## Mathematical Methods

### Maharashtra Board-Class-11-Science-Physics-Chapter-2

### Solutions

**Question 1. Choose the correct option.**

**(i) 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**

(D) 5 N along - *x-*axis

**(ii) 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**

(C) same magnitude and direction

**(iii) The magnitude of scalar product of two unit vectors perpendicular to each other is**

**(A) zero **

**(B) 1**

**(C) -1 **

**(D) 2**

(A) zero

**(iv) The magnitude of vector product of two unit vectors making an angle of 60****° ****with each other is**

**(A) 1 **

**(B) 2**

**(C) 3/2 **

**(D) ** ** \(\sqrt{3}/2\)**

(D) \(\sqrt{3}/2\)

**(v) If \(\vec{A},\vec{B}\)**** ****and \(\vec{C}\)** * *** ****are three vectors, then which of the following is not correct?**

**(A)** **\(\vec{A}.(\vec{B}+\vec{C})\) = \(\vec{A}.\vec{B}+\vec{A}.\vec{C}\)**

**(B) \(\vec{A}.\vec{B}\) = \(\vec{B}.\vec{A}\)**

**(C) \(\vec{A}×\vec{B}\) = \(\vec{B}×\vec{A}\)**

**(D) \(\vec{A}×(\vec{B}+\vec{C})\) = \(\vec{A}×\vec{B}+\vec{B}×\vec{C}\)**

(C) \(\vec{A}×\vec{B}\) = \(\vec{B}×\vec{A}\)

**Question 2. Answer the following questions.**

**(i) Show that \(\vec{a}=\frac{i-j}{\sqrt{2}}\)**** ****is a unit vector.**

a = |\(\vec{a}\)| = \(\sqrt{(\frac{1}{\sqrt{2}})^2+(-\frac{1}{\sqrt{2}})^2}\)

= \(\sqrt{\frac{1}{2} +\frac{1}{2}}\) = 1

**Hence, \(\vec{a}\)** ** is a unit vector.**

**(ii) If \(\vec{v_1}\) = \(3\hat{i} + 4\hat{j} + k\) and \(\vec{v_2}\) = \(\hat{i} - \hat{j} - k\) , determine the magnitude of \(\vec{v_1}+\vec{v_2}\) **

\(\vec{v_1}+\vec{v_2}\) = \((3\hat{i} + 4\hat{j} + k) + (\hat{i} - \hat{j} - k)\)

= (3+1)\(\hat{i}\) + (4 - 1)\(\hat{j}\) + + (1 - 1)\(\hat{k}\)

= 4\(\hat{i}\) + 3\(\hat{j}\)

∴ |\(\vec{v_1}+\vec{v_2}\)| = |4\(\hat{i}\) + 3\(\hat{j}\)| = \(\sqrt{(4^2+3^2)}\) = 5

**5 is the required magnitude**

**(iii) 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 \(\vec{v_1}+\vec{v_2}\)**

\(\vec{v}\) = \(\vec{v_1}+\vec{v_2}\) ** = **(2\(\hat{i}\) − 3\(\hat{j}\) ) *+ *(−6\(\hat{i}\) + 5\(\hat{j}\) )

= (2 - 6)\(\hat{i}\) + (-3 + 5)\(\hat{j}\)

= −4\(\hat{i}\) + 2\(\hat{j}\) = v_{x}\(\hat{i}\) + v_{y}\(\hat{i}\)

∴ |\(\vec{v}\)| = \(\sqrt{(-4)^2+(2)^2}\) = \(\sqrt{(16+4)}\)

= \(\sqrt{20}\) = 2\(\sqrt{5}\)

tan θ = v_{y}/v_{x} = 2/−4 = −1/2 = tan (180^{o} − α) = −tan α

∴ α = tan^{−1}\((\frac{1}{2})\) = 26^{o}34’

∴ θ = 180^{o} − 26^{o}34’ = 153^{o}26’

