Maharashtra Board Class 11 Physics Chapter 2 Mathematical Methods Solutions

1. Choose The Correct Option.

Question 1. The resultant of two forces 10 N and 15 N acting along + x and - x-axes respectively, is
(A) 25 N along + x-axis
(B) 25 N along - x-axis
(C) 5 N along + x-axis
(D) 5 N along - x-axis
Answer: (D) 5 N along - x-axis
In simple words: When forces act in opposite directions, their resultant is the difference between their magnitudes, acting in the direction of the larger force. Here, 15N in -x direction is larger than 10N in +x direction.

🎯 Exam Tip: Remember to consider both magnitude and direction when calculating the resultant of forces acting along an axis. Pay attention to the signs (+ or -) for directions.

Question 2. For two vectors to be equal, they should have the
(A) same magnitude
(B) same direction
(C) same magnitude and direction
(D) same magnitude but opposite direction
Answer: (C) same magnitude and direction
In simple words: Two vectors are considered identical only if they possess both the same length (magnitude) and point in the exact same orientation (direction).

🎯 Exam Tip: This is a fundamental definition in vector algebra. Ensure you clearly understand that both magnitude and direction are essential for vector equality, unlike scalars which only require magnitude.

Question 3. The magnitude of scalar product of two unit vectors perpendicular to each other is
(A) zero
(B) 1
(C) -1
(D) 2
Answer: (A) zero
In simple words: The scalar (dot) product of two vectors is given by \( |\vec{A}||\vec{B}|\cos\theta \). For perpendicular unit vectors, their magnitudes are 1 and the angle \( \theta \) is 90°, so \( \cos 90^\circ = 0 \), making the scalar product zero.

🎯 Exam Tip: A key property of the dot product is that it's zero for perpendicular vectors. This concept is crucial for solving problems involving orthogonality.

Question 4. The magnitude of vector product of two unit vectors making an angle of 60° with each other is
(A) 1
(B) 2
(C) \( \frac{3}{2} \)
(D) \( \frac{\sqrt{3}}{2} \)
Answer: (D) \( \frac{\sqrt{3}}{2} \)
In simple words: The vector (cross) product's magnitude is given by \( |\vec{A}||\vec{B}|\sin\theta \). For unit vectors, magnitudes are 1, and with \( \theta = 60^\circ \), \( \sin 60^\circ = \frac{\sqrt{3}}{2} \), so the magnitude is \( \frac{\sqrt{3}}{2} \).

🎯 Exam Tip: Remember the formula for the magnitude of the cross product. For unit vectors, it simplifies to \( \sin\theta \). Know common trigonometric values like \( \sin 60^\circ \).

Question 5. If \( \vec{A} \), \( \vec{B} \), and \( \vec{C} \) are three vectors, then which of the following is not correct?
(A) \( \vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} \)
(B) \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \)
(C) \( \vec{A} \times \vec{B} = \vec{B} \times \vec{A} \)
(D) \( \vec{A} \times (\vec{B} \times \vec{C}) = \vec{A} \times \vec{B} + \vec{B} \times \vec{C} \)
Answer: (C) \( \vec{A} \times \vec{B} = \vec{B} \times \vec{A} \)
In simple words: The cross product is anti-commutative, meaning \( \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \). Therefore, option (C) is incorrect because it implies commutativity. Option (A) shows distributivity of dot product, (B) shows commutativity of dot product, and (D) is an incorrect expansion for the triple cross product.

🎯 Exam Tip: Understand the properties of dot and cross products: dot product is commutative and distributive, while the cross product is anti-commutative and distributive. The given option (C) incorrectly states the commutativity of the cross product.

2. Answer The Following Questions.

Question 1. Show that \( \vec{A} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} \) is a unit vector. Solution: Let \( \hat{a} \) be unit vector of \( \hat{a} \). \( \therefore \quad \hat{a} = \frac{\vec{a}}{|\vec{a}|} \) Now, \( |\hat{a}| = \sqrt{a_x^2 + a_y^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{-1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1 \)
\( \therefore \quad \hat{a} = \frac{\vec{a}}{1} \implies \vec{a} \) itself is a unit vector.
Answer: The magnitude of the vector \( \vec{A} \) is 1, thus it is a unit vector.
In simple words: A unit vector has a magnitude of 1. By calculating the magnitude of the given vector using the formula \( \sqrt{x^2 + y^2} \), we find it is 1, confirming it's a unit vector.

🎯 Exam Tip: To show a vector is a unit vector, simply calculate its magnitude. If the magnitude is 1, it's a unit vector. Remember the formula \( |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \).

Question 2. If \( \vec{v_1} = 3\hat{i} + 4\hat{j} + \hat{k} \) and \( \vec{v_2} = \hat{i} - \hat{j} - \hat{k} \), determine the magnitude of \( \vec{v_1} + \vec{v_2} \). Solution: \( \vec{v_1} + \vec{v_2} = (3\hat{i} + 4\hat{j} + \hat{k}) + (\hat{i} - \hat{j} - \hat{k}) \) \( = 3\hat{i} + \hat{i} + 4\hat{j} - \hat{j} + \hat{k} - \hat{k} \) \( = 4\hat{i} + 3\hat{j} \) \( \therefore \) Magnitude of \( (\vec{v_1} + \vec{v_2}) \), \( |\vec{v_1} + \vec{v_2}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
Answer: Magnitude of \( \vec{v_1} + \vec{v_2} = 5 \) units.
In simple words: First, add the two vectors component-wise (i.e., add the \( \hat{i} \) components, then the \( \hat{j} \), then the \( \hat{k} \)). Then, calculate the magnitude of the resulting vector using the Pythagorean theorem in 3D (or 2D if one component becomes zero).

🎯 Exam Tip: Vector addition involves adding corresponding components. The magnitude of a vector \( A\hat{i} + B\hat{j} + C\hat{k} \) is \( \sqrt{A^2 + B^2 + C^2} \). Be careful with signs during addition.

Question 3. For \( \vec{v_1} = 2\hat{i} - 3\hat{j} \) and \( \vec{v_2} = -6\hat{i} + 5\hat{j} \), determine the magnitude and direction of \( \vec{v_1} + \vec{v_2} \). Answer: \( \vec{v_1} + \vec{v_2} = (2\hat{i} - 3\hat{j}) + (-6\hat{i} + 5\hat{j}) \) \( = (2\hat{i} - 6\hat{i}) + (-3\hat{j} + 5\hat{j}) \) \( = -4\hat{i} + 2\hat{j} \) \( \therefore |\vec{v_1} + \vec{v_2}| = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \)
Comparing \( \vec{v_1} + \vec{v_2} \), with \( \vec{R} = R_x\hat{i} + R_y\hat{j} \)
\( \implies R_x = -4 \) and \( R_y = 2 \)
Taking \( \theta \) to be angle made by \( \vec{R} \) with X-axis,
\( \therefore \quad \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{2}{-4}\right) \) \( = \tan^{-1}\left(-\frac{1}{2}\right) \) with X - axis
Answer: Magnitude and direction of \( \vec{v_1} + \vec{v_2} \), is respectively \( 2\sqrt{5} \) and \( \tan^{-1}\left(-\frac{1}{2}\right) \) with X - axis.
In simple words: First, add the vectors component-wise to get a resultant vector. Then, calculate its magnitude using the Pythagorean theorem. For direction, use the arctangent of the ratio of the y-component to the x-component, being mindful of the quadrant to correctly determine the angle.

