Q.The component of a vector $\vec{r}$ along $X$-axis will have maximum value if

Concept: Vector Component Extraction

The component of a vector $\vec{r}$ along the $X$-axis is given by the projection formula:

where $\theta$ is the angle between $\vec{r}$ and the positive $X$-axis.

Since the magnitude $|\vec{r}|$ is fixed, maximizing $r_x$ requires maximizing $\cos \theta$. The cosine function achieves its maximum value of $1$ when $\theta = 0°$, meaning the vector points directly along the positive $X$-axis. …

When we project a vector onto an axis, we are asking: how much of this vector "lives" in that direction? Mathematically, if $\vec{r}$ makes an angle $\theta$ with the $X$-axis and has magnitude $r$, its $X$-component is $r_x = r \cos\theta$.

The cosine function tells us everything. It reaches its maximum value of $1$ when $\theta = 0°$, meaning the vector points directly along the positive $X$-axis. At any other angle, $\cos\theta < 1$, so the component shrinks.

Let me walk through each option to see this clearly:

1. Option (A): $\vec{r}$ along positive $Y$-axis

   Here $\theta = 90°$, so $r_x = r\cos(90°) = r \cdot 0 = 0$. The component vanishes entirely because the vector is perpendicular to the $X$-axis.

2. Option (B): $\vec{r}$ along positive $X$-axis

   Now $\theta = 0°$, giving $r_x = r\cos(0°) = r \cdot 1 = r$. The entire magnitude of the vector contributes to the $X$-component. This is the maximum possible value.

3. Option (C): $\vec{r}$ at $45°$ to the $X$-axis

   We get $r_x = r\cos(45°) = r \cdot \frac{1}{\sqrt{2}} = \frac{r}{\sqrt{2}} \approx 0.707r$. This is less than the full magnitude $r$.

4. Option (D): $\vec{r}$ along negative $Y$-axis

   This means $\theta = 270°$ (or $-90°$), so $r_x = r\cos(270°) = r \cdot 0 = 0$. Again, perpendicular means zero component. …

Concept: Optimizing a Trigonometric Function Instead of Comparing Four Cases

Method: Calculus Extremum Test on $f(\theta) = r\cos\theta$ (Never Evaluate the Four Given Options)

Rather than substituting each of the four listed directions into $r_x = r\cos\theta$ and comparing the four resulting numbers, this method treats $r_x$ as a function of the free angle $\theta$ and finds where that function is maximized using ordinary single-variable calculus -- the four options are then read off as special cases of a single continuous curve, not four separate calculations.

Steps

1. Write the $X$-component as an explicit function of the angle. For a vector of fixed magnitude $r$ making angle $\theta$ with the $X$-axis:

2. Differentiate and locate critical points.

Setting $f'(\theta)=0$ (with $r\neq0$) gives $\sin\theta=0$, i.e. $\theta = 0^\circ$ or $\theta=180^\circ$ -- these are the only two candidates for an extremum on the whole circle, without ever mentioning $90^\circ$ or $45^\circ$.

3. Classify each critical point with the second-derivative test.

• At $\theta=0^\circ$: $f''(0^\circ) = -r < 0$ (since $r>0$) $\Rightarrow$ local maximum.
• At $\theta=180^\circ$: $f''(180^\circ) = +r > 0$ $\Rightarrow$ local minimum.

4. Evaluate the function at the maximum.

This is the single largest value $r_x$ can ever take, for any direction of $\vec r$ -- it is a global maximum since it's the only local max and $f$ is bounded by $\pm r$. …