**∴** **\(\vec{v_1}+\vec{v_2}\) has a magnitude of 2\(\sqrt{5}\) and makes an angle of θ = tan ^{−1}\((\frac{1}{2})\) = 153^{o}26’ with the positive x-axis**

**(iv) Find a vector which is parallel to \(\vec{v}=\hat{i}-2\hat{j}\)** **and has a magnitude 10.**

*Let * \(\vec{u}\) *be a vector of magnitude 10 and parallel to * \(\vec{v}=\hat{i}-2\hat{j}\)

∴ \(\vec{u}=λ\vec{v}\)*, *where the scalar multiple λ *= u/v*

v = |\(\vec{v}\)| = \(\sqrt{v_x^2+v_y^2}\) = \(\sqrt{(1)^2+(-2)^2}\) = \(\sqrt{5}\)

∴ λ *= *\(\frac{10}{\sqrt{5}}=2\sqrt{5}\) as u = 10

∴ \(\vec{u}\) = \(2\sqrt{5}(\hat{i}-2\hat{j})\) **= \(2\sqrt{5}\hat{i}-4\sqrt{5}\hat{j}\)** **is the required vector**.

**(v) 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.**

\(\vec{a}\) = 2\(\hat{i}\) + 5\(\hat{j}\) − 6\(\hat{k}\) = \(2(\hat{i}\) + \(\frac{5}{2}\hat{j}\) − 3\(\hat{k})\) = \(2\vec{b}\)

**Since \(\vec{a}\)**** ****is scalar multiple of** **\(\vec{b}\)**** ****the vectors are parallel**

**Question 3. Solve the following problems.**

**(i) Determine \(\vec{a}\) x \(\vec{b}\), given \(\vec{a}\) = 2\(\hat{i}\) + 3\(\hat{j}\) and \(\vec{b}\) = 3\(\hat{i}\) + 5\(\hat{j}\).**

\(\vec{a}\) = 2\(\hat{i}\) + 3\(\hat{j}\), \(\vec{b}\) = 3\(\hat{i}\) + 5\(\hat{j}\)

\(\vec{a}\) x \(\vec{b}\)**= **(2\(\hat{i}\) + 3\(\hat{j}\) ) x (3\(\hat{i}\) + 5\(\hat{j}\))

= 2\(\hat{i}\) x 3\(\hat{i}\) + 2\(\hat{i}\) x 5\(\hat{j}\) + 3\(\hat{j}\) x 3\(\hat{i}\) + 3\(\hat{j}\) x 5\(\hat{j}\)

= 0 + 10\(\hat{k}\) − 9\(\hat{k}\) + 0 = **\(\hat{k}\)**

**(ii) 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.**

\(\vec{a}\) = 2\(\hat{i}\) + 3\(\hat{j}\) + 6\(\hat{k}\)*, * \(\vec{b}\) = 3\(\hat{i}\) − 6\(\hat{j}\) + 2\(\hat{k}\), \(\vec{c}\) = 6\(\hat{i}\) + 2\(\hat{j}\) − 3\(\hat{k}\)

a ≠ 0, ** **b ≠ 0, c ≠ 0

\(\vec{a}\).\(\vec{b}\) = a_{x}b_{x} + a_{y}b_{y} + a_{z}b_{z} = (2)(3) + (3)(−6) + (6)(2) = 6 – 18 + 12 = 0

∴ \(\vec{b}\)⊥\(\vec{a}\)

\(\vec{a}\).\(\vec{c}\) = a_{x}c_{x} + a_{y}c_{y} + a_{z}c_{z} = (2)(6) + (3)(2) + (6)( −3) = 12 + 6 – 18 = 0

∴ \(\vec{c}\)⊥\(\vec{a}\)

\(\vec{b}\).\(\vec{c}\) = b_{x}c_{x} + b_{y}c_{y} + b_{z}c_{z} = (3)(6) + (−6)(2) + (2)( −3) = 18 – 12 – 6 = 0

