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Q3:
A girl walks 4 km towards west, then she walks 3 km in a direction $30^{\circ}$ east of north and stops. Determine the girl’s displacement from her initial point of departure.
Solution :
Step 1: Establishing the Coordinate System
To analyze the displacement, we define a Cartesian coordinate system where the origin $O(0, 0)$ represents the initial point of departure. Let the positive $x$-axis represent the East direction and the positive $y$-axis represent the North direction.
- The unit vector along the East is $\hat{i}$.
- The unit vector along the North is $\hat{j}$.
- The unit vector along the West is $-\hat{i}$.
Step 2: Vector Representation of Displacements
Vector $\vec{d_1}$ (First displacement): The girl walks 4 km towards the West.
$\vec{d_1} = -4\hat{i} + 0\hat{j}$ [Per Cartesian mapping of cardinal directions]
Vector $\vec{d_2}$ (Second displacement): The girl walks 3 km at an angle of $30^{\circ}$ East of North.
In this orientation, "North" corresponds to the positive $y$-axis ($90^{\circ}$). Moving $30^{\circ}$ East of North shifts the angle towards the positive $x$-axis. Thus, the angle with the positive $y$-axis is $30^{\circ}$, making the angle with the positive $x$-axis $\theta = 90^{\circ} - 30^{\circ} = 60^{\circ}$.
Using trigonometric resolution: $\vec{d_2} = 3(\cos 60^{\circ})\hat{i} + 3(\sin 60^{\circ})\hat{j}$
$\vec{d_2} = 3(\frac{1}{2})\hat{i} + 3(\frac{\sqrt{3}}{2})\hat{j} = 1.5\hat{i} + 1.5\sqrt{3}\hat{j}$
Step 3: Calculation of Net Displacement
The total displacement $\vec{R}$ is the vector sum of $\vec{d_1}$ and $\vec{d_2}$ [By Triangle Law of Vector Addition].
$\vec{R} = \vec{d_1} + \vec{d_2}$
$\vec{R} = (-4\hat{i} + 0\hat{j}) + (1.5\hat{i} + 1.5\sqrt{3}\hat{j})$
$\vec{R} = (-4 + 1.5)\hat{i} + (1.5\sqrt{3})\hat{j}$
$\vec{R} = -2.5\hat{i} + 1.5\sqrt{3}\hat{j}$
Step 4: Magnitude and Direction Analysis
The magnitude of displacement $|\vec{R}|$ is given by $\sqrt{x^2 + y^2}$.
$|\vec{R}| = \sqrt{(-2.5)^2 + (1.5\sqrt{3})^2}$
$|\vec{R}| = \sqrt{6.25 + (2.25 \times 3)} = \sqrt{6.25 + 6.75} = \sqrt{13}$
$|\vec{R}| \approx 3.605 \text{ km}$
Final Solution: The girl's displacement from the initial point is $\sqrt{13} \text{ km}$ at an angle of $\tan^{-1}\left(\frac{1.5\sqrt{3}}{-2.5}\right)$ relative to the West axis.
More Questions from Class 12 Mathematics Vector Algebra Miscellaneous Exercise on Chapter 10
- Q1: Write down a unit vector in XY-plane, making an angle of $30^{\circ}$ with the positive direction of x-axis.
- Q2: Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.
- Q4: If $\vec{a} = \vec{b} + \vec{c}$, then is it true that $|\vec{a}| = |\vec{b}| + |\vec{c}|$? Justify your answer.
- Q5: Find the value of $x$ for which $x(\hat{i} + \hat{j} + \hat{k})$ is a unit vector.
- Q6: Find a vector of magnitude 5 units, and parallel to the resultant of the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$.
- Q7: If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} - \vec{b} + 3\vec{c}$.
- Q8: Show that the points A (1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7) are collinear, and find the ratio in which B divides AC.
- Q9: Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $2\vec{a} + \vec{b}$ and $\vec{a} - 3\vec{b}$ externally in the ratio 1 : 2. Also, show that P is the mid point of the line segment RQ.
- Q10: The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} + 5\hat{k}$ and $\hat{i} - 2\hat{j} - 3\hat{k}$. Find the unit vector parallel to its diagonal. Also, find its area.
- Q11: Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.
- Q12: Let $\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}$, $\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}$. Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$, and $\vec{c} \cdot \vec{d} = 15$.
- Q13: The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.
- Q14: If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.
- Q15: Prove that $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$, if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$.
- Q16: If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \ge 0$ only when
- Q17: Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if
- Q18: The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is
- Q19: If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to
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