Solve the following math questions: 1A. Prove by induction that for n ∈ Z+, Σ(r=1 to n) r(r+3) = (1/3)n(n+1)(n+5) 1B. Prove that √3 is irrational. 2A. Simplify (cos30 + i sin30)^4... Solve the following math questions: 1A. Prove by induction that for n ∈ Z+, Σ(r=1 to n) r(r+3) = (1/3)n(n+1)(n+5) 1B. Prove that √3 is irrational. 2A. Simplify (cos30 + i sin30)^4 (cos40 - i sin40)^5 / (cos40 + i sin40)^3 (cos50 + i sin50)^4 2B. Given that z = x + iy, find the value of x and the value of y such that z + 3iz̄ = -1 + 13i, where z̄ is the complex conjugate of z. 3. With respect to a fixed origin O, the lines L1 and L2 are given by the equations l1 = [6, -3, -2] + λ[-1, 2, 3] and l2 = [-5,15,3] + μ[2, -3, 1], where λ and μ are scalar parameters. (a) Show that L1 and L2 meet and find the position vector of their point of intersection A. (b) Find, to the nearest 0.1°, the acute angle between L1 and L2. (c) Show that the point B with position vector [5, -1, 1] lies on L1. (d) Find the shortest distance from B to the line L2, giving your answer to 3 significant figures. Read more

Steps to Solve

1. 1A. Base Case (n=1) Check if the formula holds for the smallest value of $n$, which is $n=1$. Left-hand side (LHS): $\sum_{r=1}^{1} r(r+3) = 1(1+3) = 4$ Right-hand side (RHS): $\frac{1}{3}(1)(1+1)(1+5) = \frac{1}{3}(1)(2)(6) = 4$ Since LHS = RHS, the formula holds for $n=1$.

2. 1A. Inductive Hypothesis Assume the formula holds for some positive integer $k$. That is, assume $\sum_{r=1}^{k} r(r+3) = \frac{1}{3}k(k+1)(k+5)$ is true.

3. 1A. Inductive Step Prove that the formula also holds for $n=k+1$. We need to show that $\sum_{r=1}^{k+1} r(r+3) = \frac{1}{3}(k+1)(k+2)(k+6)$. Starting with the LHS: $\sum_{r=1}^{k+1} r(r+3) = \sum_{r=1}^{k} r(r+3) + (k+1)(k+1+3)$ Using the inductive hypothesis: $= \frac{1}{3}k(k+1)(k+5) + (k+1)(k+4)$ $= \frac{1}{3}k(k+1)(k+5) + 3\frac{(k+1)(k+4)}{3}$ $= \frac{1}{3}(k+1)[k(k+5) + 3(k+4)]$ $= \frac{1}{3}(k+1)[k^2 + 5k + 3k + 12]$ $= \frac{1}{3}(k+1)[k^2 + 8k + 12]$ $= \frac{1}{3}(k+1)(k+2)(k+6)$ This is the RHS with $n=k+1$, so the formula holds for $n=k+1$.

4. 1A. Conclusion By the principle of mathematical induction, the formula $\sum_{r=1}^{n} r(r+3) = \frac{1}{3}n(n+1)(n+5)$ is true for all positive integers $n$.

5. 1B. Proof by Contradiction Assume, for the sake of contradiction, that $\sqrt{3}$ is rational. Then, $\sqrt{3}$ can be written as a fraction $\sqrt{3} = \frac{a}{b}$, where $a$ and $b$ are integers with no common factors other than 1 (i.e., the fraction is in its simplest form), and $b \ne 0$.

6. 1B. Algebraic Manipulation Square both sides: $3 = \frac{a^2}{b^2}$. Multiply both sides by $b^2$: $3b^2 = a^2$. This means that $a^2$ is divisible by 3, and therefore $a$ must also be divisible by 3. So, we can write $a = 3k$ for some integer $k$.

7. 1B. Substitution Substitute $a = 3k$ into the equation $3b^2 = a^2$: $3b^2 = (3k)^2 = 9k^2$. Divide both sides by 3: $b^2 = 3k^2$. This means that $b^2$ is divisible by 3, and therefore $b$ must also be divisible by 3.

8. 1B. Contradiction We have shown that both $a$ and $b$ are divisible by 3. This contradicts our initial assumption that $a$ and $b$ have no common factors other than 1.

9. 1B. Conclusion Therefore, our initial assumption that $\sqrt{3}$ is rational must be false. Hence, $\sqrt{3}$ is irrational.