🎯 Exam Tip: When finding direction, calculate \( \tan^{-1}(R_y/R_x) \) and then adjust the angle based on the signs of \( R_x \) and \( R_y \) to place it in the correct quadrant (e.g., if \( R_x \) is negative and \( R_y \) is positive, the angle is in the second quadrant). Magnitude is always positive.

Question 4. Find a vector which is parallel to \( \vec{v} = \hat{i} - 2\hat{j} \) and has a magnitude 10. Answer: Let the vector be \( \vec{w} = w_x\hat{i} + w_y\hat{j} \) \( |\vec{w}| = \sqrt{w_x^2 + w_y^2} = 10 \)...(Given)
\( w_x^2 + w_y^2 = 100 \)....(i)
Also, \( \vec{v} \cdot \vec{w} = vw \)
...(: \( |\vec{v}| \) and \( |\vec{w}| \) are parallel vectors)
\( \implies (\hat{i}-2\hat{j}) \cdot (w_x\hat{i} + w_y\hat{j}) = \sqrt{(1)^2 + (-2)^2} \times 10 \)
\( w_x - 2w_y = \sqrt{1+4} \times 10 \)
\( w_x - 2w_y = 10\sqrt{5} \)....(ii)
Substituting for \( w_x \) in (i) using equation (ii),
\( (10\sqrt{5} + 2w_y)^2 + w_y^2 = 100 \)
\( 500 + 40\sqrt{5}w_y + 4w_y^2 - 100 = 0 \)
\( 5w_y^2 + 40\sqrt{5}w_y + 400 = 0 \)
\( w_y^2 + 8\sqrt{5}w_y + 80 = 0 \)
Using factorisation formula,
\( w_y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( w_y = \frac{-8\sqrt{5} \pm \sqrt{(8\sqrt{5})^2 - 4 \times 1 \times 80}}{2 \times 1} \)
\( w_y = \frac{-8\sqrt{5} \pm \sqrt{320 - 320}}{2} = \frac{-8\sqrt{5} \pm 0}{2} = -4\sqrt{5} \)
Using equation (ii),
\( w_x = 10\sqrt{5} + 2(-4\sqrt{5}) \)
\( w_x = 10\sqrt{5} - 8\sqrt{5} \)
\( w_x = 2\sqrt{5} \)
\( \therefore \quad \vec{w} = w_x\hat{i} + w_y\hat{j} = 2\sqrt{5}\hat{i} - 4\sqrt{5}\hat{j} \)
Answer: Required vector is \( 2\sqrt{5}\hat{i} - 4\sqrt{5}\hat{j} \).
In simple words: A vector parallel to \( \vec{v} \) will be of the form \( n\vec{v} \). First, find the unit vector of \( \vec{v} \). Then, multiply this unit vector by the desired magnitude (10) to get the required vector.

🎯 Exam Tip: For problems involving parallel vectors, the desired vector can be found by normalizing the given vector (finding its unit vector) and then multiplying it by the required magnitude. This avoids complex algebraic equations if done systematically.

Alternate method: When two vectors are parallel, one vector is scalar multiple of another, i.e., if \( \vec{v} \) and \( \vec{w} \) are parallel then, \( \vec{w} = n\vec{v} \) where, n is scalar.
This means, \( \vec{w} = n(1)\hat{i} + n(-2)\hat{j} \) \( = n\hat{i} - 2n\hat{j} \)
\( |\vec{w}| = \sqrt{(n)^2 + (-2n)^2} = \sqrt{n^2 + 4n^2} = \sqrt{5n^2} = \sqrt{5}|n| \)
Given: \( |\vec{w}| = 10 \)
\( 10 = \sqrt{5}|n| \implies |n| = \frac{10}{\sqrt{5}} = \frac{10\sqrt{5}}{5} = 2\sqrt{5} \)
Since \( \vec{w} \) is parallel to \( \vec{v} \), we can take \( n = 2\sqrt{5} \) (or \( -2\sqrt{5} \) for anti-parallel).
\( \vec{w} = 2\sqrt{5}\hat{i} - 2(2\sqrt{5})\hat{j} \) \( = 2\sqrt{5}\hat{i} - 4\sqrt{5}\hat{j} \)

Question 5. Show that vectors \( \vec{a} = 2\hat{i} + 5\hat{j} - 6\hat{k} \) and \( \vec{b} = \hat{i} + \frac{5}{2}\hat{j} - 3\hat{k} \) are parallel. Answer: Let angle between two vectors be \( \theta \).
\( \therefore \quad \cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \)
\( = \frac{(2\hat{i} + 5\hat{j} - 6\hat{k}) \cdot (\hat{i} + \frac{5}{2}\hat{j} - 3\hat{k})}{\sqrt{2^2 + 5^2 + (-6)^2} \times \sqrt{1^2 + \left(\frac{5}{2}\right)^2 + (-3)^2}} \)
\( = \frac{2 \times 1 + 5 \times \frac{5}{2} + (-6) \times (-3)}{\sqrt{4 + 25 + 36} \times \sqrt{1 + \frac{25}{4} + 9}} \)
\( = \frac{2 + \frac{25}{2} + 18}{\sqrt{65} \times \sqrt{\frac{4+25+36}{4}}} \)
\( = \frac{\frac{4 + 25 + 36}{2}}{\sqrt{65} \times \sqrt{\frac{65}{4}}} \)
\( = \frac{\frac{65}{2}}{\sqrt{65} \times \frac{\sqrt{65}}{2}} \)
\( = \frac{\frac{65}{2}}{\frac{65}{2}} = 1 \)
\( \implies \theta = \cos^{-1}(1) = 0^\circ \)
\( \implies \) Two vectors are parallel.
In simple words: Two vectors are parallel if the angle between them is 0° or 180°. This means their dot product's magnitude equals the product of their magnitudes (for 0°). Alternatively, if one vector is a scalar multiple of the other, they are parallel.

🎯 Exam Tip: To prove vectors are parallel, either show that their dot product yields \( \theta = 0^\circ \) (or \( 180^\circ \)) or demonstrate that one vector is a scalar multiple of the other (e.g., \( \vec{a} = k\vec{b} \)). The scalar multiple method is often simpler for direct calculation.

