Let's address each part of the problem based on the image provided.

First Problem (Vector Equation and Plane Intersection):

The line equation is given as: $\mathbf{r} = \mathbf{i} + \mathbf{j} - 2\mathbf{k} + \mu (\mathbf{i} - \mathbf{j} + \mathbf{k})$ The plane equation is: $\mathbf{r} \cdot (\mathbf{i} + 2\mathbf{j} - \mathbf{k}) = 2$

Part (a) - Finding the coordinates of Point $P$

To find the point where the line intersects the plane, we first express the vector form of the line as coordinates:

• The point on the line is $(1, 1, -2)$.
• The direction of the line is $(1, -1, 1)$.

The equation for the parametric point on the line is: $(1 + \mu, 1 - \mu, -2 + \mu)$

Substitute these coordinates into the plane equation: $(1 + \mu) \cdot 1 + (1 - \mu) \cdot 2 + (-2 + \mu) \cdot (-1) = 2$ Simplifying: $(1 + \mu) + 2(1 - \mu) + (-2 + \mu) = 2$ $1 + \mu + 2 - 2\mu - 2 + \mu = 2$ $1 + \mu = 2$ $\mu = 1$

Now, substitute $\mu = 1$ into the parametric form of the line: $x = 1 + 1 = 2, \quad y = 1 - 1 = 0, \quad z = -2 + 1 = -1$ Thus, the coordinates of point $P$ are $(2, 0, -1)$.

Part (b) - Calculate the angle between the line and the plane

The angle between the line and the plane is determined by the angle between the direction vector of the line and the normal vector of the plane.

• The direction vector of the line is $\mathbf{d} = (1, -1, 1)$.
• The normal vector of the plane is $\mathbf{n} = (1, 2, -1)$.

The angle $\theta$ between two vectors is given by: $\cos \theta = \frac{\mathbf{d} \cdot \mathbf{n}}{|\mathbf{d}| |\mathbf{n}|}$

First, calculate the dot product $\mathbf{d} \cdot \mathbf{n}$: $\mathbf{d} \cdot \mathbf{n} = 1 \cdot 1 + (-1) \cdot 2 + 1 \cdot (-1) = 1 - 2 - 1 = -2$

Next, calculate the magnitudes of $\mathbf{d}$ and $\mathbf{n}$: $|\mathbf{d}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}, \quad |\mathbf{n}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$

Thus, the cosine of the angle is: $\cos \theta = \frac{-2}{\sqrt{3} \cdot \sqrt{6}} = \frac{-2}{\sqrt{18}} = \frac{-2}{3\sqrt{2}}$

The angle $\theta$ is: $\theta = \cos^{-1} \left(\frac{-2}{3\sqrt{2}}\right)$

Second Problem (Plane Parallel to Line):

We are given the plane equation: $2x + 3y + az - 8 = 0$ and the line equation: $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{1}$

For the plane to be parallel to the line, the normal vector of the plane must be perpendicular to the direction vector of the line. The direction vector of the line is $\mathbf{d} = (2, 3, 1)$.

The normal vector of the plane is $\mathbf{n} = (2, 3, a)$. These vectors must be perpendicular, so their dot product must be zero: $\mathbf{d} \cdot \mathbf{n} = 2 \cdot 2 + 3 \cdot 3 + 1 \cdot a = 0$ $4 + 9 + a = 0$ $a = -13$

Thus, the value of $a$ is $-13$.

Would you like further details or have any questions? Here are five related questions to consider:

1. How do you compute the distance between a point and a plane in 3D?
2. What is the geometric interpretation of the dot product in vector algebra?
3. How would you modify the process if the line does not intersect the plane but instead is parallel to it?
4. Can you find the shortest distance between two skew lines in 3D space?
5. What is the significance of the magnitude of a vector in 3D geometry?

Tip: When solving problems involving vectors and planes, it's helpful to visualize the geometric relationships in 3D space.