Q.Find the scalar and vector components of the vector with initial point $(2, 1)$ and terminal point $(-5, 7)$.

Concept and Intuition

Any 2D vector's magnitude and its two perpendicular (x and y) components form a right triangle, so $|\vec p|^2 = p_x^2+p_y^2$. Knowing the magnitude and one component lets you solve for the other.

Step-by-Step Solution

1. Given $|\vec p| = 25$, $p_y = 7$.
2. $p_x^2 = |\vec p|^2 - p_y^2 = 625 - 49 = 576$.
3. $p_x = \sqrt{576} = 24$ units. …

Concept and Intuition

Any vector decomposes uniquely into a component along a given direction plus a component perpendicular to it; the two given pieces ARE that decomposition, so $\bar a$ is simply their vector sum, no projection formula needed.

Step-by-Step Solution

1. $\bar a = \dfrac{7}{25}(4\bar i+5\bar j+3\bar k) + \dfrac1{25}(47\bar i-10\bar j-46\bar k)$.
2. $\bar i$-component: $\dfrac{7\times4+47}{25}=\dfrac{28+47}{25}=\dfrac{75}{25}=3$.
3. $\bar j$-component: $\dfrac{7\times5-10}{25}=\dfrac{35-10}{25}=\dfrac{25}{25}=1$.
4. $\bar k$-component: $\dfrac{7\times3-46}{25}=\dfrac{21-46}{25}=\dfrac{-25}{25}=-1$. …

Concept and Intuition

The component of $\vec P$ along a direction is the projection of $\vec P$ onto the unit vector of that direction: $\vec P\cdot\hat n$, where $\hat n$ is obtained by normalizing the given direction vector.

Step-by-Step Solution

1. Direction vector: $\hat i+2\hat j$, magnitude $=\sqrt{1^2+2^2}=\sqrt5$.
2. Unit vector: $\hat n=\dfrac{\hat i+2\hat j}{\sqrt5}$.
3. $\vec P=3\hat i+4\hat j$. …

Concept and Intuition

If three vectors sum to a known resultant, the unknown one is just the resultant minus the sum of the known ones — vector subtraction is done component-by-component, independently along $\hat i$, $\hat j$, $\hat k$.

Step-by-Step Solution

1. Given $\vec A = -\hat i+2\hat j+3\hat k$ and $\vec B=-2\hat i-\hat j-4\hat k$.
2. Add them: $\vec A+\vec B = (-1-2)\hat i+(2-1)\hat j+(3-4)\hat k = -3\hat i+\hat j-\hat k$.
3. The resultant $\vec A+\vec B+\vec C$ is a vector along $+\hat z$ with magnitude 2, i.e. $\vec A+\vec B+\vec C=2\hat k$. …

Concept and Intuition

Rectangular components of a force are mutually perpendicular, so the magnitude of the resultant is the hypotenuse of a right triangle whose legs are the two components: $F_{resultant}^2=F_x^2+F_y^2$.

Step-by-Step Solution

1. Let the resultant force be $F=40\,N$, one component $F_1=20\sqrt3\,N$, and the other component $F_2$ unknown.
2. By Pythagoras: $F^2=F_1^2+F_2^2 \Rightarrow F_2^2 = F^2-F_1^2$.
3. Compute $F_1^2=(20\sqrt3)^2=400\times3=1200$.
4. Compute $F^2=40^2=1600$. …

Concept and Intuition

Vector addition in components: if the sum of two vectors points purely along one axis, the components along the other axis must cancel exactly. This gives one equation; combined with the given magnitude of $\vec Q$, we get a second equation — enough to solve for $P$.

Step-by-Step Solution

1. Let $\vec P = (P, 0)$ since it's along the x-axis, and $\vec Q=(Q_x,Q_y)$ with $|\vec Q|=10\Rightarrow Q_x^2+Q_y^2=100$.
2. Resultant $\vec R=\vec P+\vec Q=(P+Q_x,\ Q_y)$. Since $\vec R$ is along the y-axis, its x-component is zero: $P+Q_x=0\Rightarrow Q_x=-P$.
3. $\vec R$'s magnitude is given as $2P$ (twice that of $\vec P$), and since $\vec R$ is purely along y: $|R|=|Q_y|=2P$. …

Concept and Intuition

For any vector in 3D space, the direction cosines with respect to the three coordinate axes always satisfy $l^2+m^2+n^2=1$ where $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$. This single identity is the key to relating any combination of these angles' sines and cosines.