Alternate method: \( \vec{a} = 2\hat{i} + 5\hat{j} - 6\hat{k} = 2(\hat{i} + \frac{5}{2}\hat{j} - 3\hat{k}) \)
Since \( \vec{a} \) is a scalar multiple of \( \vec{b} \), the vectors are parallel.

3. Solve The Following Problems.

Question 1. Determine \( \vec{a} \times \vec{b} \), given \( \vec{a} = 2\hat{i} + 3\hat{j} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} \). Answer: Using determinant for vectors in two dimensions, \[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} \]
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 3 & 5 & 0 \end{vmatrix} \]
\( = \hat{i}((3 \times 0) - (0 \times 5)) - \hat{j}((2 \times 0) - (0 \times 3)) + \hat{k}((2 \times 5) - (3 \times 3)) \)
\( = \hat{i}(0) - \hat{j}(0) + \hat{k}(10 - 9) \)
\( = 0\hat{i} - 0\hat{j} + 1\hat{k} \)
\( = \hat{k} \)
Answer: \( \vec{a} \times \vec{b} \) gives \( \hat{k} \).
In simple words: The cross product of two vectors can be found using a determinant. For 2D vectors (lying in the XY plane), their cross product will always be a vector along the Z-axis.

🎯 Exam Tip: For 2D vectors, the cross product \( \vec{A} \times \vec{B} \) will always be in the \( \hat{k} \) direction (or \( -\hat{k} \)). Remember to set the z-components to zero in the determinant for 2D vectors.

Question 2. Show that vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k} \), \( \vec{b} = 3\hat{i} - 6\hat{j} + 2\hat{k} \) and \( \vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k} \) are mutually perpendicular. Solution: As dot product of two perpendicular vectors is zero. Taking dot product of \( \vec{a} \) and \( \vec{b} \)
\( \vec{a} \cdot \vec{b} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) \) \( = (2 \times 3) + (3 \times -6) + (6 \times 2) \)
... \( (\therefore \hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1 \text{ and } \hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0) \)
\( = 6 - 18 + 12 \) \( = 0 \)
Similarly, \( \vec{b} \cdot \vec{c} = (3\hat{i} - 6\hat{j} + 2\hat{k}) \cdot (6\hat{i} + 2\hat{j} - 3\hat{k}) \) \( = (3 \times 6) + (-6 \times 2) + (2 \times -3) \)
... \( (\therefore \hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1 \text{ and } \hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0) \)
\( = 18 - 12 - 6 \) \( = 0 \)
And \( \vec{a} \cdot \vec{c} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot (6\hat{i} + 2\hat{j} - 3\hat{k}) \) \( = (2 \times 6) + (3 \times 2) + (6 \times -3) \) \( = 12 + 6 - 18 \) \( = 0 \)
Combining two results, we can say that given three vectors \( \vec{a} \), \( \vec{b} \), and \( \vec{c} \) are mutually perpendicular to each other.
In simple words: To show that vectors are mutually perpendicular, calculate the dot product for every pair of vectors. If all dot products result in zero, then the vectors are perpendicular to each other.

🎯 Exam Tip: The most direct way to prove mutual perpendicularity of vectors is to show that the dot product of every possible pair of vectors is zero. Remember the property \( \vec{A} \cdot \vec{B} = 0 \) if \( \vec{A} \perp \vec{B} \).

Question 3. Determine the vector product of \( \vec{v_1} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \vec{v_2} = \hat{i} + 2\hat{j} - 3\hat{k} \). Solution: As \( \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} \)
Using determinant to find vector product, \[ \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 2 & -3 \end{vmatrix} \]
\( = \hat{i}((3 \times -3) - (-1 \times 2)) - \hat{j}((2 \times -3) - (-1 \times 1)) + \hat{k}((2 \times 2) - (3 \times 1)) \)
\( = \hat{i}(-9 + 2) - \hat{j}(-6 + 1) + \hat{k}(4 - 3) \)
\( = -7\hat{i} - (-5)\hat{j} + 1\hat{k} \)
\( = -7\hat{i} + 5\hat{j} + \hat{k} \)
Answer: Required vector product is \( -7\hat{i} + 5\hat{j} + \hat{k} \).
In simple words: The vector product (cross product) of two 3D vectors is found by setting up a determinant with the unit vectors in the first row and the components of the two vectors in the subsequent rows, then evaluating the determinant.

🎯 Exam Tip: Practice evaluating 3x3 determinants for cross products carefully, paying close attention to signs. The resulting vector is perpendicular to both original vectors.

Question 4. Given \( \vec{v_1} = 5\hat{i} + 2\hat{j} \) and \( \vec{v_2} = a\hat{i} - 6\hat{j} \) are perpendicular to each other, determine the value of a. Solution: As \( \vec{v_1} \) and \( \vec{v_2} \) are perpendicular to each other, \( \theta = 90^\circ \)
\( \vec{v_1} \cdot \vec{v_2} = 0 \)
\( \therefore \quad (5\hat{i} + 2\hat{j}) \cdot (a\hat{i} - 6\hat{j}) = 0 \)
\( (5\hat{i} \cdot a\hat{i}) + (2\hat{j} \cdot -6\hat{j}) = 0 \)
... \( (\therefore \hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{i} = 0) \)
\( 5a + (-12) = 0 \)
... \( (\therefore \hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = 1) \)
\( 5a - 12 = 0 \)
\( 5a = 12 \)
\( a = \frac{12}{5} \)
Answer: Value of a is \( \frac{12}{5} \).
In simple words: If two vectors are perpendicular, their scalar (dot) product is zero. Set the dot product of the given vectors to zero and solve the resulting equation for the unknown variable 'a'.

🎯 Exam Tip: The condition for perpendicular vectors is \( \vec{A} \cdot \vec{B} = 0 \). This is a very common concept used to find unknown components or to verify orthogonality.