10. 2A. Simplify using De Moivre's Theorem Rewrite the expression using Euler's formula: $e^{ix} = \cos x + i \sin x$ Given expression: $\frac{(\cos 30 + i \sin 30)^4 (\cos 40 - i \sin 40)^5}{(\cos 40 + i \sin 40)^3 (\cos 50 + i \sin 50)^4}$ Using De Moivre's Theorem $(\cos x + i \sin x)^n = \cos(nx) + i \sin(nx)$, we have: $= \frac{(\cos (4 \cdot 30) + i \sin (4 \cdot 30))(\cos (5 \cdot -40) + i \sin (5 \cdot -40))}{(\cos (3 \cdot 40) + i \sin (3 \cdot 40))(\cos (4 \cdot 50) + i \sin (4 \cdot 50))}$ $= \frac{(\cos 120 + i \sin 120)(\cos (-200) + i \sin (-200))}{(\cos 120 + i \sin 120)(\cos 200 + i \sin 200)}$ $= \frac{\cos 120 + i \sin 120}{\cos 120 + i \sin 120} \cdot \frac{\cos (-200) + i \sin (-200)}{\cos 200 + i \sin 200}$ $= 1 \cdot (\cos(-200 - 200) + i \sin(-200-200))$ $= \cos(-400) + i \sin(-400)$ $= \cos(400) - i \sin(400)$ Since $400 = 360 + 40$, we have $= \cos(40) - i \sin(40)$.

11. 2B. Complex Conjugate Given $z = x + iy$, the complex conjugate is $\bar{z} = x - iy$. The equation is $z + 3i\bar{z} = -1 + 13i$.

12. 2B. Substitution and Simplification Substitute $z = x + iy$ and $\bar{z} = x - iy$ into the equation: $(x + iy) + 3i(x - iy) = -1 + 13i$ $x + iy + 3ix - 3i^2y = -1 + 13i$ Since $i^2 = -1$, we have: $x + iy + 3ix + 3y = -1 + 13i$ $(x + 3y) + (y + 3x)i = -1 + 13i$

13. 2B. Equating Real and Imaginary Parts Equate the real parts: $x + 3y = -1$ Equate the imaginary parts: $y + 3x = 13$

14. 2B. Solving the System of Equations We now have a system of two linear equations with two variables: $x + 3y = -1$ $3x + y = 13$ Multiply the first equation by -3: $-3x - 9y = 3$ Add this to the second equation: $(3x + y) + (-3x - 9y) = 13 + 3$ $-8y = 16$ $y = -2$ Substitute $y = -2$ into the first equation: $x + 3(-2) = -1$ $x - 6 = -1$ $x = 5$

15. 3A. Showing Lines Meet $L_1: \begin{bmatrix} 6 \ -3 \ -2 \end{bmatrix} + \lambda \begin{bmatrix} -1 \ 2 \ 3 \end{bmatrix}$ and $L_2: \begin{bmatrix} -5 \ 15 \ 3 \end{bmatrix} + \mu \begin{bmatrix} 2 \ -3 \ 1 \end{bmatrix}$ For the lines to meet, there must exist $\lambda$ and $\mu$ such that: $6 - \lambda = -5 + 2\mu$ $-3 + 2\lambda = 15 - 3\mu$ $-2 + 3\lambda = 3 + \mu$ From the first equation: $\lambda + 2\mu = 11$ (1) From the second equation: $2\lambda + 3\mu = 18$ (2) From the third equation: $3\lambda - \mu = 5$ (3) Multiply (3) by 2: $6\lambda - 2\mu = 10$ (4) Add (1) and (4): $7\lambda = 21$, so $\lambda = 3$. Substitute $\lambda = 3$ into (1): $3 + 2\mu = 11$, $2\mu = 8$, so $\mu = 4$. Now, we check if these values satisfy all three equations. (1): $3 + 2(4) = 3 + 8 = 11$. Correct. (2): $2(3) + 3(4) = 6 + 12 = 18$. Correct. (3): $3(3) - 4 = 9 - 4 = 5$. Correct. Since we found values for $\lambda$ and $\mu$ that satisfy all three equations, the lines meet.

