Q.If $|\vec{A}| = 2$ and $|\vec{B}| = 4$, then match the relations in column I with the angle $\theta$ between $\vec{A}$ and $\vec{B}$ in column II. Column I | Column II

Concept: Dot Product Angle — The dot product $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$ directly relates the scalar product to the cosine of the angle between the vectors.

Given $|\vec{A}| = 2$, $|\vec{B}| = 4$, so $|\vec{A}||\vec{B}| = 8$.

1. For $\vec{A} \cdot \vec{B} = 0$: $8\cos\theta = 0 \Rightarrow \cos\theta = 0 \Rightarrow \theta = 90^\circ$. Matches (ii).
2. For $\vec{A} \cdot \vec{B} = +8$: $8\cos\theta = 8 \Rightarrow \cos\theta = 1 \Rightarrow \theta = 0^\circ$. Matches (i). …

The dot product is not just a number — it’s a geometric probe. When you take $\vec{A} \cdot \vec{B}$, you’re really measuring how much one vector stretches along the other. The formula $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$ tells you that the result depends only on the magnitudes and the cosine of the angle between them. That cosine is the key: it ranges from $+1$ (aligned, $\theta = 0^\circ$) through $0$ (perpendicular, $\theta = 90^\circ$) down to $-1$ (opposite, $\theta = 180^\circ$). Every other angle gives a value in between.

Here, $|\vec{A}| = 2$ and $|\vec{B}| = 4$, so $|\vec{A}||\vec{B}| = 8$. That means the dot product can only be $8\cos\theta$. So each given dot product directly forces a specific $\cos\theta$, and from that we read off $\theta$.

1. For (a): $\vec{A} \cdot \vec{B} = 0$

   $0 = 8\cos\theta \implies \cos\theta = 0$. The angle whose cosine is $0$ is $\theta = 90^\circ$. That matches column II entry (ii).

2. For (b): $\vec{A} \cdot \vec{B} = +8$

   $+8 = 8\cos\theta \implies \cos\theta = 1$. The only angle in the usual $0^\circ$ to $180^\circ$ range with cosine $1$ is $\theta = 0^\circ$. That’s column II entry (i).

3. For (c): $\vec{A} \cdot \vec{B} = 4$ …

Concept: Fix One Vector Along an Axis (WLOG) Instead of Using the Coordinate-Free Dot-Product Formula

Method: Place $\vec A$ Along the $x$-Axis and Resolve $\vec B$ by Its Angle

The direct method plugs magnitudes straight into $\vec A\cdot\vec B=|\vec A||\vec B|\cos\theta$. This method instead chooses a convenient coordinate system -- since only relative orientation matters, $\vec A$ can always be placed along the $x$-axis without loss of generality -- and computes the dot product from components.

Steps

1. Place $\vec A$ along the $x$-axis (WLOG): $\vec A = (2,0)$, since $|\vec A|=2$.

2. Resolve $\vec B$ at the unknown angle $\theta$ from $\vec A$: $\vec B = (4\cos\theta,\,4\sin\theta)$, since $|\vec B|=4$.

3. Compute the dot product from components (using $\vec A\cdot\vec B = A_xB_x+A_yB_y$, not the magnitude-angle formula directly):

4. Match each given value against $8\cos\theta$:
   • $\vec A\cdot\vec B=0 \Rightarrow \cos\theta=0 \Rightarrow \theta=90^\circ$ (ii).
   • $\vec A\cdot\vec B=+8 \Rightarrow \cos\theta=1 \Rightarrow \theta=0^\circ$ (i).
   • $\vec A\cdot\vec B=4 \Rightarrow \cos\theta=\tfrac12 \Rightarrow \theta=60^\circ$ (iv). …