Question 5. Obtain derivatives of the following functions:
(i) x sin x
(ii) \( x^4 + \cos x \)
(iii) x/sin x Answer: (i) x sin x Solution: Using \( \frac{d}{dx}[f_1(x) \times f_2(x)] = f_1(x)\frac{df_2(x)}{dx} + \frac{df_1(x)}{dx}f_2(x) \) For \( f_1(x) = x \) and \( f_2(x) = \sin x \)
\( \frac{d}{dx}(x \sin x) = x\frac{d(\sin x)}{dx} + \frac{d(x)}{dx}\sin x \)
\( = x \cos x + 1 \times \sin x \)
\( = \sin x + x \cos x \)
(ii) \( x^4 + \cos x \) Solution: Using \( \frac{d}{dx}[f_1(x) + f_2(x)] = \frac{df_1(x)}{dx} + \frac{df_2(x)}{dx} \) For \( f_1(x) = x^4 \) and \( f_2(x) = \cos x \)
\( \frac{d}{dx}(x^4 + \cos x) = \frac{d(x^4)}{dx} + \frac{d(\cos x)}{dx} \)
\( = 4x^3 - \sin x \)
(iii) \( \frac{x}{\sin x} \) Solution: Using \( \frac{d}{dx}\left[\frac{f_1(x)}{f_2(x)}\right] = \frac{f_2(x)\frac{df_1(x)}{dx} - f_1(x)\frac{df_2(x)}{dx}}{(f_2(x))^2} \) For \( f_1(x) = x \) and \( f_2(x) = \sin x \)
\( \frac{d}{dx}\left(\frac{x}{\sin x}\right) = \frac{\sin x \frac{d(x)}{dx} - x \frac{d(\sin x)}{dx}}{\sin^2 x} \)
\( = \frac{\sin x \times 1 - x \cos x}{\sin^2 x} \)
\( = \frac{\sin x - x \cos x}{\sin^2 x} \)
[Note: As derivative of \( (\sin x) \) is \( \cos x \), negative sign that occurs in rule for differentiation for quotient of two functions gets retained in final answer]
Answer:
(i) \( \frac{d}{dx}(x \sin x) = \sin x + x \cos x \)
(ii) \( \frac{d}{dx}(x^4 + \cos x) = 4x^3 - \sin x \)
(iii) \( \frac{d}{dx}\left(\frac{x}{\sin x}\right) = \frac{\sin x - x \cos x}{\sin^2 x} \)
In simple words: (i) Use the product rule: \( (uv)' = u'v + uv' \). (ii) Use the sum/difference rule: \( (u+v)' = u' + v' \). (iii) Use the quotient rule: \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \).

🎯 Exam Tip: Master the basic differentiation rules: power rule, sum/difference rule, product rule, and quotient rule. Also, memorize the derivatives of common trigonometric functions like \( \sin x \) and \( \cos x \).

Question 6. Using the rule for differentiation for quotient of two functions, prove that \( \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \sec^2 x \) Solution: Using, \( \frac{d}{dx}\left[\frac{f_1(x)}{f_2(x)}\right] = \frac{f_2(x)\frac{df_1(x)}{dx} - f_1(x)\frac{df_2(x)}{dx}}{(f_2(x))^2} \) For \( f_1(x) = \sin x \) and \( f_2(x) = \cos x \)
\( \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x \frac{d(\sin x)}{dx} - \sin x \frac{d(\cos x)}{dx}}{\cos^2 x} \)
\( = \frac{\cos x \times \cos x - \sin x \times (-\sin x)}{\cos^2 x} \)
\( = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \)
\( = \frac{1}{\cos^2 x} \)
... \( (\therefore \sin^2 x + \cos^2 x = 1) \)
\( = \sec^2 x \)
... \( (\therefore \frac{1}{\cos x} = \sec x) \)
Answer: The derivation above shows that \( \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \sec^2 x \) using the quotient rule.
In simple words: To prove the derivative, apply the quotient rule \( \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \) to \( \frac{\sin x}{\cos x} \). Simplify the result using trigonometric identities like \( \sin^2 x + \cos^2 x = 1 \) and \( \frac{1}{\cos x} = \sec x \).

🎯 Exam Tip: This question tests your knowledge of the quotient rule and fundamental trigonometric identities. Ensure you can recall and apply both accurately for successful derivation.

Question 7. Evaluate the following integral:
(i) \( \int_0^{\pi/2} \sin x dx \)
(ii) \( \int_1^5 x dx \) Answer: (i) \( \int_0^{\pi/2} \sin x dx \) Solution: Using \( \int_a^b f(x) dx = F(x)|_a^b = F(b) - F(a) \) where \( F'(x) = f(x) \).
\( \int_0^{\pi/2} \sin x dx = [-\cos x]_0^{\pi/2} \)
\( = -[\cos(\frac{\pi}{2}) - \cos(0)] \)
Since, \( \cos(\frac{\pi}{2}) = 0 \) and \( \cos 0 = 1 \)
\( \int_0^{\pi/2} \sin x dx = -[0 - 1] = 1 \)
(ii) \( \int_1^5 x dx \) Solution: Using, \( \int_a^b f(x) dx = F(x)|_a^b = F(b) - F(a) \)
\( \int_1^5 x dx = \left[\frac{x^2}{2}\right]_1^5 \)
\( = \frac{5^2}{2} - \frac{1^2}{2} \)
\( = \frac{25}{2} - \frac{1}{2} \)
\( = \frac{24}{2} \)
\( = 12 \)
Answer:
(i) \( \int_0^{\pi/2} \sin x dx = 1 \)
(ii) \( \int_1^5 x dx = 12 \)
In simple words: (i) Find the antiderivative of \( \sin x \) which is \( -\cos x \), then evaluate it at the upper and lower limits and subtract. (ii) Find the antiderivative of \( x \) which is \( \frac{x^2}{2} \), then evaluate it at the upper and lower limits and subtract.

🎯 Exam Tip: For definite integrals, remember the fundamental theorem of calculus: evaluate the antiderivative at the upper limit and subtract its value at the lower limit. Be careful with signs, especially when integrating trigonometric functions.

Can You Recall? (Textbook Page No. 16)

Question 1. Define scalars and vectors. Answer:
1. Physical quantities which can be completely described by their magnitude (a number and unit) are called scalars.
2. Physical quantities which need magnitude as well as direction for their complete description are called vectors.
In simple words: Scalars are quantities described only by size (like temperature), while vectors need both size and direction for a full description (like velocity).

🎯 Exam Tip: This is a basic definition. Make sure to clearly state that scalars only need magnitude, while vectors need both magnitude and direction. Provide simple examples if possible.

Question 2. Which of the following are scalars or vectors? Displacements, distance travelled, velocity, speed, force, work done, energy Answer:
1. Scalars: Distance travelled, speed, work done, energy.
2. Vectors: Displacement, velocity, force.
In simple words: Quantities like distance, speed, work, and energy are scalars because they only have a size. Quantities like displacement, velocity, and force are vectors because they have both size and direction.

🎯 Exam Tip: To differentiate between scalars and vectors, ask yourself if a direction is essential for its complete understanding. If yes, it's a vector; if not, it's a scalar.

Question 3. What is the difference between a scalar and a vector? Answer:

In simple words: Scalars are defined only by their amount, combining like regular numbers. Vectors are defined by both amount and direction, requiring special rules for combination (like the head-to-tail method for addition).

🎯 Exam Tip: When listing differences, focus on the core attributes (magnitude/direction), rules of addition/subtraction, representation, and mathematical operations like division. Providing examples for each is also beneficial.

