Numerical Problems Based on Scalar or Dot Product of two Vectors for Class 11 Physics

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Numerical Problems Based on Scalar or Dot Product of two Vectors for Class 11 Physics

Numerical Problems on Dot Product and Work Done

Class 11 Physics · Motion in a Plane
SQ

Scalar Product Practice Set

Apply $\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$ and $W = \vec{F} \cdot \vec{S}$
Ch 4 · Motion in a Plane
1
Find the angle between the vectors $\vec{A} = 2\hat{i} – 4\hat{j} + 6\hat{k}$ and $\vec{B} = 3\hat{i} + \hat{j} + 2\hat{k}$. [Central Schools 17]
Answer $60^\circ$ 📝
Detailed Solution

The angle $\theta$ between two vectors is given by $\cos\theta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}$.

1. Dot Product ($\vec{A} \cdot \vec{B}$):

$$\vec{A} \cdot \vec{B} = (2)(3) + (-4)(1) + (6)(2) = 6 – 4 + 12 = 14$$

2. Magnitudes:

$$|\vec{A}| = \sqrt{2^2 + (-4)^2 + 6^2} = \sqrt{4 + 16 + 36} = \sqrt{56}$$
$$|\vec{B}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}$$

3. Calculate Angle:

$$\cos\theta = \frac{14}{\sqrt{56} \times \sqrt{14}} = \frac{14}{\sqrt{784}} = \frac{14}{28} = \frac{1}{2}$$
$$\theta = \cos^{-1}\left(\frac{1}{2}\right) = 60^\circ$$
2
Find the angles between the vectors $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{B} = -\hat{i} – \hat{j} + 2\hat{k}$. [Delhi 95]
Answer $90^\circ$ 📝
Detailed Solution

First, calculate the dot product $\vec{A} \cdot \vec{B}$:

$$\vec{A} \cdot \vec{B} = (1)(-1) + (1)(-1) + (1)(2)$$
$$\vec{A} \cdot \vec{B} = -1 – 1 + 2 = 0$$

Since the dot product of the two non-zero vectors is $0$, they are mutually perpendicular. Therefore, the angle between them is $90^\circ$.

3
For what value of $m$, is the vector $\vec{A} = 2\hat{i} + 3\hat{j} – 6\hat{k}$ perpendicular to the vector $\vec{B} = 3\hat{i} – m\hat{j} + 6\hat{k}$? [Delhi 16]
Answer $-10$ 📝
Detailed Solution

Two vectors are perpendicular if their dot product is zero ($\vec{A} \cdot \vec{B} = 0$).

$$(2)(3) + (3)(-m) + (-6)(6) = 0$$
$$6 – 3m – 36 = 0$$
$$-3m – 30 = 0 \implies -3m = 30$$
$$m = -10$$
4
For what value of $a$ are the vectors $\vec{A} = a\hat{i} – 2\hat{j} + \hat{k}$ and $\vec{B} = 2a\hat{i} + a\hat{j} – 4\hat{k}$ perpendicular to each other?
Answer $2, -1$ 📝
Detailed Solution

For the vectors to be perpendicular, $\vec{A} \cdot \vec{B} = 0$.

$$(a)(2a) + (-2)(a) + (1)(-4) = 0$$
$$2a^2 – 2a – 4 = 0$$

Divide the entire equation by 2:

$$a^2 – a – 2 = 0$$

Factorize the quadratic equation:

$$(a – 2)(a + 1) = 0$$

Therefore, $a = 2$ or $a = -1$.

5
Find the angles between the following pairs of vectors:
(i) $\vec{A} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{B} = -2\hat{i} – 2\hat{j} – 2\hat{k}$
(ii) $\vec{A} = -2\hat{i} + 2\hat{j} – \hat{k}$ and $\vec{B} = 3\hat{i} + 6\hat{j} + 2\hat{k}$
(iii) $\vec{A} = 4\hat{i} + 6\hat{j} – 3\hat{k}$ and $\vec{B} = -2\hat{i} – 5\hat{j} + 7\hat{k}$
Answer (i) $180^\circ$ (ii) $79^\circ$ (iii) $148.9^\circ$ 📝
Detailed Solution

(i) Notice that $\vec{B} = -2(\hat{i} + \hat{j} + \hat{k}) = -2\vec{A}$. Since $\vec{B}$ is a negative scalar multiple of $\vec{A}$, the vectors are anti-parallel. Thus, the angle is $\mathbf{180^\circ}$.

(ii) Calculate dot product and magnitudes:

$$\vec{A} \cdot \vec{B} = (-2)(3) + (2)(6) + (-1)(2) = -6 + 12 – 2 = 4$$
$$|\vec{A}| = \sqrt{(-2)^2 + 2^2 + (-1)^2} = \sqrt{9} = 3$$
$$|\vec{B}| = \sqrt{3^2 + 6^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$$
$$\cos\theta = \frac{4}{3 \times 7} = \frac{4}{21} \approx 0.1905 \implies \theta \approx \mathbf{79^\circ}$$

(iii) Calculate dot product and magnitudes:

$$\vec{A} \cdot \vec{B} = (4)(-2) + (6)(-5) + (-3)(7) = -8 – 30 – 21 = -59$$
$$|\vec{A}| = \sqrt{4^2 + 6^2 + (-3)^2} = \sqrt{16 + 36 + 9} = \sqrt{61}$$
$$|\vec{B}| = \sqrt{(-2)^2 + (-5)^2 + 7^2} = \sqrt{4 + 25 + 49} = \sqrt{78}$$
$$\cos\theta = \frac{-59}{\sqrt{61}\sqrt{78}} = \frac{-59}{\sqrt{4758}} \approx -0.855 \implies \theta \approx \mathbf{148.9^\circ}$$
6
Calculate the values of (i) $\hat{j} \cdot (2\hat{i} – 3\hat{j} + \hat{k})$ and (ii) $(2\hat{i} – \hat{j}) \cdot (3\hat{i} + \hat{k})$.
Answer (i) $-3$ (ii) $6$ 📝
Detailed Solution

We use the distributive property of the dot product and the orthogonal unit vector rules ($\hat{i}\cdot\hat{i} = 1$, $\hat{j}\cdot\hat{j} = 1$, $\hat{i}\cdot\hat{j} = 0$, etc.).

