Q.$\hat{i}$ and $\hat{j}$ are unit vectors along $x$- and $y$-axis respectively. What is the magnitude and direction of the vectors $\hat{i} + \hat{j}$, and $\hat{i} - \hat{j}$? What are the components of a vector $\vec{A} = 2\,\hat{i} + 3\,\hat{j}$ along the directions of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$? [You may use graphical method]

The problem involves fundamental vector operations: finding magnitudes and directions of resultant vectors, and calculating scalar components of a vector along specific directions. The key idea is Vector Operations and Scalar Projection.

1. Magnitude and Direction of $\hat{i} + \hat{j}$:

   Let $\vec{V_1} = \hat{i} + \hat{j}$.

   Its magnitude is $|\vec{V_1}| = \sqrt{1^2 + 1^2} = \sqrt{2}$.

   Its direction, $\theta_1$, with the positive x-axis is given by $\tan \theta_1 = \frac{1}{1} = 1$, so $\theta_1 = 45^\circ$.

2. Magnitude and Direction of $\hat{i} - \hat{j}$:

   Let $\vec{V_2} = \hat{i} - \hat{j}$.

   Its magnitude is $|\vec{V_2}| = \sqrt{1^2 + (-1)^2} = \sqrt{2}$.

   Its direction, $\theta_2$, with the positive x-axis is given by $\tan \theta_2 = \frac{-1}{1} = -1$, so $\theta_2 = -45^\circ$ (or $315^\circ$).

3. Component of $\vec{A} = 2\,\hat{i} + 3\,\hat{j}$ along $\hat{i} + \hat{j}$:

   The unit vector in the direction of $\hat{i} + \hat{j}$ is $\hat{u}_1 = \frac{\hat{i} + \hat{j}}{|\hat{i} + \hat{j}|} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.

   The scalar component is $\vec{A} \cdot \hat{u}_1 = (2\,\hat{i} + 3\,\hat{j}) \cdot \left(\frac{\hat{i} + \hat{j}}{\sqrt{2}}\right) = \frac{(2)(1) + (3)(1)}{\sqrt{2}} = \frac{5}{\sqrt{2}}$.

4. Component of $\vec{A} = 2\,\hat{i} + 3\,\hat{j}$ along $\hat{i} - \hat{j}$: …

Understanding vectors begins with grasping their two fundamental properties: magnitude (how long they are) and direction (where they point). When we add or subtract vectors, we are essentially finding a resultant vector that represents the combined effect of the individual vectors. For example, if you walk 1 meter east ($\hat{i}$) and then 1 meter north ($\hat{j}$), your final displacement from the start is $\hat{i} + \hat{j}$.

The problem asks us to find these properties for two specific vectors, $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$, and then to find the scalar components of another vector $\vec{A}$ along these directions. The components tell us "how much" of vector $\vec{A}$ acts in a particular direction.

We'll use the standard Cartesian coordinate system where $\hat{i}$ is a unit vector along the positive $x$-axis and $\hat{j}$ is a unit vector along the positive $y$-axis. This means $\hat{i} = (1,0)$ and $\hat{j} = (0,1)$.

Part 1: Magnitude and Direction of $\hat{i} + \hat{j}$

1. Visualizing the vector:

   Imagine starting at the origin $(0,0)$. The vector $\hat{i}$ takes you one unit along the positive $x$-axis to $(1,0)$. From there, the vector $\hat{j}$ takes you one unit along the positive $y$-axis to $(1,1)$. The resultant vector $\hat{i} + \hat{j}$ is the vector from the origin to $(1,1)$. Graphically, this forms a right-angled triangle with sides of length 1 along the $x$ and $y$ axes.

2. Calculating the magnitude:

   The magnitude of a vector $\vec{V} = x\,\hat{i} + y\,\hat{j}$ is given by the Pythagorean theorem: $|\vec{V}| = \sqrt{x^2 + y^2}$.

   For $\vec{V}_1 = \hat{i} + \hat{j}$, we have $x=1$ and $y=1$.

3. Calculating the direction: The direction of a vector is typically given by the angle it makes with the positive $x$-axis, measured counter-clockwise. If $\vec{V} = x\,\hat{i} + y\,\hat{j}$, the angle $\theta$ can be found using trigonometry: $\tan \theta = \frac{y}{x}$. For $\hat{i} + \hat{j}$, we have $x=1$ and $y=1$.

Since both $x$ and $y$ components are positive, the vector lies in the first quadrant.