Internet My Friend (Textbook Page No. 28)

Answer: 1. hyperphysics.phy-astr.gsu.edu/hbase/vect. html#veccon 2. hyperphysics.phy-astr.gsu.edu/hbase/ hframe.html [Students can use links given above as reference and collect information about mathematical methods]

1. Choose The Correct Option.

Question 1. The resultant of two forces 10 N and 15 N acting along + x and - x-axes respectively, is
(A) 25 N along + x-axis
(B) 25 N along - x-axis
(C) 5 N along + x-axis
(D) 5 N along - x-axis
Answer: (D) 5 N along - x-axis
In simple words: When two forces act in opposite directions, their resultant is the difference between their magnitudes, pointing in the direction of the larger force. Here, 15N in -x direction is larger than 10N in +x direction.

🎯 Exam Tip: Understand vector addition for forces acting along the same line but in opposite directions. The resultant is the difference, and the direction follows the larger force.

Question 2. For two vectors to be equal, they should have the
(A) same magnitude
(B) same direction
(C) same magnitude and direction
(D) same magnitude but opposite direction
Answer: (C) same magnitude and direction
In simple words: Two vectors are considered equal if they have the exact same length (magnitude) and point in the exact same direction.

🎯 Exam Tip: Equality of vectors requires both magnitude and direction to be identical. Remember that even if magnitudes are equal, different directions mean different vectors.

Question 3. The magnitude of scalar product of two unit vectors perpendicular to each other is
(A) zero
(B) 1
(C) -1
(D) 2
Answer: (A) zero
In simple words: The scalar product (dot product) of two perpendicular vectors is zero because the cosine of 90 degrees is zero. For unit vectors, their magnitude is 1.

🎯 Exam Tip: Recall the definition of the dot product: \( \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \theta \). For perpendicular vectors, \( \theta = 90^\circ \), so \( \cos 90^\circ = 0 \), making the dot product zero.

Question 4. The magnitude of vector product of two unit vectors making an angle of 60° with each other is
(A) 1
(B) 2
(C) \( \frac{3}{2} \)
(D) \( \frac{\sqrt{3}}{2} \)
Answer: (D) \( \frac{\sqrt{3}}{2} \)
In simple words: The magnitude of the vector product (cross product) of two unit vectors is found using the sine of the angle between them. For a 60° angle, this is \( \sin 60^\circ = \frac{\sqrt{3}}{2} \).

🎯 Exam Tip: Remember the definition of the cross product magnitude: \( |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \). For unit vectors, magnitudes are 1.

Question 5. If \( \vec{A}, \vec{B}, \) and \( \vec{C} \) are three vectors, then which of the following is not correct?
(A) \( \vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C} \)
(B) \( \vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A} \)
(C) \( \vec{A} \times \vec{B} = \vec{B} \times \vec{A} \)
(D) \( \vec{A} \times (\vec{B} \times \vec{C}) = \vec{A} \times \vec{B} + \vec{B} \times \vec{C} \)
Answer: (C) \( \vec{A} \times \vec{B} = \vec{B} \times \vec{A} \)
In simple words: The cross product is anti-commutative, meaning the order matters and \( \vec{A} \times \vec{B} \) is not equal to \( \vec{B} \times \vec{A} \). Instead, \( \vec{A} \times \vec{B} = -(\vec{B} \times \vec{A}) \).

🎯 Exam Tip: Differentiate between the commutative properties of dot products and the anti-commutative properties of cross products. The scalar triple product and vector triple product rules are also important to remember.

2. Answer The Following Questions.

Question 1. Show that \( \vec{A} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} \) is a unit vector.
Solution:
Let \( \vec{a} \) be unit vector.
\( |\vec{a}| = \sqrt{a_x^2 + a_y^2} = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{-1}{\sqrt{2}}\right)^2} = \sqrt{\frac{1}{2} + \frac{1}{2}} = \sqrt{1} = 1 \)
\( \implies \) Since the magnitude is 1, \( \vec{a} \) itself is a unit vector.
Answer: The vector \( \vec{A} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} \) is a unit vector because its magnitude is 1.
In simple words: A unit vector is any vector that has a magnitude (length) of exactly one. By calculating the magnitude of the given vector, we found it to be 1, thus confirming it is a unit vector.

🎯 Exam Tip: To prove a vector is a unit vector, always calculate its magnitude. If the magnitude is 1, it's a unit vector. This concept is fundamental for normalization.

Question 2. If \( \vec{v_1} = 3\hat{i} + 4\hat{j} + \hat{k} \) and \( \vec{v_2} = \hat{i} - \hat{j} - \hat{k} \), determine the magnitude of \( \vec{v_1} + \vec{v_2} \).
Solution:
\( \vec{v_1} + \vec{v_2} = (3\hat{i} + 4\hat{j} + \hat{k}) + (\hat{i} - \hat{j} - \hat{k}) \)
\( = 3\hat{i} + \hat{i} + 4\hat{j} - \hat{j} + \hat{k} - \hat{k} \)
\( = 4\hat{i} + 3\hat{j} \)
Magnitude of \( (\vec{v_1} + \vec{v_2}) \),
\( |\vec{v_1} + \vec{v_2}| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 \) units.
Answer: Magnitude of \( \vec{v_1} + \vec{v_2} = 5 \) units.
In simple words: First, we add the corresponding components of the two vectors to get a resultant vector. Then, we find the magnitude of this resultant vector using the Pythagorean theorem.

🎯 Exam Tip: Vector addition is done component-wise. The magnitude of a 3D vector \( \vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} \) is \( |\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2} \). For 2D, one component is zero.

Question 3. For \( \vec{v_1} = 2\hat{i} - 3\hat{j} \) and \( \vec{v_2} = -6\hat{i} + 5\hat{j} \), determine the magnitude and direction of \( \vec{v_1} + \vec{v_2} \).
Answer:
\( \vec{v_1} + \vec{v_2} = (2\hat{i} - 3\hat{j}) + (-6\hat{i} + 5\hat{j}) \)
\( = (2\hat{i} - 6\hat{i}) + (-3\hat{j} + 5\hat{j}) \)
\( = -4\hat{i} + 2\hat{j} \)
\( |\vec{v_1} + \vec{v_2}| = \sqrt{(-4)^2 + 2^2} = \sqrt{16 + 4} = \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \)
Comparing \( \vec{v_1} + \vec{v_2} \) with \( \vec{R} = R_x\hat{i} + R_y\hat{j} \)
\( \implies R_x = -4 \) and \( R_y = 2 \)
Taking \( \theta \) to be angle made by \( \vec{R} \) with X-axis,
\( \theta = \tan^{-1}\left(\frac{R_y}{R_x}\right) = \tan^{-1}\left(\frac{2}{-4}\right) \)
\( = \tan^{-1}\left(-\frac{1}{2}\right) \) with X-axis
Answer: Magnitude and direction of \( \vec{v_1} + \vec{v_2} \) is respectively \( 2\sqrt{5} \) and \( \tan^{-1}\left(-\frac{1}{2}\right) \) with X-axis.
In simple words: First, add the vectors component-wise. Then, calculate the magnitude of the resultant vector using the formula \( \sqrt{R_x^2 + R_y^2} \). Finally, find the direction using \( \theta = \tan^{-1}(R_y/R_x) \).