Step-by-Step Solution

1. Direction cosine identity: $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$.
2. Use $\sin^2\theta=1-\cos^2\theta$ for each of $\alpha,\beta$:

3. From the identity, $\cos^2\alpha+\cos^2\beta=1-\cos^2\gamma$.
4. Substitute: $\sin^2\alpha+\sin^2\beta=2-(1-\cos^2\gamma)=1+\cos^2\gamma$.

Common Mistakes …

Concept and Intuition

The component of any vector $\vec V$ along a direction $\hat n$ is $\vec V\cdot\hat n$. Here the natural vector to project is $(\vec A-\vec B)$ along $(\vec A+\vec B)$. Since $\vec A\perp\vec B$, we know $\vec A\cdot\vec B=0$, which simplifies both the dot product and the magnitude of $\vec A+\vec B$ nicely (Pythagoras-like).

Step-by-Step Solution

1. Unit vector along $\vec A+\vec B$: $\hat n=\dfrac{\vec A+\vec B}{|\vec A+\vec B|}$.
2. Since $\vec A\perp\vec B$, $\vec A\cdot\vec B=0$, so $|\vec A+\vec B|^2=|A|^2+|B|^2+2\vec A\cdot\vec B=|A|^2+|B|^2$, giving $|\vec A+\vec B|=\sqrt{|A|^2+|B|^2}$.
3. Component of $(\vec A-\vec B)$ along $\hat n$: $(\vec A-\vec B)\cdot\hat n=\dfrac{(\vec A-\vec B)\cdot(\vec A+\vec B)}{|\vec A+\vec B|}$.
4. Expand numerator: $(\vec A-\vec B)\cdot(\vec A+\vec B)=|A|^2-\vec A\cdot\vec B+\vec B\cdot\vec A-|B|^2=|A|^2-|B|^2$ (using $\vec A\cdot\vec B=0$). …

The problem involves a classic constrained motion setup: two blocks connected by a string that passes over a fixed pulley. The string’s length doesn’t change, so the rate at which the string shortens on one side must equal the rate at which it lengthens on the other. Here, block A moves horizontally, pulling the string along the 60° direction; block B moves vertically. The trick is to relate A’s horizontal speed to the speed of the string segment that actually moves B.

1. Identify the constraint

   The string is inextensible. Therefore, the speed of block B (which moves straight up or down) equals the speed at which the string is being pulled along its own direction from block A’s side. That is, the component of A’s velocity parallel to the string is what matters.

2. Resolve A’s velocity along the string

   Block A moves horizontally to the right at $10\ \text{m/s}$. The string makes a $60^\circ$ angle with the horizontal (measured at A). The component of A’s velocity along the string is:

This is the speed at which the string is being pulled from A’s side.

3. Relate to block B’s speed …

Concept and Intuition

$\vec{OP}\cdot\bar i$ is just the $x$-component of the position vector $\vec{OP}$ (the dot product with the unit vector $\bar i$ picks out the $x$-coordinate). So these conditions directly specify where on the curve $P$ and $Q$ sit.

Step-by-Step Solution

1. $\vec{OP}\cdot\bar i=-1 \Rightarrow$ $P$ has $x=-1$. Since $P$ lies on $y=2^{x+2}$: $y_P = 2^{-1+2}=2^1=2$. So $P=(-1,2)$.
2. $\vec{OQ}\cdot\bar i=2 \Rightarrow$ $Q$ has $x=2$. Then $y_Q=2^{2+2}=2^4=16$. So $Q=(2,16)$.
3. $\vec{OQ}=2\bar i+16\bar j$, $\vec{OP}=-\bar i+2\bar j$. …