Iii. Projection of a Figure on a Plane

Projection of a figure on a plane is the set of all points which are the projections of all points of the figure on the plane.

Some figures may have projections different from themselves. For example, projection of a circle may be an ellipse, also it may be a line segment.

Theorems:

1. The projection of a line, not perpendicular to a plane, is also a line.

2. The projections of parallel lines not perpendicular to a plane are also parallel.

3. The projections of parallel and equal line segments on a plane are also parallel and equal.

4. The projection of a right angle on a plane, whose one arm is parallel and the other arm is not perpendicular to that plane, is also a right angle.

Proofs:

1. Given a line l and a plane α such that l ⊥ α (Figure 2.7).

All perpendiculars drawn from line l to plane α are parallel to each other.

Let β be the plane determined by l and one of these perpendiculars.

Then all perpendiculars drawn from l to α will be in β, so the feet of all these perpendiculars will lie on the intersection line of planes α and β which is l '.

Therefore, Proj_α l = l '.

2. Given lines d₁, d₂ and plane α such that d₁ // d₂,

d₁ ⊥ α and d₂ ⊥ α (Figure 2.9).

Let d₁' = Proj_α d₁ and d₂' = Proj_α d₂.

Take A ∈ d₁ and B ∈ d₂.

If A' = Proj_α A and B' = Proj_α B then AA' ⊥ α and BB' ⊥ α.

AA' // BB' because, two lines perpendicular to the same plane are parallel.

Since AA' // BB' and d₁ // d₂ then plane formed by AA' and d₁ is parallel to the plane formed by BB' and d₂.

Hence their intersection lines d₁' and d₂' with plane α are also parallel to each other.

3. We proved that the projections of two parallel lines on a given plane are parallel.

So if two line segments are parallel, their projections on a given plane are parallel.

Now we will prove that the length of projections are equal.

Take two line segments AB and CD such that

AB // CD and AB = CD.

Let A'B' = Proj_α AB and C'D' = Proj_α CD.

The quadrilateral ABCD is a parallelogram because AB // CD and AB = CD.

In a parallelogram opposite sides are parallel and equal. Therefore AD // BC and AD = BC. (I)

AA' and BB', are perpendicular to α. So AA' // BB' (II).

From (I) and (II), (AA'D'D) // (BB'C'C). So A'D' // B'C'.

We know that A'B' // D'C'. Therefore quadrilateral A'B'C'D' will be a parallelogram.

Finally, we get A'B' = C'D'.

4. Given an angle BAC where ∠BAC = 90° and AB // α.

Let ∠B'A'C' = Proj_α ∠BAC (Figure 2.11).

We need to prove that ∠B'A'C' = 90°.

Since AB // α, AB // A'B'.

Since AB // A'B' and AB ⊥ AC, AC ⊥ A'B'.

So, A'B' is perpendicular to the plane formed by AA' and AC.

Therefore, A'B' ⊥ A'C' which means ∠B'A'C' = 90°.

Example 48: Given a triangle ABC and its orthogonal projection A₁B₁C₁. The distances between corresponding vertices of these triangles are a, b and c. Show that the distance between their centroids is .

Solution:

Let AM and A₁M₁ be the medians to CB and C₁B₁ in ΔABC and ΔA₁B₁C₁ respectively.

Let O and O₁ be centroids of ΔABC and ΔA₁B₁C₁, respectively.

CC₁ // BB₁ (we are taking projections so they are

perpendicular to projection plane which means they are parallel to each other).

So CBB₁C₁ will be a trapezoid and MM1 is its middle base.

and MM₁ // CC₁. So M₁M₁ // A₁A which means that AA₁M₁M is a trapezoid.

Draw AK ⊥ MM₁.

Let OO₁ ∩ AK = T.

ΔAMK ∼ ΔAOT so . (O is centroid)

M₁K = O₁T = AA₁ = a.

MK = MM₁ – M₁K = .

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Example 49: Given two intersecting planes α and β where l is their intersection

line. Line b is perpendicular to plane β. Show that projection of line b on plane α is perpendicular to line l.

Solution:

If b // α then the projection of b on α will be parallel to b.

Then projection will be perpendicular to β too.

Since l is a line in β the projection will be perpendicular to l.

If b // α then b intersects α at a point B.

Let A be the intersection of b and β.

Let A₁ be the projection of A on α. Then BA₁ will be the projection of b on α.

Let C be the intersection of BA₁ and l.

Since b ⊥ β, b ⊥ l and since AA1 ⊥ α, AA₁ ⊥ l.

So l ⊥ AA₁, l ⊥ b, that means l ⊥ (BCA).

Therefore, l ⊥ BC.

Example 50: Given an isosceles ΔABC (AB = AC). Through BC passes a plane α. Show that projection of ΔABC on plane α is also an isosceles triangle.

Solution: Let A₁ be the projection of A on α.

AB = AC is given.

We need to prove that A₁B = A₁C. Because the projection of ΔABC on α is ΔA₁BC.

Since AA₁ ⊥ α, AA₁ ⊥ BA₁ and AA₁ ⊥ CA₁.

Then and .

Since AB = AC, A₁B = A₁C.

Example 51:

Side AD of parallelogram ABCD lies in plane α.

AB = 15 cm, BC = 19 cm. The lengths of projections of diagonals of the parallelogram on plane α are 20 cm and 22 cm. Find the distance from side BC to plane α.

(Hint: In a parallelogram e² + f² = 2 (a² + b²) where a, and b are the sides of parallelogram and e, f are the diagonals of it.)

Solution:

Since CC₁ ⊥ α then CC₁ ⊥ AC₁, and since BB₁ ⊥ α,

BB₁ ⊥ B₁D.

Since BC // AD, BC // α and CC₁ = BB₁.

So BD² = DB₁² + BB₁² = DB₁² + CC₁²

AC² = AC₁² + CC₁²

Add both sides of these equations:

BD² + AC² = DB₁² + AC₁² + 2CC₁²

One of DB₁ and AC₁ is 20 cm and the other is 22 cm. So

BD² + AC² = 20² + 22² + 2CC₁²

We know that BD² + AC² = 2(AB² + BC²) = 2 · (15² + 19²).

So 20² + 22² + 2CC₁² = 2 · (15² + 19²).

Then 2CC₁² = 288, CC₁² =144 and CC₁ = 12 cm.

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