1. **Problem 1(a):** Find the equation of the line which lies in the plane $\pi$ and intersects the line $l$ at right angles. Given: $$\vec{r} = 5\vec{i} - 3\vec{j} - \vec{k} + \lambda(\vec{i} - 2\vec{j} + \vec{k})$$ Plane $\pi$: $$(\vec{r} - \vec{i} - 2\vec{j}) \cdot (3\vec{i} + \vec{j} + \vec{k}) = 0$$ **Step 1:** Find the point of intersection of line $l$ with the plane $\pi$. Let $\vec{r} = \vec{a} + \lambda \vec{d}$ where $\vec{a} = 5\vec{i} - 3\vec{j} - \vec{k}$ and $\vec{d} = \vec{i} - 2\vec{j} + \vec{k}$. Substitute into plane equation: $$((\vec{a} + \lambda \vec{d}) - \vec{i} - 2\vec{j}) \cdot (3\vec{i} + \vec{j} + \vec{k}) = 0$$ Simplify: $$((5 - 1)\vec{i} + (-3 - 2)\vec{j} + (-1 - 0)\vec{k} + \lambda(\vec{i} - 2\vec{j} + \vec{k})) \cdot (3\vec{i} + \vec{j} + \vec{k}) = 0$$ $$ (4\vec{i} - 5\vec{j} - \vec{k} + \lambda(\vec{i} - 2\vec{j} + \vec{k})) \cdot (3\vec{i} + \vec{j} + \vec{k}) = 0$$ Calculate dot product: $$ (4)(3) + (-5)(1) + (-1)(1) + \lambda[(1)(3) + (-2)(1) + (1)(1)] = 0$$ $$ 12 - 5 - 1 + \lambda(3 - 2 + 1) = 0$$ $$ 6 + 2\lambda = 0 \implies \lambda = -3$$ **Step 2:** Find the point of intersection $P$: $$\vec{r} = 5\vec{i} - 3\vec{j} - \vec{k} - 3(\vec{i} - 2\vec{j} + \vec{k}) = (5 - 3)\vec{i} + (-3 + 6)\vec{j} + (-1 - 3)\vec{k} = 2\vec{i} + 3\vec{j} - 4\vec{k}$$ **Step 3:** Find direction vector of the required line. The required line lies in plane $\pi$ and is perpendicular to line $l$. Direction vector of $l$ is $\vec{d} = \vec{i} - 2\vec{j} + \vec{k}$. Let direction vector of required line be $\vec{m} = a\vec{i} + b\vec{j} + c\vec{k}$. Since the line lies in plane $\pi$, $\vec{m}$ is perpendicular to the plane's normal vector: $$\vec{m} \cdot (3\vec{i} + \vec{j} + \vec{k}) = 0 \implies 3a + b + c = 0$$ Since the line is perpendicular to $l$: $$\vec{m} \cdot \vec{d} = 0 \implies a - 2b + c = 0$$ **Step 4:** Solve the system: $$\begin{cases} 3a + b + c = 0 \\ a - 2b + c = 0 \end{cases}$$ Subtract second from first: $$(3a - a) + (b + 2b) + (c - c) = 0 \implies 2a + 3b = 0 \implies 2a = -3b \implies a = -\frac{3}{2}b$$ From second equation: $$a - 2b + c = 0 \implies c = 2b - a = 2b - \left(-\frac{3}{2}b\right) = 2b + \frac{3}{2}b = \frac{7}{2}b$$ Choose $b = 2$ for simplicity: $$a = -3, \quad b = 2, \quad c = 7$$ Direction vector: $$\vec{m} = -3\vec{i} + 2\vec{j} + 7\vec{k}$$ **Step 5:** Equation of required line: $$\vec{r} = 2\vec{i} + 3\vec{j} - 4\vec{k} + \mu(-3\vec{i} + 2\vec{j} + 7\vec{k})$$ --- 2. **Problem 1(b):** Solve for $z$ if $z = x + iy$ satisfies $$(1 + i)z^2 - (4 + 3i)z + 5 + i = 0$$ **Step 1:** Write quadratic