Distance Between Two Lines

The given lines are:

and

.

The formula for the shortest distance between two skew lines is:

Shortest distance (S.D.) =

Here:

Step 1: Compute

The cross product is given by:

Expanding the determinant:

Simplifying:

So:

Step 2: Compute

The magnitude of is:

Simplifying:

Step 3: Compute

The vector is:

15

The dot product is:

Simplifying:

Step 4: Compute the shortest distance

Using the formula:

Substituting the values:

Finally, rationalizing the denominator gives:

Correct Option: 3 (2)

The shortest distance between two skew lines is:

Here:

Calculate , simplify , and compute:

Step 1: The shortest distance between the lines is given by: This is the cross product of the direction vectors and . Now, compute : Step 2: The distance formula gives: Simplifying the equation: The solutions for are 9 and -15.
Step 3: From this, we find: Thus, the equation becomes: This gives the final result for , which can be either -13 or 35, depending on the value of .

Let the equation in the new position be:

Simplifying, we solve for :

Expand and simplify:

Substitute back into the equation:

Simplify the equation:

From this equation, we find:

Step 1: Find the direction of the line of shortest distance.
The two given lines can be written in vector form as Hence their direction vectors are The line of shortest distance between two skew lines is parallel to the cross product . Compute: So .

Step 2: Parametric form from the point to the plane along .
We want the distance from the point to the plane in the direction . Set up the line We must find such that lies on the plane. Let be the plane’s normal. Then Explicitly: Compute the dot product: Thus

Step 3: Distance in that direction.
Hence the intersection point with the plane is The vector from to is which is indeed . Its magnitude is Therefore, the required distance along the direction from the point to the plane is

The image of the point in the plane is given as P. We are tasked to find given that the distance from point Q(6, -2, a) to point P is 13.

1. Finding the Image Point P:

Given Point:

Plane Equation:

Let the image point P be .

1. Midpoint: The midpoint of AP lies on the plane.

Midpoint

Substitute M into the plane equation:

2. Line AP is Perpendicular to the Plane:

The direction vector of AP is

The normal vector of the plane is

Since AP is perpendicular to the plane, is parallel to . Therefore, for some scalar .

3. Solve for x, y, z:

Substitute these into the midpoint equation:

Then:

So,

2. Calculating alpha:

P = (8/3, -1/3, 11/3)

Q = (6, -2, a)

Distance PQ = 13

Correct final answer:

To find the shortest distance between the given lines and , we will use the formula for the shortest distance between two skew lines in vector form. The given lines are:

First, extract the direction vectors and a point from each line:

• Direction vector of :
• Direction vector of :

Let be a point on when and be a point on when .

The vector connecting these points is: .

Now, the formula for the shortest distance between two skew lines is given by:

First, calculate the cross product :

Calculating the determinant, we find:

Now, calculate the dot product with the cross product:

Now, find the magnitude of the cross product:

Finally, the shortest distance is:

Given that the shortest distance is , we identify and . The greatest common divisor of 32 and 355 is 1, thus the ratio is already in its simplest form.

Therefore, the value of .

Given lines:

Direction vectors:
The direction vector of the first line is:
The direction vector of the second line is:
Shortest Distance Formula:
The shortest distance between skew lines with direction vectors and is given by:

where is the vector between points on the lines.

Cross Product:
Calculating :

Simplifying:


Magnitude:

Given distance:

Integral Evaluation:
Consider:
Substituting the value of :
Comparing with , we find:
Calculating :

To find the shortest distance between two skew lines, we use the formula for the distance between two skew lines:

Let's consider the lines:

The direction vectors for these lines are:

• For :
• For :

The formula for the distance between two skew lines is:

Substitute the initial points on each line: and .

First, find the vector between points on the lines:

Next, calculate the cross product of and :

Now, calculate the dot product :

Next, find the magnitude of the cross product :

Finally, calculate the shortest distance :

=

Therefore, the shortest distance between the given lines is .

Let the line of the shortest distance between the lines and intersect and at and , respectively. If is the midpoint of the line segment , then is equal to ____.

Concept Used:

The line of shortest distance between two skew lines is perpendicular to both lines. The points and on and respectively can be found using the condition that the vector is perpendicular to both direction vectors. The midpoint coordinates are then used to find the required expression.

Step-by-Step Solution:

Step 1: Write the general points on and .

For : ⇒

For : ⇒

Step 2: Find the vector .

Step 3: Apply the condition that is perpendicular to both direction vectors.

Direction vector of :

Direction vector of :

Condition 1:

Condition 2:

Step 4: Solve equations (1) and (2) to find and .

From (1):

Substitute into (2):

Then

Step 5: Find coordinates of P and Q.

Step 6: Find the midpoint of PQ.

So

Step 7: Compute .

Hence, is equal to 21.

The problem asks for the largest possible value of given the equations of two lines in 3D space and the shortest distance between them. The shortest distance between two skew lines is given by a standard formula involving vector operations.

Concept Used:

The shortest distance (SD) between two skew lines, and , is given by the formula:

This formula calculates the length of the projection of the vector connecting a point on each line ( ) onto the vector normal to both lines ( ).

Step-by-Step Solution:

Step 1: Identify the position vectors and direction vectors from the given line equations.

For the first line, :

• A point on the line is , so the position vector is .
• The direction vector is .

For the second line, :

• A point on the line is , so the position vector is .
• The direction vector is .

Step 2: Calculate the vector .

Step 3: Calculate the cross product of the direction vectors, .

Step 4: Calculate the magnitude of the cross product, .

Step 5: Calculate the scalar triple product, .

Step 6: Substitute these results into the shortest distance formula and solve for .

The shortest distance is given as .

We can cancel from both sides:

This absolute value equation leads to two possible cases:

Case 1:

Case 2:

Step 7: Determine the largest possible value of .

The two possible values for are 1 and -43. We find their absolute values:

The largest of these two values is 43.

Therefore, the largest possible value of is 43.