🎯 Exam Tip: When determining the direction of a vector using \( \tan^{-1}(R_y/R_x) \), pay attention to the signs of \( R_x \) and \( R_y \) to correctly identify the quadrant of the angle.

Question 4. Find a vector which is parallel to \( \vec{v} = \hat{i} - 2\hat{j} \) and has a magnitude 10.
Answer:
Let the vector be \( \vec{w} = w_x\hat{i} + w_y\hat{j} \)
\( |\vec{w}| = \sqrt{w_x^2 + w_y^2} = 10 \) ....(Given)
\( \implies w_x^2 + w_y^2 = 100 \) ....(i)
Also, if \( \vec{v} \) and \( \vec{w} \) are parallel, then \( \vec{w} = n\vec{v} \) for some scalar \( n \).
So, \( \vec{w} = n(\hat{i} - 2\hat{j}) = n\hat{i} - 2n\hat{j} \)
Thus, \( w_x = n \) and \( w_y = -2n \).
Substitute these into equation (i):
\( n^2 + (-2n)^2 = 100 \)
\( n^2 + 4n^2 = 100 \)
\( 5n^2 = 100 \)
\( n^2 = 20 \)
\( n = \pm\sqrt{20} = \pm 2\sqrt{5} \)
Using \( n = 2\sqrt{5} \):
\( w_x = 2\sqrt{5} \)
\( w_y = -2(2\sqrt{5}) = -4\sqrt{5} \)
So, \( \vec{w} = 2\sqrt{5}\hat{i} - 4\sqrt{5}\hat{j} \)
This can also be written as \( \frac{10}{\sqrt{5}}\hat{i} - \frac{20}{\sqrt{5}}\hat{j} \).
Using \( n = -2\sqrt{5} \):
\( w_x = -2\sqrt{5} \)
\( w_y = -2(-2\sqrt{5}) = 4\sqrt{5} \)
So, \( \vec{w} = -2\sqrt{5}\hat{i} + 4\sqrt{5}\hat{j} \)
Answer: The required vector is \( 2\sqrt{5}\hat{i} - 4\sqrt{5}\hat{j} \) or \( -2\sqrt{5}\hat{i} + 4\sqrt{5}\hat{j} \). (The provided solution only gives one, but two parallel vectors exist)
In simple words: A vector parallel to another means it's a scalar multiple of that vector. We use this relationship and the given magnitude to find the scalar multiple, which then allows us to determine the components of the desired vector. There are typically two such vectors, one in each direction.

🎯 Exam Tip: When finding a parallel vector with a specific magnitude, remember that if \( \vec{w} \) is parallel to \( \vec{v} \), then \( \vec{w} = n\vec{v} \) for some scalar \( n \). Since \( n \) can be positive or negative, there will generally be two such vectors (in opposite directions) that satisfy the magnitude condition.

Question 5. Show that vectors \( \vec{a} = 2\hat{i} + 5\hat{j} - 6\hat{k} \) and \( \vec{b} = \hat{i} + \frac{5}{2}\hat{j} - 3\hat{k} \) are parallel.
Answer:
Let angle between two vectors be \( \theta \).
\( \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \)
\( \cos \theta = \frac{(2\hat{i} + 5\hat{j} - 6\hat{k}) \cdot (\hat{i} + \frac{5}{2}\hat{j} - 3\hat{k})}{\sqrt{2^2 + 5^2 + (-6)^2} \times \sqrt{1^2 + (\frac{5}{2})^2 + (-3)^2}} \)
\( \cos \theta = \frac{(2 \times 1) + (5 \times \frac{5}{2}) + (-6 \times -3)}{\sqrt{4 + 25 + 36} \times \sqrt{1 + \frac{25}{4} + 9}} \)
\( \cos \theta = \frac{2 + \frac{25}{2} + 18}{\sqrt{65} \times \sqrt{\frac{4 + 25 + 36}{4}}} \)
\( \cos \theta = \frac{\frac{4+25+36}{2}}{\sqrt{65} \times \sqrt{\frac{65}{4}}} \)
\( \cos \theta = \frac{\frac{65}{2}}{\sqrt{65} \times \frac{\sqrt{65}}{2}} \)
\( \cos \theta = \frac{\frac{65}{2}}{\frac{65}{2}} = 1 \)
\( \implies \theta = \cos^{-1}(1) = 0^\circ \)
\( \implies \) Two vectors are parallel.
Alternate method:
\( \vec{a} = 2\hat{i} + 5\hat{j} - 6\hat{k} \)
\( \vec{b} = \hat{i} + \frac{5}{2}\hat{j} - 3\hat{k} \)
Notice that \( \vec{a} = 2 (\hat{i} + \frac{5}{2}\hat{j} - 3\hat{k}) = 2\vec{b} \)
Since \( \vec{a} \) is a scalar multiple of \( \vec{b} \), the vectors are parallel.
In simple words: Two vectors are parallel if the angle between them is 0 degrees, which means their dot product divided by the product of their magnitudes is 1. Alternatively, if one vector is a scalar multiple of the other, they are parallel.

🎯 Exam Tip: To prove vectors are parallel, you can either show that \( \theta = 0^\circ \) using the dot product formula or demonstrate that one vector is a scalar multiple of the other (e.g., \( \vec{A} = k\vec{B} \)). The scalar multiple method is often simpler if applicable.

3. Solve The Following Problems.

Question 1. Determine \( \vec{a} \times \vec{b} \), given \( \vec{a} = 2\hat{i} + 3\hat{j} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} \).
Answer:
Using determinant for vectors in two dimensions,
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ 3 & 5 & 0 \end{vmatrix} \]
(Since the vectors are in the XY-plane, \( a_z = 0 \) and \( b_z = 0 \))
\( \vec{a} \times \vec{b} = \hat{i}(3 \times 0 - 0 \times 5) - \hat{j}(2 \times 0 - 0 \times 3) + \hat{k}(2 \times 5 - 3 \times 3) \)
\( = \hat{i}(0) - \hat{j}(0) + \hat{k}(10 - 9) \)
\( = 1\hat{k} = \hat{k} \)
Answer: \( \vec{a} \times \vec{b} \) gives \( \hat{k} \).
In simple words: The cross product of two vectors in a 2D plane results in a vector perpendicular to that plane. Using the determinant method helps systematically calculate the components of this resultant vector.

🎯 Exam Tip: For cross products of 2D vectors (lying in the xy-plane), the resultant vector will always be along the z-axis (either \( \hat{k} \) or \( -\hat{k} \)). Use the determinant method for reliable calculation.