in $z$: $$ (1 + i)z^2 - (4 + 3i)z + (5 + i) = 0$$ **Step 2:** Use quadratic formula: $$z = \frac{(4 + 3i) \pm \sqrt{(4 + 3i)^2 - 4(1 + i)(5 + i)}}{2(1 + i)}$$ Calculate discriminant: $$(4 + 3i)^2 = 16 + 24i + 9i^2 = 16 + 24i - 9 = 7 + 24i$$ Calculate: $$4(1 + i)(5 + i) = 4(5 + i + 5i + i^2) = 4(5 + 6i - 1) = 4(4 + 6i) = 16 + 24i$$ Discriminant: $$7 + 24i - (16 + 24i) = (7 - 16) + (24i - 24i) = -9 + 0 = -9$$ **Step 3:** Square root of discriminant: $$\sqrt{-9} = 3i$$ **Step 4:** Compute denominator: $$2(1 + i) = 2 + 2i$$ **Step 5:** Calculate numerator for both roots: $$4 + 3i \pm 3i$$ Root 1: $$4 + 3i + 3i = 4 + 6i$$ Root 2: $$4 + 3i - 3i = 4$$ **Step 6:** Divide numerator by denominator: For root 1: $$z_1 = \frac{4 + 6i}{2 + 2i}$$ Multiply numerator and denominator by conjugate of denominator: $$\frac{(4 + 6i)(2 - 2i)}{(2 + 2i)(2 - 2i)} = \frac{8 - 8i + 12i - 12i^2}{4 + 4} = \frac{8 + 4i + 12}{8} = \frac{20 + 4i}{8} = \frac{20}{8} + \frac{4}{8}i = 2.5 + 0.5i$$ For root 2: $$z_2 = \frac{4}{2 + 2i} = \frac{4(2 - 2i)}{(2 + 2i)(2 - 2i)} = \frac{8 - 8i}{8} = 1 - i$$ **Answer:** $$z = 2.5 + 0.5i \quad \text{or} \quad z = 1 - i$$ --- 3. **Problem 2(a):** Find $a$ if lines $l$ and $m$ pass through the same point. Lines: $$l: \vec{r} = a\vec{i} + 2\vec{j} + 3\vec{k} + \lambda(\vec{i} - 2\vec{j} + 3\vec{k})$$ $$m: \vec{r} = 2\vec{i} + \vec{j} + 2\vec{k} + \mu(2\vec{i} - \vec{j} + \vec{k})$$ **Step 1:** Equate points on $l$ and $m$: $$a + \lambda = 2 + 2\mu$$ $$2 - 2\lambda = 1 - \mu$$ $$3 + 3\lambda = 2 + \mu$$ **Step 2:** Solve system: From second: $$2 - 2\lambda = 1 - \mu \implies \mu = 1 - 2 + 2\lambda = -1 + 2\lambda$$ From third: $$3 + 3\lambda = 2 + \mu \implies \mu = 1 + 3\lambda$$ Equate $\mu$ expressions: $$-1 + 2\lambda = 1 + 3\lambda \implies -1 + 2\lambda - 1 - 3\lambda = 0 \implies -2 - \lambda = 0 \implies \lambda = -2$$ Find $\mu$: $$\mu = 1 + 3(-2) = 1 - 6 = -5$$ Find $a$ from first: $$a + (-2) = 2 + 2(-5) \implies a - 2 = 2 - 10 = -8 \implies a = -6$$ --- 4. **Problem 2(b):** Find equation of plane containing $l$ and $m$ for $a = -6$. **Step 1:** Direction vectors: $$\vec{d}_l = \vec{i} - 2\vec{j} + 3\vec{k}$$ $$\vec{d}_m = 2\vec{i} - \vec{j} + \vec{k}$$ **Step 2:** Find normal vector to plane: $$\vec{n} = \vec{d}_l \times \vec{d}_m$$ Calculate cross product: $$\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -2 & 3 \\ 2 & -1 & 1 \end{vmatrix} = \vec{i}((-2)(1) - 3(-1)) - \vec{j}(1(1) - 3(2)) + \vec{k}(1(-1) - (-2)(2))$$ $$= \vec{i}(-2 + 3) - \vec{j}(1 - 6) + \vec{k}(-1 + 4) = \vec{i}(1) - \vec{j}(-5) + \vec{k}(3) = \vec{i} + 