The assignment you uploaded covers a variety of problems in linear algebra, vector calculus, and complex numbers. I'll go through each problem in detail and solve it.

Question 1.1

We are given two planes $U$ and $V$ with the following equations:

• $U: \lambda x + 5y - 2\lambda z - 3 = 0$
• $V: -\lambda x + y + 2z + 1 = 0$

(a) Orthogonality:

Planes are orthogonal if their normal vectors are orthogonal. For $U$ and $V$, the normal vectors are:

• For $U$, $\mathbf{n}_U = \langle \lambda, 5, -2\lambda \rangle$
• For $V$, $\mathbf{n}_V = \langle -\lambda, 1, 2 \rangle$

Two vectors are orthogonal if their dot product is zero. Let's compute the dot product:

$\mathbf{n}_U \cdot \mathbf{n}_V = \lambda(-\lambda) + 5(1) + (-2\lambda)(2)$ $= -\lambda^2 + 5 - 4\lambda = 0$

Thus, we solve for $\lambda$: $-\lambda^2 - 4\lambda + 5 = 0$

This is a quadratic equation: $\lambda^2 + 4\lambda - 5 = 0$

Factoring the quadratic: $(\lambda + 5)(\lambda - 1) = 0$

Thus, $\lambda = -5$ or $\lambda = 1$.

(b) Parallelism:

Planes are parallel if their normal vectors are proportional. This means there must be a constant $k$ such that:

$\mathbf{n}_U = k \mathbf{n}_V$

In component form, this gives us the system of equations: $\lambda = -k\lambda, \quad 5 = k \times 1, \quad -2\lambda = k \times 2$

From $5 = k$, we get $k = 5$. Substituting into the first equation: $\lambda = -5\lambda$

Solving for $\lambda$, we get $\lambda = 0$.

Question 1.2

We need to find an equation for a plane that passes through the origin and is parallel to the plane: $-x + 3y - 2z = 6$

A plane parallel to another has the same normal vector. The normal vector for the given plane is $\mathbf{n} = \langle -1, 3, -2 \rangle$. Since the required plane passes through the origin, its equation will be:

$-x + 3y - 2z = 0$

Question 1.3

We are asked to find the distance between the point $(-1, -2, 0)$ and the plane $3x - y + 4z = -2$. The formula for the distance between a point $(x_1, y_1, z_1)$ and a plane $Ax + By + Cz + D = 0$ is:

$d = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}$

Substituting the given values:

$d = \frac{|3(-1) - (-2) + 4(0) + 2|}{\sqrt{3^2 + (-1)^2 + 4^2}}$ $= \frac{|-3 + 2 + 2|}{\sqrt{9 + 1 + 16}} = \frac{|1|}{\sqrt{26}} = \frac{1}{\sqrt{26}}$

Thus, the distance is $\frac{1}{\sqrt{26}}$.

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Would you like to continue solving the next questions in the assignment, or need further clarifications on the steps so far? Here are 5 related questions to extend your understanding:

1. How do you find the distance between two parallel planes?
2. What is the geometric interpretation of the dot product of vectors?
3. How do you determine if two planes are coincident (i.e., the same)?
4. Can a plane be both orthogonal and parallel to another plane?
5. How would you find the intersection line of two planes?

Tip:

When solving for the distance between a point and a plane, always ensure the plane equation is in the standard form $Ax + By + Cz + D = 0$ before applying the distance formula.