# Let P(1,0,1) A(0,1,1) B( 1,1,0) . Find the distance between the point P and the line passing through A and B ?

Hi, I really need help with this problem. Please show me steps.

### 3 Answers

- SqdancefanLv 73 years agoFavorite Answer
The cross product is useful for this.

|AP×AB| = |AP|*|AB|*sin(θ)

where θ is the angle between AP and AB.

The distance you seek is |AP|*sin(θ), so will be the cross product of AP with a unit vector in the direction of AB.

.. AP = (1, 0, 1) -(0, 1, 1) = (1, -1, 0)

.. AB = (1, 1, 0) -(0, 1, 1) = (1, 0, -1)

.. |AB| = √(1^2 +0^2 +(-1)^2) = √2

AP×AB = (1, 1, 1)

|AP×AB| = √(1^2 +1^2 +1^2) = √3

The distance you seek is

.. |AP×AB|/|AB| = (√3)/√2 = √(3/2)

- Login to reply the answers

- Φ² = Φ+1Lv 73 years ago
By observation, △PAB is equilateral with a side of length √2.

The distance between the point P and the line passing through A and B is the height of an equilateral triangle with a side of length √2, which is h=s½√3, which is √³∕₂.

- Login to reply the answers

- ted sLv 73 years ago
turns out to be fairly easy....| w x m | = | w | | m | sin Θ....

let w = < AB> and m = <AP>...then the distance is | m | sin Θ...

Θ is the angle between w and m

- Login to reply the answers