5\vec{j} + 3\vec{k}$$ **Step 3:** Use point on $l$ or $m$ (e.g., $\lambda = -2$ on $l$): $$\vec{r}_0 = -6\vec{i} + 2\vec{j} + 3\vec{k} - 2(\vec{i} - 2\vec{j} + 3\vec{k}) = (-6 - 2)\vec{i} + (2 + 4)\vec{j} + (3 - 6)\vec{k} = -8\vec{i} + 6\vec{j} - 3\vec{k}$$ **Step 4:** Plane equation: $$\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$$ $$ (\vec{i} + 5\vec{j} + 3\vec{k}) \cdot ((x + 8)\vec{i} + (y - 6)\vec{j} + (z + 3)\vec{k}) = 0$$ Simplify: $$ (x + 8) + 5(y - 6) + 3(z + 3) = 0$$ $$ x + 8 + 5y - 30 + 3z + 9 = 0$$ $$ x + 5y + 3z - 13 = 0$$ --- 5. **Problem 3:** Find equation of plane $q$ containing line $$\vec{r} = \vec{i} + 3\vec{j} - 2\vec{k} + \lambda(\vec{i} - 2\vec{j} + 3\vec{k})$$ and perpendicular to plane $p: 2x + y - 3z = 5$. **Step 1:** Normal vector of plane $p$ is $$\vec{n}_p = 2\vec{i} + \vec{j} - 3\vec{k}$$ **Step 2:** Plane $q$ is perpendicular to $p$, so normal vector of $q$, $\vec{n}_q$, is perpendicular to $\vec{n}_p$. **Step 3:** Since $q$ contains the line with direction vector $$\vec{d} = \vec{i} - 2\vec{j} + 3\vec{k}$$ Normal vector $\vec{n}_q$ is perpendicular to $\vec{d}$ and $\vec{n}_p$. So, $$\vec{n}_q = \vec{n}_p \times \vec{d}$$ Calculate cross product: $$\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & -3 \\ 1 & -2 & 3 \end{vmatrix} = \vec{i}(1 \cdot 3 - (-3)(-2)) - \vec{j}(2 \cdot 3 - (-3)(1)) + \vec{k}(2 \cdot (-2) - 1 \cdot 1)$$ $$= \vec{i}(3 - 6) - \vec{j}(6 + 3) + \vec{k}(-4 - 1) = -3\vec{i} - 9\vec{j} - 5\vec{k}$$ **Step 4:** Use point on line: $$\vec{r}_0 = \vec{i} + 3\vec{j} - 2\vec{k}$$ **Step 5:** Plane $q$ equation: $$\vec{n}_q \cdot (\vec{r} - \vec{r}_0) = 0$$ $$(-3, -9, -5) \cdot (x - 1, y - 3, z + 2) = 0$$ Simplify: $$-3(x - 1) - 9(y - 3) - 5(z + 2) = 0$$ $$-3x + 3 - 9y + 27 - 5z - 10 = 0$$ $$-3x - 9y - 5z + 20 = 0$$ Or $$3x + 9y + 5z = 20$$ --- 6. **Problem 4:** Find $\sqrt{-3 - 2\sqrt{10}i}$ in form $a + ib$. **Step 1:** Let $$a + ib = \sqrt{-3 - 2\sqrt{10}i}$$ Square both sides: $$a^2 - b^2 + 2abi = -3 - 2\sqrt{10}i$$ Equate real and imaginary parts: $$a^2 - b^2 = -3$$ $$2ab = -2\sqrt{10} \implies ab = -\sqrt{10}$$ **Step 2:** From $ab = -\sqrt{10}$, express $b = -\frac{\sqrt{10}}{a}$. Substitute into first: $$a^2 - \left(-\frac{\sqrt{10}}{a}\right)^2 = -3$$ $$a^2 - \frac{10}{a^2} = -3$$ Multiply both sides by $a^2$: $$a^4 + 3a^2 - 10 = 0$$ Let $x = a^2$: $$x^2 + 3x - 10 = 0$$ Solve quadratic: $$x = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm 7}{2}$$ Positive root: $$x = \frac{4}{2} = 2$$ So, $$a^2 = 2 \implies a = \pm \sqrt{2}$$ **Step 3:** Find $b$: $$b = -\frac{\sqrt{10}}{a} = -\frac{\sqrt{10}}{\pm \sqrt{2}} = \mp \sqrt{5}$$ Choose $a = \sqrt{2}$, then $b = -\sqrt{5}$. **Answer:** $$\sqrt{-3 - 2\sqrt{10}i} = \sqrt{2} - i\sqrt{5}$$ --- 7. **Problem 5:** Sketch region in Argand diagram satisfying $$|z - 3 - i| \leq 3, \quad \arg(z) \geq \frac{\pi}{4}, \quad \text{and} \quad \text{Im}(z) \geq 2$$ **Step 1:** $|z - 3 - i| \leq 3$ is a circle centered at $(3,1)$ with radius 3. **Step 2:** $\arg(z) \geq \frac{\pi}{4}$ is the region above the line from origin at $45^\circ$. **Step 3:** $\text{Im}(z) \geq 2$ is the half-plane above the horizontal line $y=2$. **Answer:** The shaded region is the intersection of the circle, the sector above $\pi/4$ line, and above $y=2$. --- 8. **Problem 6:** Find point $P$ where $$\arg(z - 2i) = \frac{\pi}{6}$$ and $$|z - 3| = |z - 3i|$$ **Step 1:** Let $z = x + iy$. Condition 1: $$\arg(z - 2i) = \frac{\pi}{6} \implies \tan^{-1}\left(\frac{y - 2}{x}\right) = \frac{\pi}{6} \implies \frac{y - 2}{x} = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$ So, $$y = \frac{x}{\sqrt{3}} + 2$$ Condition 2: $$|z - 3| = |z - 3i|$$ Square both sides: $$(x - 3)^2 + y^2 = x^2 + (y - 3)^2$$ Expand: $$x^2 - 6x + 9 + y^2 = x^2 + y^2 - 6y + 9$$ Simplify: $$-6x = -6y \implies y = x$$ **Step 2:** Solve system: $$y = x$$ $$y = \frac{x}{\sqrt{3}} + 2$$ Set equal: $$x = \frac{x}{\sqrt{3}} + 2 \implies x - \frac{x}{\sqrt{3}} = 2 \implies x\left(1 - \frac{1}{\sqrt{3}}\right) = 2$$ Calculate: $$1 - \frac{1}{\sqrt{3}} = \frac{\sqrt{3} - 1}{\sqrt{3}}$$ So, $$x = \frac{2\sqrt{3}}{\sqrt{3} - 1}$$ Rationalize denominator: $$x = \frac{2\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{2\sqrt{3}(\sqrt{3} + 1)}{3 - 1} = \frac{2\sqrt{3}(\sqrt{3} + 1)}{2} = \sqrt{3}(\sqrt{3} + 1) = 3 + \sqrt{3}$$ Then, $$y = x = 3 + \sqrt{3}$$ **Step 3:** Express $P = x + iy$ in polar form: $$r = \sqrt{x^2 + y^2} = \sqrt{(3 + \sqrt{3})^2 + (3 + \sqrt{3})^2} = \sqrt{2(3 + \sqrt{3})^2} = (3 + \sqrt{3})\sqrt{2}$$ Calculate numerically: $$3 + \sqrt{3} \approx 3 + 1.732 = 4.732$$ $$r \approx 4.732 \times 1.414 = 6.69$$ Angle: $$\theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}(1) = \frac{\pi}{4} = 0.785$$ **Answer:** $$P = 6.69 e^{i 0.785}$$ --- **Final answers:** 1(a): $$\vec{r} = 2\vec{i} + 3\vec{j} - 4\vec{k} + \mu(-3\vec{i} + 2\vec{j} + 7\vec{k})$$ 1(b): $$z = 2.5 + 0.5i \quad \text{or} \quad z = 1 - i$$ 2(a): $$a = -6$$ 2(b): $$x + 5y + 3z = 13$$ 3: $$3x + 9y + 5z = 20$$ 4: $$\sqrt{2} - i\sqrt{5}$$ 5: Region inside circle centered at $(3,1)$ radius 3, above line $\arg(z) = \pi/4$, and above $y=2$. 6: $$P = 6.69 e^{i 0.785}$$