Part 2: Magnitude and Direction of $\hat{i} - \hat{j}$

1. Visualizing the vector:

   Starting at the origin, $\hat{i}$ takes you one unit along the positive $x$-axis to $(1,0)$. The vector $-\hat{j}$ takes you one unit along the negative $y$-axis from $(1,0)$ to $(1,-1)$. The resultant vector $\hat{i} - \hat{j}$ is the vector from the origin to $(1,-1)$. This also forms a right-angled triangle, but in the fourth quadrant.

2. Calculating the magnitude:

   Using the same formula, for $\vec{V}_2 = \hat{i} - \hat{j}$, we have $x=1$ and $y=-1$.

Notice that the magnitude is the same as for $\hat{i} + \hat{j}$. This makes sense because the lengths of the components are the same, just their directions differ.

3. Calculating the direction:

For $\hat{i} - \hat{j}$, we have $x=1$ and $y=-1$.

Since $x$ is positive and $y$ is negative, the vector lies in the fourth quadrant. The angle whose tangent is $-1$ is typically given as $-45^\circ$ or $315^\circ$ (which is $360^\circ - 45^\circ$). Both are valid ways to express the direction. We will use $-45^\circ$.

Part 3: Components of $\vec{A} = 2\,\hat{i} + 3\,\hat{j}$ along the directions of $\hat{i} + \hat{j}$ and $\hat{i} - \hat{j}$

The component of a vector $\vec{A}$ along the direction of another vector $\vec{B}$ is the scalar projection of $\vec{A}$ onto $\vec{B}$. This value tells us how much of $\vec{A}$ "points" in the direction of $\vec{B}$.

1. Find the unit vectors for the given directions:
   • For the direction of $\vec{V}_1 = \hat{i} + \hat{j}$: We found $|\vec{V}_1| = \sqrt{2}$. So, the unit vector is $\hat{u}_1 = \frac{\hat{i} + \hat{j}}{\sqrt{2}}$.
   • For the direction of $\vec{V}_2 = \hat{i} - \hat{j}$: We found $|\vec{V}_2| = \sqrt{2}$. …

Concept: Treat $\hat i+\hat j$ and $\hat i-\hat j$ as a Rotated Coordinate Frame

Method: Standard Axis-Rotation Transformation Formulas (Not Dot-Product-with-Unit-Vector)

Both existing solutions find the component of $\vec A$ along a direction by taking the dot product of $\vec A$ with the unit vector in that direction. This method instead recognizes that $\dfrac{\hat i+\hat j}{\sqrt2}$ and $\dfrac{\hat i-\hat j}{\sqrt2}$ are two mutually perpendicular unit vectors forming a new, rotated coordinate system — tilted $45^\circ$ from the original $x$-$y$ axes — and finds $\vec A$'s components in that new frame using the standard rotation-of-axes transformation formula from vector algebra, rather than a projection formula.

Step 1 — Magnitude and direction of the two given vectors

$\hat i+\hat j$ points at $\theta=45^\circ$ from the $+x$-axis (equal positive $x,y$ parts); $\hat i-\hat j$ points at $\theta=-45^\circ$ (equal positive $x$, negative $y$).

Step 2 — Recognize these as a rotated pair of axes

Since $\hat i+\hat j$ and $\hat i-\hat j$ are perpendicular (their dot product $1\cdot1+1\cdot(-1)=0$) and equal in length, $\hat u_1=\dfrac{\hat i+\hat j}{\sqrt2}$ and $\hat u_2=\dfrac{\hat i-\hat j}{\sqrt2}$ form a valid orthonormal basis — a new set of axes ($x',y'$) obtained by rotating the original axes by $\theta=45^\circ$.

Step 3 — Apply the standard rotation-transformation formula

For axes rotated by an angle $\theta$, a vector's component along the new $x'$-axis (the direction at angle $\theta$ from the original $x$-axis) is given by the standard transformation rule:

With $\vec A = 2\,\hat i+3\,\hat j$ (so $A_x=2$, $A_y=3$) and $\theta=45^\circ$ (the direction of $\hat i+\hat j$):

Step 4 — Repeat for the second (perpendicular) rotated axis, at $\theta=-45^\circ$

The direction of $\hat i-\hat j$ is at $\theta'=-45^\circ$ from the original $x$-axis. Using the same transformation formula with $\theta'=-45^\circ$ (so $\cos\theta'=\tfrac{1}{\sqrt2}$, $\sin\theta'=-\tfrac{1}{\sqrt2}$): …