Given points P(2,1,3), Q(1,2,1), R(-1,-2,-2) and S(1,-4,0), find the shortest distance between lines PQ and RS.

To find the shortest distance between two skew lines, like lines PQ and RS in three-dimensional space, we can use a systematic approach involving vector mathematics. Let's break this down step by step.

Identifying Points and Vectors

First, we need to determine the direction vectors of each line. The direction vector for line PQ can be found by subtracting the coordinates of point P from point Q:

Direction vector for line PQ:

Vector PQ = Q - P = (1 - 2, 2 - 1, 1 - 3) = (-1, 1, -2).

Next, we calculate the direction vector for line RS:

Direction vector for line RS:

Vector RS = S - R = (1 - (-1), -4 - (-2), 0 - (-2)) = (2, -2, 2).

Finding the Vector Between the Lines

To find the shortest distance between the two lines, we also need a vector that connects a point on line PQ to a point on line RS. We can take points P and R for this purpose:

Connecting vector:

Vector PR = R - P = (-1 - 2, -2 - 1, -2 - 3) = (-3, -3, -5).

Calculating the Cross Product

The next step is to compute the cross product of the direction vectors of the two lines. This gives us a vector that is perpendicular to both direction vectors:

Cross product of PQ and RS:

Let A = (-1, 1, -2) and B = (2, -2, 2). The cross product A × B is calculated as follows:

Cross product formula:

• A × B = |i j k|
• |-1 1 -2|
• |2 -2 2|

Calculating this, we find:

• i component: (1 * 2) - (-2 * -2) = 2 - 4 = -2
• j component: -((-1 * 2) - (-2 * 2)) = 2 - 4 = -2
• k component: (-1 * -2) - (1 * 2) = 2 - 2 = 0

Thus, the cross product A × B = (-2, -2, 0).

Finding the Magnitude of the Cross Product

The magnitude of the cross product helps us in calculating the distance between the two lines:

Magnitude:

|A × B| = √((-2)² + (-2)² + 0²) = √(4 + 4) = √8 = 2√2.

Calculating the Distance

Finally, we can find the shortest distance (d) between the two lines using the formula:

Distance formula:

d = |(PR · (A × B))| / |A × B|.

First, we calculate the dot product PR · (A × B):

Dot product:

PR · (A × B) = (-3, -3, -5) · (-2, -2, 0) = (-3 * -2) + (-3 * -2) + (-5 * 0) = 6 + 6 + 0 = 12.

Now, we can substitute this back into our distance formula:

d = |12| / |2√2| = 12 / (2√2) = 6 / √2 = 3√2.

Final Result

The shortest distance between lines PQ and RS is 3√2 units.