3d Geometry

Step 1: Standardize the equations of the lines
The given line is: The line parallel to and passing through is: Step 2: Vector between the lines
Let . The direction vector . Step 3: Find the shortest distance
The shortest distance is given by: Compute : The magnitude:

Step 1: Find
The direction ratios of the first line are and for the second line are . Since the lines are perpendicular: Step 2: Find the vector equation of the perpendicular line
The direction vectors are: The perpendicular vector is: The required equation is:

The equation of the mirror is given as the line: This is a parametric equation of the line. Let the parameter be , so the parametric equations are: The point lies at a distance from the mirror. The reflection of a point across a plane can be calculated by finding the perpendicular distance from the point to the line and then determining the symmetric point on the opposite side of the mirror.
By applying the reflection formula and solving for the coordinates of the reflected point, we find the image point and the distance between and .

To check if two lines are perpendicular, we compare their direction vectors. If the dot product of direction vectors is zero, the lines are perpendicular. From , the direction ratios (DRs) are: From , the direction ratios are: Now, for the lines to be perpendicular: Wait — correction. On close inspection of the signs: But the correct approach is to just treat both in terms of direction vectors derived from denominators: From : DRs =
From : DRs =
Dot product: Oops! There’s a misinterpretation in calculation — actually the correct DRs must come directly from the denominators in symmetric form (each line is in symmetric form). So the correct DRs are: - For :
- For :
Dot product: Final boxed value:

We are given two lines: For the lines to intersect, there must exist values of and such that the coordinates of points on both lines are equal. Equating the , , and components of the two lines:
- -component: ,
- -component: ,
- -component: .
From the -component equation, we have . Substituting into the -component equation: Substituting and into the -component equation: This results in a contradiction, so the lines do not intersect.

The given line is in symmetric form: Let this be equal to a parameter . Then any point on the line can be written as: Let be the given point. The foot of the perpendicular will be the point such that vector is perpendicular to the direction vector . So, : Taking dot product with : Expanding: Substitute into : So the foot of the perpendicular is . Now compute the length of perpendicular : % Final Answer Foot of the Perpendicular:
Length of the Perpendicular: \begin{center} \includegraphics[width=0.55\textwidth]{ig5.png} \end{center}

To find the equation of the line passing through point and parallel to line , we first need the direction vector of line . The direction vector of is: Thus, the direction vector of the line passing through and parallel to is . The parametric equations of the line passing through and parallel to are: Thus, the parametric equation of the line is: The Cartesian equation of the line can be found by eliminating the parameter from the parametric equations. From the equations and , solve for in terms of and : Equating these expressions for : Cross-multiply: which simplifies to: Thus, the Cartesian equation of the line is:

We are given the point and the equation of the line in symmetric form: Let the parametric form of the line be: where is a parameter. Let be a point on the line and let the direction vector of the line be: The vector from point to point is: The distance from a point to a line is given by: First, we compute the cross product : Expanding this determinant: This results in: Simplifying: Thus, Next, we compute the magnitude of : Now, we compute the magnitude of : Finally, the distance is: Thus, the distance from the point to the line is .

Step 1: Convert the line into parametric form.
Let Then Thus the coordinates of any point on the line are Step 2: Use perpendicular condition.
The foot of the perpendicular from the origin will satisfy that the vector from origin to this point is perpendicular to the direction of the line.
Direction ratios of the line are Vector Perpendicular condition: Step 3: Compute the dot product.
Step 4: Substitute the value of .
Step 5: Final conclusion.
Thus the coordinates of the foot of perpendicular are Final Answer:

Step 1: Write the equation of line AB.
Direction vector of line : Thus parametric form: Step 2: Write the second line in parametric form.
Given Thus Step 3: Find intersection point.
At intersection both coordinates must match. From : Substitute into : Now Step 4: Coordinates of intersection.
Substitute : Thus point of intersection is Step 5: Find a line perpendicular to both lines.
Direction vectors: First line: Second line: Required direction is Thus direction vector Step 6: Write equation of required line.
Passing through Equation of required line: Final Answer:
Point of intersection: Equation of required line: