Solutions-Class-11-Science-Physics-Chapter-2-Mathematical Methods-Maharashtra Board

(D) 5 N along - x-axis

(C) same magnitude and direction

(A) zero

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

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

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.

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

= (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

\(\vec{v}\) = \(\vec{v_1}+\vec{v_2}\)  = (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^o34’

∴ θ = 180^o − 26^o34’ = 153^o26’

∴ \(\vec{v_1}+\vec{v_2}\) has a magnitude of 2\(\sqrt{5}\)  and makes an angle of θ = tan⁻¹\((\frac{1}{2})\) = 153^o26’ with the positive x-axis

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.

\(\vec{a}\) = 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

\(\vec{a}\)  = 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 = \(\hat{k}\)

\(\vec{a}\) = 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_xb_x + a_yb_y + a_zb_z = (2)(3) + (3)(−6) + (6)(2) = 6 – 18 + 12 = 0

∴ \(\vec{b}\)⊥\(\vec{a}\)

\(\vec{a}\).\(\vec{c}\) = a_xc_x + a_yc_y + a_zc_z = (2)(6) + (3)(2) + (6)( −3) = 12 + 6 – 18  = 0

∴ \(\vec{c}\)⊥\(\vec{a}\)

\(\vec{b}\).\(\vec{c}\) = b_xc_x + b_yc_y + b_zc_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.

\(\vec{v_1}\) = 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\(\hat{j}\)  + \(\hat{k}\)

\(\vec{v_1}\)   = 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

(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)(x⁴+cos x) = 4x² − sin x

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

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

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

= \(\frac{cos^2x+sin^2x}{cos^2x}\) = \(\frac{1}{cos^2x}\)  sec²x

(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