Numerical Problems Based on Expressing the Vectors in terms of Base Vectors and Rectangular Components of Vectors for Class 11 Physics

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Numerical Problems Based on Expressing the Vectors in terms of Base Vectors and Rectangular Components of Vectors for Class 11 Physics

Numerical Problems on Vector Addition and Resolution

Class 11 Physics · Motion in a Plane
SQ

Vector Algebra Practice Set

Apply vector addition, unit vectors, and resolution of vectors
Ch 4 · Motion in a Plane
1
If $\vec{A} = 3\hat{i} + 2\hat{j}$ and $\vec{B} = \hat{i} – 2\hat{j} + 3\hat{k}$, find the magnitudes of $\vec{A} + \vec{B}$ and $\vec{A} – \vec{B}$.
Answer 5, $\sqrt{29}$ 📝
Detailed Solution

Addition ($\vec{A} + \vec{B}$):

$$\vec{A} + \vec{B} = (3\hat{i} + 2\hat{j} + 0\hat{k}) + (\hat{i} – 2\hat{j} + 3\hat{k})$$
$$= (3 + 1)\hat{i} + (2 – 2)\hat{j} + (0 + 3)\hat{k} = 4\hat{i} + 0\hat{j} + 3\hat{k}$$

Magnitude:

$$|\vec{A} + \vec{B}| = \sqrt{4^2 + 0^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$

Subtraction ($\vec{A} – \vec{B}$):

$$\vec{A} – \vec{B} = (3\hat{i} + 2\hat{j} + 0\hat{k}) – (\hat{i} – 2\hat{j} + 3\hat{k})$$
$$= (3 – 1)\hat{i} + (2 – (-2))\hat{j} + (0 – 3)\hat{k} = 2\hat{i} + 4\hat{j} – 3\hat{k}$$

Magnitude:

$$|\vec{A} – \vec{B}| = \sqrt{2^2 + 4^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$$
2
Find the unit vector parallel to the resultant of the vectors $\vec{A} = 2\hat{i} – 6\hat{j} – 3\hat{k}$ and $\vec{B} = 4\hat{i} + 3\hat{j} – \hat{k}$.
Answer $\frac{1}{\sqrt{61}}(6\hat{i} – 3\hat{j} – 4\hat{k})$ 📝
Detailed Solution

The resultant vector $\vec{R}$ is the sum of vectors $\vec{A}$ and $\vec{B}$:

$$\vec{R} = \vec{A} + \vec{B} = (2+4)\hat{i} + (-6+3)\hat{j} + (-3-1)\hat{k}$$
$$\vec{R} = 6\hat{i} – 3\hat{j} – 4\hat{k}$$

The magnitude of the resultant vector is:

$$|\vec{R}| = \sqrt{(6)^2 + (-3)^2 + (-4)^2} = \sqrt{36 + 9 + 16} = \sqrt{61}$$

The unit vector $\hat{R}$ parallel to $\vec{R}$ is given by $\hat{R} = \frac{\vec{R}}{|\vec{R}|}$:

$$\hat{R} = \frac{1}{\sqrt{61}}(6\hat{i} – 3\hat{j} – 4\hat{k})$$
3
Determine the vector which when added to the resultant of $\vec{A} = 2\hat{i} – 4\hat{j} – 6\hat{k}$ and $\vec{B} = 4\hat{i} + 3\hat{j} + 3\hat{k}$ gives the unit vector along z axis.
Answer $-6\hat{i} + \hat{j} + 4\hat{k}$ 📝
Detailed Solution

(Note: The original question text contained minor typos in the notation for $\vec{A}$ and $\vec{B}$, which have been corrected here based on standard vector notation.)

First, find the resultant $\vec{R}$ of $\vec{A}$ and $\vec{B}$:

$$\vec{R} = \vec{A} + \vec{B} = (2\hat{i} – 4\hat{j} – 6\hat{k}) + (4\hat{i} + 3\hat{j} + 3\hat{k})$$
$$\vec{R} = (2+4)\hat{i} + (-4+3)\hat{j} + (-6+3)\hat{k} = 6\hat{i} – \hat{j} – 3\hat{k}$$

Let the required vector be $\vec{C}$. We are given that adding $\vec{C}$ to $\vec{R}$ gives the unit vector along the z-axis, which is $\hat{k}$:

$$\vec{R} + \vec{C} = \hat{k}$$
$$\vec{C} = \hat{k} – \vec{R}$$
$$\vec{C} = \hat{k} – (6\hat{i} – \hat{j} – 3\hat{k})$$
$$\vec{C} = -6\hat{i} + \hat{j} + (1 – (-3))\hat{k} = -6\hat{i} + \hat{j} + 4\hat{k}$$
4
Find the value of $\lambda$ in the unit vector $0.4\hat{i} + 0.8\hat{j} + \lambda\hat{k}$.
Answer $\sqrt{0.2}$ 📝
Detailed Solution

By definition, the magnitude of a unit vector is exactly 1.

$$\sqrt{(0.4)^2 + (0.8)^2 + (\lambda)^2} = 1$$

Squaring both sides:

$$0.16 + 0.64 + \lambda^2 = 1^2$$
$$0.80 + \lambda^2 = 1$$
$$\lambda^2 = 1 – 0.80 = 0.20$$
$$\lambda = \sqrt{0.2}$$
5
Given three coplanar vectors $\vec{a} = 4\hat{i} – \hat{j}$, $\vec{b} = -3\hat{i} + 2\hat{j}$ and $\vec{c} = -3\hat{j}$. Find the magnitude of the sum of the three vectors.
Answer $\sqrt{5}$ 📝
Detailed Solution

Let the sum of the three vectors be $\vec{S}$:

$$\vec{S} = \vec{a} + \vec{b} + \vec{c}$$
$$\vec{S} = (4\hat{i} – \hat{j}) + (-3\hat{i} + 2\hat{j}) + (-3\hat{j})$$

Combine the $\hat{i}$ and $\hat{j}$ components:

$$\vec{S} = (4 – 3)\hat{i} + (-1 + 2 – 3)\hat{j}$$
$$\vec{S} = 1\hat{i} – 2\hat{j}$$

The magnitude of this sum is:

$$|\vec{S}| = \sqrt{(1)^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$$
6
A force is inclined at $30^\circ$ to the horizontal. If its rectangular component in the horizontal direction is $50\text{ N}$, find the magnitude of the force and its vertical component.
Answer $57.74\text{ N}, 28.87\text{ N}$ 📝
Detailed Solution

Let the magnitude of the force be $F$. The angle $\theta = 30^\circ$.

The horizontal component of the force ($F_x$) is given by:

$$F_x = F \cos\theta$$

Substitute the given values ($F_x = 50\text{ N}$):

$$50 = F \cos(30^\circ) = F \left( \frac{\sqrt{3}}{2} \right)$$
$$F = \frac{100}{\sqrt{3}} \approx \frac{100}{1.732} \approx 57.74\text{ N}$$

The vertical component of the force ($F_y$) is given by:

$$F_y = F \sin\theta = 57.74 \sin(30^\circ)$$
$$F_y = 57.74 \times 0.5 = 28.87\text{ N}$$

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