I. The point is on the line.

Let d be a line and A be a point on d.

Let us take two distinct planes α and β containing d.

(Figure 1.38)

Let m be the line lying in α and perpendicular to d at A and n be the line lying in β and perpendicular to d at A.

Then m and n determine a unique plane λ perpendicular to d.

I I. The point is not on the line.

Let d be a line and A be a point in space not on d.

Line d and point A determine a plane α (Figure 1.39).

In α there can be drawn a line m passing through A and perpendicular to d at a point B.

Let β be another plane containing line d.

In β, through point B, let us draw a line n perpendicular to d.

Since m and n are intersecting lines, they determine a plane λ.

Since d is perpendicular to two intersecting lines in λ (those are m and n), d is perpendicular to λ.

Now let us prove that this plane is unique.

Let d be a line and A be a point not on d.

Assume that through A there can be drawn more than one plane perpendicular to d (Figure1.40).

Let α and β be two of these planes.

Let B be the intersection of d and α, and C be the intersection of d and β.

Then d ⊥ AB and d ⊥ AC which is impossible because ABC cannot have two right angles.

3. Let m and d be two parallel lines and α be a plane such that m ⊥ α. We need to show that d ⊥ α. (Figure 1.41)

Since m ⊥ α, m is perpendicular to every line in α.

It is proven that if one of two parallel lines is

perpendicular to a third line then the other one is perpendicular to the same line.

So d is perpendicular to every line in α too.

Therefore d ⊥ α.

4 . Let lines d and b be perpendicular to a plane α (Figure 1.42).

On b, take a point B which is not in α and through B, draw a line b₁ parallel to d.

If b and b₁ are not coincident they determine a

plane β.

Let m be the intersection of β and α.

Since b₁ // d and d ⊥ α, b₁ ⊥ α. So b₁ ⊥ m.

Since b is perpendicular to α, b is also perpendicular to m.

However through B there can be drawn only one line perpendicular to α.

Hence b₁ and b are coincident.

That means lines d and b are parallel.

5. Let α be a plane and A be a point.

In α take a line m (if A is in α take m in such a way that A is not on m) (Figure 1.44).

Through A there can be drawn a plane β perpendicular to m.

α and β are intersecting planes because m intersects β at a point B and m is in α.

So α and β have a common line b.

In β through A there can be drawn line c perpendicular to b.

Then since c ⊥ b and c ⊥ m, it can be concluded that c ⊥ α.

Now let us prove that this line is unique.

Let α be a plane and A be a point not in α.

Assume that through A there can be drawn more than one line perpendicular to α.

Let c and c' be two of these lines.

Since c and c' perpendicular to α, they intersect α.

Let B and C be the intersection points. B and C determine a line n. (Figure 1.45)

Since c ⊥ α, c ⊥ n and since c' ⊥ α, c' ⊥ n. Then ABC is a triangle with two right angles which is impossible.

So through A we can draw only one line perpendicular to α.

6 . Let α and β be two parallel planes and m be a line perpendicular to α. (Figure 1.46)

Since m is perpendicular to α, it intersects α at a point A.

If a line intersects one of two parallel

planes, it intersects the other too. So m intersects β at a point B.

In α, let us take two lines d and b intersecting at A.

Let d' be the intersection of β and the plane determined by d and m, and let

b' be the intersection of β and the plane determined by m and b.

When two parallel planes are intersected by a plane the intersections are parallel. d' // d and b' // b.

Since m ⊥ α, m ⊥ d and m ⊥ b. So m ⊥ d' and m ⊥ b'.

Since m is perpendicular to two intersecting lines in β, m is perpendicular to β.

Theorem: (Three Perpendiculars Theorem)

A line drawn in a plane through the foot of an inclined line is perpendicular to the projection of the inclined line if and only if it is perpendicular to the inclined line itself.

Proof:

In this theorem we need to prove that

1. If a line drawn in a plane through the foot of an inclined line is perpendicular to the projection of the inclined line, it is perpendicular to the inclined line.

2. If a line drawn in a plane through the foot of an inclined line is perpendicular to the inclined line, it is perpendicular to the projection of the inclined line.

1. Let m be the line lying in a plane α perpendicular to the projection of an inclined line d onto α at point B which is the intersection

of d and α.

Let A be any point on d and let the perpendicular

drawn through A intersect α at C (Figure 1.43).

Point C is on the projection of d.

CB and d determine a plane β. Since AC ⊥ α,

AC ⊥ m.

Given that m ⊥ CB, so m ⊥ β and m ⊥ d.

2. Referring to the same figure, we need to prove that if m ⊥ d then m ⊥ CB. Since m is perpendicular to two intersecting lines in β which are d and AC, it is perpendicular to β.

So it is perpendicular to CB.

Example 27: ABC is an isosceles right triangle such that AB = BC = 4 cm. P is a point not in plane (ABC) and M is the midpoint of AC. If PB ⊥ AB, PB ⊥ BC and PB = cm, find PM.