Question 2. Show that vectors \( \vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k} \), \( \vec{b} = 3\hat{i} - 6\hat{j} + 2\hat{k} \) and \( \vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k} \) are mutually perpendicular.
Solution:
As dot product of two perpendicular vectors is zero. Taking dot product of \( \vec{a} \) and \( \vec{b} \)
\( \vec{a} \cdot \vec{b} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot (3\hat{i} - 6\hat{j} + 2\hat{k}) \)
\( = (2 \times 3) + (3 \times -6) + (6 \times 2) \)
(Since \( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1 \) and \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0 \))
\( = 6 - 18 + 12 \)
\( = 0 \)
Similarly, \( \vec{b} \cdot \vec{c} = (3\hat{i} - 6\hat{j} + 2\hat{k}) \cdot (6\hat{i} + 2\hat{j} - 3\hat{k}) \)
\( = (3 \times 6) + (-6 \times 2) + (2 \times -3) \)
\( = 18 - 12 - 6 \)
\( = 0 \)
And \( \vec{a} \cdot \vec{c} = (2\hat{i} + 3\hat{j} + 6\hat{k}) \cdot (6\hat{i} + 2\hat{j} - 3\hat{k}) \)
\( = (2 \times 6) + (3 \times 2) + (6 \times -3) \)
\( = 12 + 6 - 18 \)
\( = 0 \)
Combining two results, we can say that given three vectors \( \vec{a}, \vec{b}, \) and \( \vec{c} \) are mutually perpendicular to each other.
Answer: Since the dot product of every pair of vectors (\( \vec{a} \cdot \vec{b} \), \( \vec{b} \cdot \vec{c} \), \( \vec{a} \cdot \vec{c} \)) is zero, the vectors are mutually perpendicular.
In simple words: To prove vectors are mutually perpendicular, calculate the dot product for every unique pair of vectors. If all dot products are zero, then the vectors are perpendicular to each other.

🎯 Exam Tip: The scalar product (dot product) being zero is the defining condition for two non-zero vectors to be perpendicular. Ensure you check all unique pairs for mutual perpendicularity.

Question 3. Determine the vector product of \( \vec{v_1} = 2\hat{i} + 3\hat{j} - \hat{k} \) and \( \vec{v_2} = \hat{i} + 2\hat{j} - 3\hat{k} \). If they are perpendicular to each other, determine the value of a. (Note: The second part of the question "If they are perpendicular to each other, determine the value of a" seems out of context as 'a' is not present in the given vectors. Assuming the question intended only to calculate the vector product.)
Solution:
Using determinant to find vector product,
\[ \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 2 & -3 \end{vmatrix} \]
\( = \hat{i}((3 \times -3) - (-1 \times 2)) - \hat{j}((2 \times -3) - (-1 \times 1)) + \hat{k}((2 \times 2) - (3 \times 1)) \)
\( = \hat{i}(-9 + 2) - \hat{j}(-6 + 1) + \hat{k}(4 - 3) \)
\( = -7\hat{i} - (-5)\hat{j} + 1\hat{k} \)
\( = -7\hat{i} + 5\hat{j} + \hat{k} \)
Answer: Required vector product is \( -7\hat{i} + 5\hat{j} + \hat{k} \).
In simple words: The vector product (cross product) is calculated using the determinant of a matrix formed by the unit vectors and the components of the two given vectors. This results in a new vector perpendicular to both original vectors.

🎯 Exam Tip: Master the determinant method for calculating cross products in 3D. Be careful with signs, especially when expanding along the middle (\( \hat{j} \)) component. Double-check your arithmetic.

Question 4. Given \( \vec{v_1} = 5\hat{i} + 2\hat{j} \) and \( \vec{v_2} = a\hat{i} - 6\hat{j} \) are perpendicular to each other, determine the value of a.
Solution:
As \( \vec{v_1} \) and \( \vec{v_2} \) are perpendicular to each other, \( \theta = 90^\circ \).
Thus, their dot product must be zero: \( \vec{v_1} \cdot \vec{v_2} = 0 \)
\( (5\hat{i} + 2\hat{j}) \cdot (a\hat{i} - 6\hat{j}) = 0 \)
\( (5\hat{i} \cdot a\hat{i}) + (2\hat{j} \cdot -6\hat{j}) = 0 \)
(Since \( \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = 1 \) and \( \hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{i} = 0 \))
\( 5a + (-12) = 0 \)
\( 5a = 12 \)
\( a = \frac{12}{5} \)
Answer: Value of a is \( \frac{12}{5} \).
In simple words: If two vectors are perpendicular, their dot product is zero. We use this property to set up an equation with the unknown 'a' and solve for it.

🎯 Exam Tip: The condition for perpendicular vectors is \( \vec{A} \cdot \vec{B} = 0 \). This is a crucial property for solving problems involving orthogonality. Ensure you correctly apply the dot product rules (like \( \hat{i} \cdot \hat{i} = 1 \), \( \hat{i} \cdot \hat{j} = 0 \)).

Question 5. Obtain derivatives of the following functions:
(i) \( x \sin x \)
(ii) \( x^4 + \cos x \)
(iii) \( x/\sin x \)
Answer:
(i) \( x \sin x \)
Solution:
Using \( \frac{d}{dx}[f_1(x) \times f_2(x)] = f_1(x)\frac{df_2(x)}{dx} + f_2(x)\frac{df_1(x)}{dx} \)
For \( f_1(x) = x \) and \( f_2(x) = \sin x \)
\( \frac{d}{dx}(x \sin x) = x\frac{d(\sin x)}{dx} + \sin x\frac{d(x)}{dx} \)
\( = x \cos x + \sin x \times 1 \)
\( = \sin x + x \cos x \)
Answer: The derivative of \( x \sin x \) is \( \sin x + x \cos x \).
In simple words: To differentiate a product of two functions, apply the product rule: \( (uv)' = u'v + uv' \). Take the derivative of each function separately and combine them as per the rule.

🎯 Exam Tip: Remember the product rule for differentiation. It's often used when functions are multiplied together. Make sure to correctly identify \( u \) and \( v \) and their derivatives.

(ii) \( x^4 + \cos x \)
Solution:
Using \( \frac{d}{dx}[f_1(x) + f_2(x)] = \frac{df_1(x)}{dx} + \frac{df_2(x)}{dx} \)
For \( f_1(x) = x^4 \) and \( f_2(x) = \cos x \)
\( \frac{d}{dx}(x^4 + \cos x) = \frac{d(x^4)}{dx} + \frac{d(\cos x)}{dx} \)
\( = 4x^3 - \sin x \)
Answer: The derivative of \( x^4 + \cos x \) is \( 4x^3 - \sin x \).
In simple words: For a sum of functions, the derivative is the sum of the derivatives of each individual function. We apply power rule for \( x^4 \) and standard derivative for \( \cos x \).

