3d Geometry

To find the image of the point in the plane , we need to follow these steps:

1. First, find the normal vector of the plane. The equation of the plane is given by . The normal vector to the plane can be extracted from the coefficients of , , and , which is .
2. We use the formula to find the perpendicular distance from a point to the plane :

D = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}

1. Substituting and , we calculate:

D = \frac{|2(1) - 1(3) + 1(4) + 3|}{\sqrt{2^2 + (-1)^2 + 1^2}} = \frac{|2 - 3 + 4 + 3|}{\sqrt{4 + 1 + 1}} = \frac{|6|}{\sqrt{6}} = \frac{6}{\sqrt{6}} = \sqrt{6}

1. Find the coordinates of the foot of the perpendicular from point on the plane. The coordinates of the foot of the perpendicular are computed using:

(x_2, y_2, z_2) = (x_1, y_1, z_1) - D \cdot \mathbf{n}

1. where is the perpendicular distance calculated previously, and is a unit normal vector to the plane:

\mathbf{n}_{unit} = \left( \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right)

1. Compute the coordinates:

(x_2, y_2, z_2) = \left( 1 - \sqrt{6} \cdot \frac{2}{\sqrt{6}}, 3 - \sqrt{6} \cdot \frac{-1}{\sqrt{6}}, 4 - \sqrt{6} \cdot \frac{1}{\sqrt{6}} \right) = (1 - 2, 3 + 1, 4 - 1) = (-1,4,3)

1. The image of the point with respect to the plane is the symmetric point about the foot of the perpendicular. Use the midpoint formula between and its image such that:

(x_2, y_2, z_2) = \frac{(x + 1)}{2}, \frac{(y + 3)}{2}, \frac{(z + 4)}{2}

1. Solving the above:

\begin{align*} x + 1 & = -2 \times (-1), \\ y + 3 & = 2 \times 4, \\ z + 4 & = 2 \times 3, \\ \end{align*}

(x, y, z) = \left ( -3, 5, 2 \right )

Thus, the image of point in the given plane is , which matches the correct option.

To find the equation of the plane that contains the given lines, we first need to understand the components of the problem. Let's break it down:

1. Both given lines pass through the common point .
2. The direction vector of the first line is .
3. The direction vector of the second line is .

The plane containing these two lines will have a normal vector that is perpendicular to both and . Thus, the normal vector can be found using the cross product:

Computing the cross product:

Thus, the normal vector , which can be simplified to .

Now, using the point which lies on the plane, the equation of the plane is:

This can also be expressed as .

We can now conclude that the correct answer is:

To find the distance of a point from a line measured parallel to a plane, we need to calculate the perpendicular distance from the point to the line and ensure that this perpendicular distance is measured along a direction parallel to the given plane. Let's solve this step-by-step:

1. Identify the given data:
   • Point
   • Line (direction ratios are )
   • Plane
2. Find a direction vector parallel to the plane:
3. Find a direction vector that is perpendicular to both and using the cross product:
4. Calculate the cross product:
   • Determinant calculation results in:
5. Find the distance using the vector :
6. Substitute the point and direction vector into the formula:
7. Compute the dot product and magnitude:
   • Dot product:
8. Calculate the distance:
   • Upon simplification, we find .

Therefore, the distance from the point to the line, measured parallel to the given plane, is 7.

Step 1: Understanding the Concept:
Area of parallelogram when diagonals are given .
Step 2: Detailed Explanation:
.
.
Area .
Step 3: Final Answer:
Area .