Q.The angle between $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$ is

Concept: Dot product — the angle $\theta$ between two vectors satisfies $\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta$.

Compute the dot product:

Magnitudes are $|\vec{A}| = \sqrt{1^2+1^2} = \sqrt{2}$ and $|\vec{B}| = \sqrt{1^2+(-1)^2} = \sqrt{2}$.

Since $\vec{A} \cdot \vec{B} = 0$, we have $\cos\theta = 0$, so $\theta = 90^\circ$.

The key to finding the angle between two vectors is the dot product — it directly connects the geometric idea of "how much one vector points along the other" to a simple algebraic calculation. For any two vectors $\vec{A}$ and $\vec{B}$, the dot product is defined as:

where $\theta$ is the angle between them. If you can compute the dot product and the magnitudes, you can solve for $\cos\theta$, and then $\theta$ itself.

Here, the vectors are given in component form: $\vec{A} = \hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$. Notice that $\vec{B}$ is just $\vec{A}$ with the $y$-component flipped — that suggests they might be perpendicular, but let's verify.

1. Compute the dot product. For vectors in $\hat{i}, \hat{j}$ components, multiply corresponding components and add:

2. Interpret the result. Since $\vec{A} \cdot \vec{B} = 0$, the equation $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos\theta$ gives:

Neither $\vec{A}$ nor $\vec{B}$ is the zero vector (each has magnitude $\sqrt{2}$), so we can divide by $|\vec{A}| |\vec{B}|$ to get:

3. Find the angle. The cosine of an angle is zero at $90^\circ$ (and also at $270^\circ$, but the angle between vectors is conventionally taken between $0^\circ$ and $180^\circ$). So:

Concept: Find the Angle from the Cross Product (via $\sin\theta$), Not the Dot Product

Method: $|\vec A\times\vec B| = |\vec A||\vec B|\sin\theta$ — a Different Vector Tool Entirely

Both existing solutions use the dot product ($\vec A\cdot\vec B=|\vec A||\vec B|\cos\theta$) to get $\cos\theta=0$. This method uses the cross product instead, which gives $\sin\theta$ rather than $\cos\theta$ — a genuinely different vector operation, not just a different arrangement of the same numbers.

Step 1 — Write the two vectors in full 3D form (needed for a cross product)

Step 2 — Compute the cross product

Step 3 — Magnitude of the cross product

Step 4 — Apply the cross-product magnitude formula to solve for $\theta$

With $|\vec A|=|\vec B|=\sqrt{1^2+1^2}=\sqrt2$:

(In the conventional range $0^\circ\le\theta\le180^\circ$ used for the angle between two vectors, $\sin\theta=1$ has exactly one solution, $\theta=90^\circ$ — no ambiguity.)

Why the cross-product route is a genuinely different check

The dot product isolates how much $\vec A$ and $\vec B$ point the same way (via $\cos\theta$); the cross product instead isolates how much they point in genuinely different directions (via $\sin\theta$, and its direction along $\hat k$ also reveals the sense of rotation from $\vec A$ to $\vec B$). That both operations independently point to exactly $90^\circ$ is a strong cross-check that neither computation has an arithmetic slip.

Final Answer

$|\vec A\times\vec B|=2=|\vec A||\vec B|\sin\theta \implies \sin\theta=1 \implies \theta=\boxed{90^\circ}$ — option (b).