∴ \(\vec{c}\)⊥\(\vec{b}\)

**It follows that \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular.**

**(iii) 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\(\hat{k}\) **

\(\vec{v_1}\) = 2\(\hat{i}\) + 3\(\hat{j}\) − \(\hat{k}\), \(\vec{v_2}\) = \(\hat{i}\) + 2\(\hat{j}\) − 3\(\hat{k}\)

\(\vec{v_1}\) x \(\vec{v_2}\) = \(\begin{vmatrix} \hat{i} & \hat{j}& \hat{k}\cr 2 & 3 & -1\cr 1 & 2 & -3 \end{vmatrix}\)

= \(\hat{i}\)[(3)( −3) − (−1)(2)] + \(\hat{j}\)[(−1)(1) − (2)( −3)] + \(\hat{k}\)[(2)(2) − (3)(1)]

= **-7\(\hat{i}\)** **+ 5\(\hat{j}\)** ** + \(\hat{k}\)**

**(iv) Given \(\vec{v_1}\) = 5\(\hat{i}\) + 2\(\hat{j}\) and \(\vec{v_2}\) = a\(\hat{i}\) − 6\(\hat{i}\) are perpendicular to each other, determine the value of a.**

\(\vec{v_1}\) = 5\(\hat{i}\) + 2\(\hat{j}\) perpendicular to \(\vec{v_2}\) = a\(\hat{i}\) − 6\(\hat{i}\)

\(\vec{v_1}\).\(\vec{v_2}\) ** **= 5a + (2)(−6) = 0

5a = 12

∴ a = 12/5 = **2.4**

**(v) Obtain derivatives of the following functions:**

**(i) ***x ***sin ***x ***(ii) ***x ^{4}+*

**cos**

*x***(iii) \(\frac{x}{sin\,x}\)**

(i) \(\frac{d}{dx}\)(x sin x) = \(x(\frac{d\,sin\,x}{dx})+sin\,x\frac{dx}{dx}\)(x sin x) = **x cos x + sin x**

(i) \(\frac{d}{dx}\)(x sin x)(x^{4}+cos x) = **4x ^{2} **

**−**

**sin x**

(iii) \(\frac{d}{dx}\frac{x}{sin\,x}\) = \(\frac{1}{sin\,x}.\frac{dx}{dx}-\frac{1}{sin^2x}.x\frac{d(sin\,x)}{dx}\) = **\(\frac{1}{sin\,x}-\frac{x\,cox\,x}{sin^2x}\)**

**(vi) Using the rule for differentiation for quotient of two functions, prove that \(\frac{d}{dx}(\frac{sin\,x}{cos\,x})\) ****= sec ^{2}x**

\(\frac{d}{dx}(\frac{sin\,x}{cos\,x})\) = \(\frac{1}{cos\,x}.\frac{d\,sin\,x}{dx}-\frac{1}{cos^2x}.sin\,x\frac{d}{dx}(cox\,x)\)

= \(\frac{cos^2x}{cos\,x}+\frac{sin^2x}{cos^2x}\) = \(1+\frac{sin^2x}{cos^2x}\)

= \(\frac{cos^2x+sin^2x}{cos^2x}\) = \(\frac{1}{cos^2x}\) **sec ^{2}x **

**(vii) Evaluate the following integral:**

**(i) \(\int_{0}^{π/2} sin\,x\,dx\)**** ****(ii) \(\int_{1}^{5} x\,dx\) **

(i) \(\int_{0}^{π/2} sin\,x\,dx\) = − cos x \(|_o^{π/2}\)

= − cos (π/2) + cos 0 = 0 + 1 = 1

(ii) \(\int_{1}^{5} x\,dx\) = \(\frac{x^2}{2}|_1^5\)

= \(\frac{5^2}{2}-\frac{1^2}{2}\)

= \(\frac{25-1}{2}\) = 24/2 = **12**

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