🎯 Exam Tip: Remember basic derivative rules: power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \), and derivatives of trigonometric functions like \( \frac{d}{dx}(\sin x) = \cos x \) and \( \frac{d}{dx}(\cos x) = -\sin x \).

(iii) \( \frac{x}{\sin x} \)
Solution:
Using \( \frac{d}{dx}\left[\frac{f_1(x)}{f_2(x)}\right] = \frac{f_2(x)\frac{df_1(x)}{dx} - f_1(x)\frac{df_2(x)}{dx}}{[f_2(x)]^2} \)
For \( f_1(x) = x \) and \( f_2(x) = \sin x \)
\( \frac{d}{dx}\left(\frac{x}{\sin x}\right) = \frac{\sin x \frac{d(x)}{dx} - x \frac{d(\sin x)}{dx}}{\sin^2 x} \)
\( = \frac{\sin x \times 1 - x \times \cos x}{\sin^2 x} \)
\( = \frac{\sin x - x \cos x}{\sin^2 x} \)
[Note: As derivative of \( (\sin x) \) is \( \cos x \), negative sign that occurs in rule for differentiation for quotient of two functions gets retained in final answer]
Answer: The derivative of \( \frac{x}{\sin x} \) is \( \frac{\sin x - x \cos x}{\sin^2 x} \).
In simple words: To find the derivative of a quotient of two functions, use the quotient rule: \( (g f' - f g') / g^2 \). Carefully apply the derivatives of each function and simplify.

🎯 Exam Tip: The quotient rule is critical for derivatives of fractional functions. Be careful with the order of terms in the numerator (\( \text{low } d(\text{high}) - \text{high } d(\text{low}) \)) and the square of the denominator.

Question 6. Using the rule for differentiation for quotient of two functions, prove that \( \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \sec^2 x \).
Solution:
Using, \( \frac{d}{dx}\left[\frac{f_1(x)}{f_2(x)}\right] = \frac{f_2(x)\frac{df_1(x)}{dx} - f_1(x)\frac{df_2(x)}{dx}}{[f_2(x)]^2} \)
For \( f_1(x) = \sin x \) and \( f_2(x) = \cos x \)
\( \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \frac{\cos x \frac{d(\sin x)}{dx} - \sin x \frac{d(\cos x)}{dx}}{\cos^2 x} \)
\( = \frac{\cos x (\cos x) - \sin x (-\sin x)}{\cos^2 x} \)
\( = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} \)
\( = \frac{1}{\cos^2 x } \) (Since \( \sin^2 x + \cos^2 x = 1 \))
\( = \sec^2 x \)
Answer: It is proven that \( \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right) = \sec^2 x \).
In simple words: The derivative of \( \tan x \) is \( \sec^2 x \). We proved this by applying the quotient rule to \( \frac{\sin x}{\cos x} \), using the derivatives of sine and cosine, and simplifying with a trigonometric identity.

🎯 Exam Tip: This proof is a classic application of the quotient rule combined with trigonometric identities. Remember that \( \frac{\sin x}{\cos x} = \tan x \) and \( \frac{1}{\cos x} = \sec x \).

Question 7. Evaluate the following integral:
(i) \( \int_{0}^{\pi/2} \sin x \, dx \)
(ii) \( \int_{1}^{5} x \, dx \)
Answer:
(i) \( \int_{0}^{\pi/2} \sin x \, dx \)
Solution:
Using \( \int_{a}^{b} f(x) \, dx = [F(x)]_a^b \)
\( \int_{0}^{\pi/2} \sin x \, dx = [-\cos x]_{0}^{\pi/2} \)
\( = -[\cos(\frac{\pi}{2}) - \cos 0] \)
Since, \( \cos(\frac{\pi}{2}) = 0 \) and \( \cos 0 = 1 \)
\( = -[0 - 1] = 1 \)
Answer: The value of the integral \( \int_{0}^{\pi/2} \sin x \, dx \) is 1.
In simple words: To evaluate a definite integral, first find the antiderivative of the function, then evaluate it at the upper and lower limits, and subtract the lower limit result from the upper limit result.

🎯 Exam Tip: Remember the fundamental theorem of calculus for definite integrals. Also, recall standard integrals like \( \int \sin x \, dx = -\cos x \) and exact values of trigonometric functions at common angles (e.g., \( \cos(\pi/2) = 0 \), \( \cos 0 = 1 \)).

(ii) \( \int_{1}^{5} x \, dx \)
Solution:
Using, \( \int_{a}^{b} f(x) \, dx = [F(x)]_a^b \)
\( \int_{1}^{5} x \, dx = \left[\frac{x^2}{2}\right]_{1}^{5} \)
\( = \frac{5^2}{2} - \frac{1^2}{2} \)
\( = \frac{25}{2} - \frac{1}{2} \)
\( = \frac{24}{2} \)
\( = 12 \)
Answer: The value of the integral \( \int_{1}^{5} x \, dx \) is 12.
In simple words: For a definite integral of \( x \), find its antiderivative \( \frac{x^2}{2} \), then plug in the upper and lower limits and subtract to get the final numerical value.

🎯 Exam Tip: For integrating powers of x, use the power rule for integration: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \) (for indefinite integrals) or \( [\frac{x^{n+1}}{n+1}]_a^b \) for definite integrals.

Can You Recall? (Textbook Page No. 16)

Question 1. Define scalars and vectors.
Answer:
(i) Physical quantities which can be completely described by their magnitude (a number and unit) are called scalars.
(ii) Physical quantities which need magnitude as well as direction for their complete description are called vectors.
In simple words: Scalars are quantities defined by magnitude only (like mass or time), while vectors require both magnitude and direction for their full description (like force or velocity).

🎯 Exam Tip: Clearly distinguish between scalars and vectors based on their defining properties: magnitude only for scalars, magnitude and direction for vectors. Provide examples for each.

Question 2. Which of the following are scalars or vectors?
Displacements, distance travelled, velocity, speed, force, work done, energy
Answer:
(i) Scalars: Distance travelled, speed, work done, energy.
(ii) Vectors: Displacement, velocity, force.
In simple words: We categorize quantities like distance, speed, work, and energy as scalars because they only have an amount, while displacement, velocity, and force are vectors because they also have a specific direction.

🎯 Exam Tip: Practice classifying common physical quantities as either scalar or vector. Remember that "distance" is scalar, but "displacement" is vector; "speed" is scalar, but "velocity" is vector.

Question 3. What is the difference between a scalar and a vector?
Answer:

In simple words: The fundamental difference is that scalars only have magnitude and follow simple arithmetic, while vectors have both magnitude and direction, requiring specific geometric rules for operations.

🎯 Exam Tip: Be prepared to explain the distinctions between scalars and vectors, including their representation, methods of addition/subtraction, and whether division is applicable, along with clear examples.