1. **Problem:** Find the equation of the line which lies in the plane $\pi$ and intersects the line $l$ at right angles. Given: - Line $l$: $\vec{r} = 5\mathbf{i} - 3\mathbf{j} - \mathbf{k} + \lambda(\mathbf{i} - 2\mathbf{j} + \mathbf{k})$ - Plane $\pi$: $(\vec{r} - \mathbf{i} - 2\mathbf{j}) \cdot (3\mathbf{i} + \mathbf{j} + \mathbf{k}) = 0$ **Step 1:** Find the direction vector of line $l$: $$\vec{d}_l = \mathbf{i} - 2\mathbf{j} + \mathbf{k}$$ **Step 2:** The line we seek, call it $m$, lies in plane $\pi$ and intersects $l$ at right angles. Let the direction vector of $m$ be $\vec{d}_m = a\mathbf{i} + b\mathbf{j} + c\mathbf{k}$. **Step 3:** Since $m$ lies in plane $\pi$, its direction vector $\vec{d}_m$ is perpendicular to the plane's normal vector: $$\vec{n} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}$$ So, $$\vec{d}_m \cdot \vec{n} = 0 \implies 3a + b + c = 0$$ **Step 4:** Since $m$ is perpendicular to $l$, their direction vectors satisfy: $$\vec{d}_m \cdot \vec{d}_l = 0 \implies a(1) + b(-2) + c(1) = 0 \implies a - 2b + c = 0$$ **Step 5:** Solve the system: $$\begin{cases} 3a + b + c = 0 \\ a - 2b + c = 0 \end{cases}$$ Subtract second from first: $$ (3a + b + c) - (a - 2b + c) = 0 \implies 2a + 3b = 0 \implies 2a = -3b \implies a = -\frac{3}{2}b $$ Substitute $a$ into second equation: $$ -\frac{3}{2}b - 2b + c = 0 \implies c = \frac{7}{2}b $$ Choose $b=2$ for simplicity: $$ a = -3, \quad b=2, \quad c=7 $$ So, $$ \vec{d}_m = -3\mathbf{i} + 2\mathbf{j} + 7\mathbf{k} $$ **Step 6:** Find the point of intersection between $l$ and $m$. Parametric form of $l$: $$ x = 5 + \lambda, \quad y = -3 - 2\lambda, \quad z = -1 + \lambda $$ Let the point of intersection be $P$ with parameter $\lambda_0$ on $l$ and parameter $\mu$ on $m$. Since $m$ lies in plane $\pi$, and passes through $P$, write $m$ as: $$ \vec{r} = \vec{P} + \mu \vec{d}_m $$ **Step 7:** To find $P$, note that $m$ is perpendicular to $l$ and intersects it, so vector $\vec{P} - \vec{r}_0$ (where $\vec{r}_0 = 5\mathbf{i} - 3\mathbf{j} - \mathbf{k}$) is perpendicular to $\vec{d}_l$ and $\vec{d}_m$. Set $\vec{P} = (5 + \lambda_0, -3 - 2\lambda_0, -1 + \lambda_0)$. The vector from $\vec{r}_0$ to $P$ is: $$ \vec{P} - \vec{r}_0 = (\lambda_0, -2\lambda_0, \lambda_0) = \lambda_0(1, -2, 1) $$ This is parallel to $\vec{d}_l$, so the vector connecting $l$ and $m$ at $P$ is along $\vec{d}_l$. Since $m$ is perpendicular to $l$, the shortest distance vector between them is perpendicular to both. **Step 8:** The line $m$ passes through $P$ and has direction $\vec{d}_m$. Hence, the equation of $m$ is: $$ \vec{r} = (5 + \lambda_0)\mathbf{i} + (-3 - 2\lambda_0)\mathbf{j} + (-1 + \lambda_0)\mathbf{k} + \mu(-3\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}) $$ This is the required line. --- 2. **Problem:** Solve for $z$ if $z = x + iy$ satisfies: $$(1 + i)z^2 - (4 + 3i)z + 5 + i = 0$$ **Step 1:** Write the quadratic equation: $$(1 + i)z^2 - (4 + 3i)z + (5 + i) = 0$$ **Step 2:** Use quadratic formula: $$ z = \frac{(4 + 3i) \pm \sqrt{(4 + 3i)^2 - 4(1 + i)(5 + i)}}{2(1 + i)} $$ **Step 3:** Compute discriminant: $$ D = (4 + 3i)^2 - 4(1 + i)(5 + i) $$ Calculate: $$ (4 + 3i)^2 = 16 + 24i + 9i^2 = 16 + 24i - 9 = 7 + 24i $$ Calculate: $$ 4(1 + i)(5 + i) = 4(5 + i + 5i + i^2) = 4(5 + 6i - 1) = 4(4 + 6i) = 16 + 24i $$ So, $$ D = (7 + 24i) - (16 + 24i) = -9 + 0i = -9 $$ **Step 4:** Square root of $D$: $$ \sqrt{-9} = 3i $$ **Step 5:** Substitute back: $$ z = \frac{4 + 3i \pm 3i}{2(1 + i)} $$ Calculate denominator: $$ 2(1 + i) = 2 + 2i $$ **Step 6:** Simplify numerator for both cases: - Case 1: $4 + 3i + 3i = 4 + 6i$ - Case 2: $4 + 3i - 3i = 4$ **Step 7:** Divide numerator by denominator: $$ z = \frac{4 + 6i}{2 + 2i} \quad \text{or} \quad z = \frac{4}{2 + 2i} $$ Multiply numerator and denominator by conjugate $2 - 2i$: For case 1: $$ z = \frac{(4 + 6i)(2 - 2i)}{(2 + 2i)(2 - 2i)} = \frac{8 - 8i + 12i - 12i^2}{4 + 4} = \frac{8 + 4i + 12}{8} = \frac{20 + 4i}{8} = \frac{20}{8} + \frac{4i}{8} = 2.5 + 0.5i $$ For case 2: $$ z = \frac{4(2 - 2i)}{8} = \frac{8 - 8i}{8} = 1 - i $$ **Final answer:** $$ z = 2.5 + 0.5i \quad \text{or} \quad z = 1 - i $$