S olution:

AC is the hypotenuse of ABC, AC = cm.

Since BM is the median of the hypotenuse, BM = cm.

Since PB ⊥ BC and PB ⊥ AB , PB ⊥ (ABC).

So PB ⊥ BM.

Then PM² = PB² + BM² = 8 + 8 = 16, PM = 4 cm.

Example 28: Show that if two planes α and β are perpendicular to a line m

then they are parallel.

Solution:

Since m is perpendicular to α and β, it intersects both planes.

Let A and B be the intersections.

Assume that α and β are not parallel. Then they will have a common point P.

Since AP is in α, m ⊥ AP and since BP is in β, m ⊥ BP.

So APM is a triangle with two right angles which is impossible.

Therefore α and β can not have any common point which means α // β.

Example 29: In the adjacent figure, ABC is in the plane α, PC ⊥ α, and AC ⊥ CB.

If ∠CAB = 30° and ∠PBC = 45°, find the

cosine of angle ∠APC.

Solution:

Since PC ⊥ α, PC ⊥ CA and PC ⊥ CB.

If AB = a then CB = and CA = .

Since ∠PBC = 45°, CP = CB = .

So ∠APC = 60° and cos ∠APC = .

Example 30: Line m is inclined to a plane α. Show that through m, there can

be drawn one and only one plane perpendicular to α.

Solution:

L et m intersect α at A, and B be another point

on m.

There can be drawn a line d passing through B and perpendicular to α.

Then since d and m are intersecting lines they determine a plane β.

In β there is a line (line d) perpendicular to α.

Therefore, β  α.

Now, let us prove that this plane is unique.

Assume that there are planes β₁ and β₂ containing m and perpendicular to α.

Then through a point on m there can be

drawn line segments AB and BC perpendicular to

α (A and C are in α).

Then BA ⊥ AC and BC ⊥ AC which is impossible.

Therefore, there can not be drawn more than one plane perpendicular to α through m.

Example 31: In figure, PC ⊥ (ABC) ,

∠ACB = 90° ,

PM ⊥ AB, and

AM = MB.

Find m(∠CAB).

Solution:

and .

Then, since AM = MB, PA = PB.

Since PC ⊥ (ABC), PC ⊥ AC and PC ⊥ CB .

Then, and .

Since PA = PB, CA = CB.

So ABC is an isosceles right triangle and m(∠CAB) = 45° .

E xample 32: In the adjacent figure, PB ⊥ (ABC), ABC is an equilateral triangle, and M and N are midpoints of AB and BC respectively.

What is the ratio of the area of PCA to the area of PNM if PB = AB ?

S olution:

It is given that PB ⊥ (ABC).

So PB ⊥ BC and PB ⊥ BA.

Since AB = BC, PA = PC, and since BN = BM,

PM = PN.

Let L and K be the midpoints of AC and MN respectively.

Since PCA and PMN are isosceles triangles, PL ⊥ AC and PK ⊥ MN.

Hence .

Let us take CA = a.

Then AB = BC = PB = a, MN =

, Since PB ⊥ (ABC), PB ⊥ BL.

So and

.

Then .

Example 33: In the given figure, ABCD is a quadrilateral in a plane α.

P is a point not in α such that PO  α.

If PA = PB = PC = PD, find ∠DAB + ∠BCD.

S olution:

Since PO ⊥ α, PO is perpendicular to all OA, OB, OC and OD.

Then since PA = PB = PC = PD and

∠POA = ∠POB = ∠POC = ∠POD = 90° ,

OA = OB = OC = OD.

So, O is the center of the circumscribed

circle of ABCD i.e, ABCD is a cyclic quadrilateral.

Therefore ∠DAB + ∠BCD = 180°.

E xample 34: In figure, A, B, C are on d, A₁, B₁, C₁ are in α such that AA₁ ⊥ α, BB₁ ⊥ α and CC₁ ⊥ α .

Show that A₁, B₁, C₁ are collinear.

Solution: Since AA₁, BB₁ and CC₁ are perpendicular to α, they are parallel

lines.

Since AA₁ // BB₁, they determine a plane β.

Since A and B are in β, d is in β. So C is in β.

Since BB₁ // CC₁, CC₁ is in the plane containing BB1 and C.

So CC₁ is also in β.

Since A₁, B₁, C₁ are intersections of α and β, they are collinear.

Check Yourself 9

1. In the adjacent figure, PA ⊥ α ,

∠BPA = 20°,

∠CPA = 10°, and

∠APD = 30° .

Write PB, PC, PD in ascending order.

2. In the adjacent figure, M is the midpoint of CB and

PM ⊥ (ABC).

If MC = CA and PA = PB = PC,

find ∠ABC.

3. Lines a, b, c are coplanar and line d is not in the plane containing a, b, c. It is given that d is perpendicular to a and b but it is not perpendicular to c.

What can be concluded about a and b?

4. ABC is right angled at A, lies in a plane α and PA ⊥ α where P is a point not in α. If A is equidistant from points B, C and P, find ∠BPC .