Finding the Distance Between Two Skew Lines by Projection

Theorem: Distance between two skew lines is equal to the distance between their projections on the plane perpendicular to one of them.

Proof: Given two skew lines a and b with common perpendicular NM, and a plane α such that b ⊥ α.

Let M₁ be intersection of α and b, and a₁ be projection of a on α (Figure 2.37).

So, we need to prove that the length of NM is equal to the distance between M₁ and a₁.

Let N₁ ∈ a₁ be the projection of N on α.

Since b ⊥ α, b ⊥ M₁N₁.

For the common perpendicular, we can write NM ⊥ b.

So, both NM and N₁M₁ are perpendicular to the same line b.

Therefore, NM // N₁M₁.

On the order hand, MM₁ ⊥ α and NN₁ ⊥ α. So, MM₁ // NN₁, which yields the result that NMM₁N₁ is a rectangle. So we have NM = N₁M₁.

Now, we will show that N₁M₁ is the distance between a₁ and M₁. It can be done just by showing that N₁M₁ ⊥ a₁.

Since MN ⊥ a, ∠MNK = 90°.

The projection of angle MNK on α is ∠M₁N₁K₁.

Since ∠MNK= 90° and MN // M₁N₁ we can write ∠M₁N₁K₁= 90°, which means that M₁N₁ ⊥ a₁.

So, we can conclude that the distance between a and b is equal to the distance between M₁ and a₁.

Example 65: Given an equilateral ΔABC with a side of m units. If ΔA₁B₁C₁ is the corresponding projection of ΔABC on a parallel plane, find the distance between AA₁ and B₁C.

Solution: We just need to find a plane perpendicular to one of skew lines AA₁ and B₁C.

Here, AA₁ is perpendicular to ΔA₁B₁C₁.

We find the projections of AA₁ and B₁C on (A₁B₁C₁).

A₁ = Proj (AA₁)

B₁C₁ = Proj (B₁C)

The answer of our question will be the distance from A₁ to B₁C₁ which is the height to B₁C₁ in ΔA₁B₁C₁.

ΔABC is equilateral. So its projection on a parallel plane will be a congruent triangle with a side of m units.

Thus, A₁H = . So, the distance between AA₁ and B1C is units.

Example 66: Given parallelogram ABCD and point K not lying in the plane of ABCD. If DK ⊥ (ABCD), AB = 6 cm, AD = 8 cm, DK = 3 cm, and ∠BAD = 30°,

a) find the distance between DK and AB.

b) find the distance between BK and CD.

Solution:

a) DK ⊥ (ABCD), so we can take ABCD as projection plane.

Proj_ABC DK = D and Proj_ABC AB = AB.

Therefore, distance between DK and AB will be distance between D and AB, i.e. height DH of parallelogram ABCD.

m(∠BAD) = 30°, so DH = = 4 cm.

Hence, the distance betweem DK and AB is DH = 4 cm.

b) DH ⊥ DC and DK ⊥ DC, so DC ⊥ (DKH).

Proj_(DKH) DC = D.

By the three perpendiculars theorem,

BH ⊥ (DKH), so Proj_(DKH) KB = KH.

Now we need to find the distance from D to KH.

Since ΔKDH is a right triangle with sides 3, 4, 5, this distance is

Example 67: Given a square ABCD and O intersection point of its diagonals. Line segment MO is perpendicular to the plane of the square. MO = cm and one side of the square is 4 cm. Find the distance

a) between AB and MO.

b) between BD and MC.

Solution: a) (ABC) is a plane perpendicular to MO.

Proj_(ABC) AB = AB and Proj_(ABC) MO = O.

Hence, the distance between MO and AB is the distance from O to AB, that is OK.

b) We know that, in a square the diagonals are perpendicular to each other.

So, BD ⊥ AC.

BD ⊥ AC and BD ⊥ MO, so BD ⊥ (MOC).

Therefore, we can take plane MOC as the projection plane.

Proj_(MOC) BD = O, Proj_(MOC) MC = MC.

So, the distance between BD and MC is the distance between point O and MC.

In ΔMOC, ∠MOC = 90°.

Draw OT ⊥ MC. We need to find OT.

MC² = MO² + OC²; MC² = ()² + ()² = 16; MC = 4 cm.

Finally, MO ⋅ OC = MC ⋅ OT and

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