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

The key idea is the magnitude of the cross product of two vectors.

Given $|\vec{A}| = 2$ and $|\vec{B}| = 4$, the formula becomes $|\vec{A} \times \vec{B}| = (2)(4) \sin\theta = 8 \sin\theta$.

We match the relations from Column I with the corresponding angles from Column II:

1. If $|\vec{A} \times \vec{B}| = 0$, then $8 \sin\theta = 0 \implies \sin\theta = 0 \implies \theta = 0^\circ$. This matches (iv).
2. If $|\vec{A} \times \vec{B}| = 8$, then $8 \sin\theta = 8 \implies \sin\theta = 1 \implies \theta = 90^\circ$. This matches (iii). …

The cross product of two vectors, $\vec{A}$ and $\vec{B}$, results in a new vector that is perpendicular to both $\vec{A}$ and $\vec{B}$. The magnitude of this resultant vector, $|\vec{A} \times \vec{B}|$, has a specific geometric meaning: it represents the area of the parallelogram formed by the two vectors $\vec{A}$ and $\vec{B}$ when placed tail-to-tail.

The magnitude of the cross product is directly related to the magnitudes of the individual vectors and the sine of the angle between them. This relationship is crucial for solving problems like this one, where we need to determine the angle based on the cross product's magnitude.

We are given the magnitudes of the two vectors:

$|\vec{A}| = 2$

$|\vec{B}| = 4$

Substituting these values into the formula, we get:

$|\vec{A} \times \vec{B}| = (2)(4)\sin\theta = 8\sin\theta$

Now, let's use this expression to match each relation in Column I with the corresponding angle in Column II.

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

   We set the magnitude of the cross product to $0$:

   $8\sin\theta = 0$

   $\sin\theta = 0$

   This implies that $\theta = 0^\circ$ or $\theta = 180^\circ$. In the context of the given options, $\theta = 0^\circ$ is the relevant choice. This means the vectors are parallel.

   Therefore, (a) matches with (iv) $\theta = 0^\circ$.

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

   We set the magnitude of the cross product to $8$:

   $8\sin\theta = 8$

   $\sin\theta = 1$ …

Concept: Compute the Cross Product's $z$-Component From Coordinates Instead of the Magnitude Formula

Method: Place $\vec A$ Along the $x$-Axis and Use the 2D Determinant

Just as with the dot product, this method fixes $\vec A$ along the $x$-axis (WLOG) and resolves $\vec B$ at angle $\theta$, then computes the cross product's out-of-plane component directly from the determinant $A_xB_y-A_yB_x$, instead of quoting $|\vec A\times\vec B|=|\vec A||\vec B|\sin\theta$ from memory.

Steps

1. Coordinates (as before): $\vec A=(2,0,0)$, $\vec B=(4\cos\theta,4\sin\theta,0)$.

2. Compute the cross product's $z$-component using the determinant rule:

Since both vectors lie in the plane, this $z$-component is the entire cross product, so $|\vec A\times\vec B| = 8\sin\theta$ (for $\theta$ between $0^\circ$ and $180^\circ$, where $\sin\theta\ge0$).

3. Match each given magnitude against $8\sin\theta$ by recognizing the value against the standard sine table, rather than solving $\sin\theta=k$ as a fresh equation each time:
   • $|\vec A\times\vec B|=0 \Rightarrow \sin\theta=0$, the table entry for $\theta=0^\circ$ (iv).
   • $|\vec A\times\vec B|=8 \Rightarrow \sin\theta=1$, the table entry for $\theta=90^\circ$ (iii). …