16. 3A. Finding the Point of Intersection Using $\lambda = 3$ in $L_1$: $\begin{bmatrix} 6 \ -3 \ -2 \end{bmatrix} + 3 \begin{bmatrix} -1 \ 2 \ 3 \end{bmatrix} = \begin{bmatrix} 6 - 3 \ -3 + 6 \ -2 + 9 \end{bmatrix} = \begin{bmatrix} 3 \ 3 \ 7 \end{bmatrix}$ So the point of intersection A is $(3, 3, 7)$.

17. 3B. Finding the Angle Between the Lines The direction vectors of the lines are $\vec{d_1} = \begin{bmatrix} -1 \ 2 \ 3 \end{bmatrix}$ and $\vec{d_2} = \begin{bmatrix} 2 \ -3 \ 1 \end{bmatrix}$. The angle $\theta$ between the lines is given by: $\cos \theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{||\vec{d_1}|| \cdot ||\vec{d_2}||}$ $\vec{d_1} \cdot \vec{d_2} = (-1)(2) + (2)(-3) + (3)(1) = -2 - 6 + 3 = -5$ $||\vec{d_1}|| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$ $||\vec{d_2}|| = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$ $\cos \theta = \frac{|-5|}{\sqrt{14} \cdot \sqrt{14}} = \frac{5}{14}$ $\theta = \cos^{-1}\left(\frac{5}{14}\right) \approx 69.255^\circ$ To the nearest 0.1°, the acute angle is $69.3^\circ$.

18. 3C. Showing B lies on L1 $L_1: \begin{bmatrix} 6 \ -3 \ -2 \end{bmatrix} + \lambda \begin{bmatrix} -1 \ 2 \ 3 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \ -1 \ 1 \end{bmatrix}$ We need to find a value of $\lambda$ such that: $5 = 6 - \lambda$ $-1 = -3 + 2\lambda$ $1 = -2 + 3\lambda$ From the first equation: $\lambda = 6 - 5 = 1$ From the second equation: $2\lambda = -1 + 3 = 2$, so $\lambda = 1$ From the third equation: $3\lambda = 1 + 2 = 3$, so $\lambda = 1$ Since $\lambda = 1$ satisfies all three equations, the point B lies on $L_1$.

19. 3D. Finding the Shortest Distance from B to L2 $B = \begin{bmatrix} 5 \ -1 \ 1 \end{bmatrix}$ and $L_2: \begin{bmatrix} -5 \ 15 \ 3 \end{bmatrix} + \mu \begin{bmatrix} 2 \ -3 \ 1 \end{bmatrix}$ Let $\vec{a} = \begin{bmatrix} -5 \ 15 \ 3 \end{bmatrix}$ (a point on $L_2$) and $\vec{d} = \begin{bmatrix} 2 \ -3 \ 1 \end{bmatrix}$ (direction vector of $L_2$). $\vec{BA} = \vec{a} - \vec{b} = \begin{bmatrix} -5 \ 15 \ 3 \end{bmatrix} - \begin{bmatrix} 5 \ -1 \ 1 \end{bmatrix} = \begin{bmatrix} -10 \ 16 \ 2 \end{bmatrix}$ The shortest distance $d$ is given by: $d = \frac{||\vec{BA} \times \vec{d}||}{||\vec{d}||}$

20. 3D. Calculate the Cross Product

$\vec{BA} \times \vec{d} = \begin{bmatrix} -10 \ 16 \ 2 \end{bmatrix} \times \begin{bmatrix} 2 \ -3 \ 1 \end{bmatrix} = \begin{bmatrix} (16)(1) - (2)(-3) \ (2)(2) - (-10)(1) \ (-10)(-3) - (16)(2) \end{bmatrix} = \begin{bmatrix} 16 + 6 \ 4 + 10 \ 30 - 32 \end{bmatrix} = \begin{bmatrix} 22 \ 14 \ -2 \end{bmatrix}$

21. 3D. Calculate the Magnitude of the Cross Product

$||\vec{BA} \times \vec{d}|| = \sqrt{22^2 + 14^2 + (-2)^2} = \sqrt{484 + 196 + 4} = \sqrt{684} = 6\sqrt{19}$

22. 3D. Calculate the Magnitude of the Direction Vector (already calculated in 3B) $||\vec{d}|| = \sqrt{14}$

23. 3D. Calculate the Shortest Distance

$d = \frac{6\sqrt{19}}{\sqrt{14}} = 6\sqrt{\frac{19}{14}} \approx 6\sqrt{1.357} \approx 6(1.165) \approx 7.0$

The shortest distance from B to $L_2$ is approximately 7.00.