. - **(iv)** $270^\circ < \theta < 360^\circ$ — $\overrightarrow{p}$ points right, and here $B$ lies below the line $l$. The core idea is that the projection of a vector onto a directed line is itself a vector — its magnitude is $|\overrightarrow{AB}|\,|\cos\theta|$, and its direction is the same as $l$ when $\cos\theta > 0$ and opposite when $\cos\theta < 0$. The sign of $\cos\theta$ alone decides the direction of the projection vector along $l$. > [!FORMULA] > $$ \text{Projection vector of } \vec{a} \text{ on line } l = (\vec{a}\cdot\hat{p})\,\hat{p} $$ > where $\hat{p}$ is the unit vector along $l$. The scalar $\vec{a}\cdot\hat{p}$ is called the **projection** (a signed scalar), and its absolute value is the length of the projection vector. The figure makes this sign convention visual: in panels (i) and (iv), $\cos\theta$ is positive, so the projection vector points rightward along $l$; in (ii) and (iii), $\cos\theta$ is negative, so it points leftward. The dashed perpendicular from $B$ to $C$ is the geometric construction that gives $AC = |\overrightarrow{AB}|\cos\theta$ — the signed length of the projection. > [!WARNING] > A common mistake is to think the projection is always the length of the shadow on the line. The projection **vector** has a direction: it points along $l$ if the angle is acute, and opposite to $l$ if the angle is obtu

Q.If two vectors a and b are such that |a| = 2, |b| = 3 and a . b = 4, then find |a - b|.

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