(i) $\hat{j} \cdot (2\hat{i} – 3\hat{j} + \hat{k})$

$$= 2(\hat{j}\cdot\hat{i}) – 3(\hat{j}\cdot\hat{j}) + 1(\hat{j}\cdot\hat{k}) = 2(0) – 3(1) + 1(0) = -3$$

(ii) $(2\hat{i} – \hat{j}) \cdot (3\hat{i} + 0\hat{j} + \hat{k})$

$$= (2)(3) + (-1)(0) + (0)(1) = 6$$
7
A force $\vec{P} = 4\hat{i} + \hat{j} + 3\hat{k}$ newton acts on a particle and displaces it through displacement $\vec{S} = 11\hat{i} + 11\hat{j} + 15\hat{k}$ metre. Calculate the work done by the force.
Answer 100 J 📝
Detailed Solution

Work done ($W$) is the dot product of the Force vector ($\vec{P}$) and the Displacement vector ($\vec{S}$).

$$W = \vec{P} \cdot \vec{S} = (4\hat{i} + \hat{j} + 3\hat{k}) \cdot (11\hat{i} + 11\hat{j} + 15\hat{k})$$
$$W = (4)(11) + (1)(11) + (3)(15)$$
$$W = 44 + 11 + 45 = 100\text{ Joules}$$
8
Under a force of $10\hat{i} – 3\hat{j} + 6\hat{k}$ newton, a body of mass $5\text{ kg}$ is displaced from the position $6\hat{i} + 5\hat{j} – 3\hat{k}$ to the position $10\hat{i} – 2\hat{j} + 7\hat{k}$. Calculate the work done.
Answer 121 J 📝
Detailed Solution

Force $\vec{F} = 10\hat{i} – 3\hat{j} + 6\hat{k}$.

The displacement vector $\vec{S}$ is the final position vector minus the initial position vector:

$$\vec{S} = \vec{r}_f – \vec{r}_i = (10\hat{i} – 2\hat{j} + 7\hat{k}) – (6\hat{i} + 5\hat{j} – 3\hat{k})$$
$$\vec{S} = (10-6)\hat{i} + (-2-5)\hat{j} + (7-(-3))\hat{k} = 4\hat{i} – 7\hat{j} + 10\hat{k}$$

Work done ($W$) is the dot product of $\vec{F}$ and $\vec{S}$ (Mass is extra information and not needed here):

$$W = \vec{F} \cdot \vec{S} = (10)(4) + (-3)(-7) + (6)(10)$$
$$W = 40 + 21 + 60 = 121\text{ Joules}$$
9
The sum and difference of two vectors $\vec{A}$ and $\vec{B}$ are $\vec{A} + \vec{B} = 2\hat{i} + 6\hat{j} + \hat{k}$ and $\vec{A} – \vec{B} = 4\hat{i} + 2\hat{j} – 11\hat{k}$. Find the magnitude of each vector and their scalar product $\vec{A} \cdot \vec{B}$.
Answer $|\vec{A}| = 5\sqrt{2}$, $|\vec{B}| = \sqrt{41}$, $\vec{A}\cdot\vec{B} = -25$ 📝
Detailed Solution

(Note: A typo in the original prompt incorrectly labeled the sum equation. We have applied the logical intent $\vec{A} + \vec{B} = 2\hat{i} + 6\hat{j} + \hat{k}$.)

We are given two equations:

  1. $\vec{A} + \vec{B} = 2\hat{i} + 6\hat{j} + \hat{k}$
  2. $\vec{A} – \vec{B} = 4\hat{i} + 2\hat{j} – 11\hat{k}$

To find $\vec{A}$, add the two equations:

$$2\vec{A} = (2+4)\hat{i} + (6+2)\hat{j} + (1-11)\hat{k} = 6\hat{i} + 8\hat{j} – 10\hat{k}$$
$$\vec{A} = 3\hat{i} + 4\hat{j} – 5\hat{k}$$

Magnitude of $\vec{A}$: $|\vec{A}| = \sqrt{3^2 + 4^2 + (-5)^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$

To find $\vec{B}$, subtract equation 2 from equation 1:

$$2\vec{B} = (2-4)\hat{i} + (6-2)\hat{j} + (1-(-11))\hat{k} = -2\hat{i} + 4\hat{j} + 12\hat{k}$$
$$\vec{B} = -\hat{i} + 2\hat{j} + 6\hat{k}$$

Magnitude of $\vec{B}$: $|\vec{B}| = \sqrt{(-1)^2 + 2^2 + 6^2} = \sqrt{1 + 4 + 36} = \sqrt{41}$

Scalar product $\vec{A} \cdot \vec{B}$ :

$$\vec{A} \cdot \vec{B} = (3)(-1) + (4)(2) + (-5)(6)$$
$$\vec{A} \cdot \vec{B} = -3 + 8 – 30 = -25$$

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