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Parallelograms, ratios and reflections

Find missing parallelogram vertices, divide a segment in a specified ratio, locate external points and reflect a point in a line using its perpendicular foot.

Before you startPosition vectors, vector arithmetic, collinearity and completing the square.

01 / Write a position as a start plus a fraction of a displacement

p = a + t(b − a) locates points on the line through distinct A and B.

p = (1 − t)a + tb

The coefficients add to one. The same value of t applies to all three coordinates.

At t = 0, P is A; at t = 1, P is B. Between those values, P lies on the segment. Negative t or t greater than 1 extends the same line beyond an endpoint.

Locate a point along ABExplore
An affine point on a lineChange t in p = a + t(b − a). The point lies between A and B when t is between zero and one, and beyond them otherwise.ABP

A = (−1, 2, 0); B = (3, −2, 4); AB = (4, −4, 4).

P = a + t(b − a) = (1, 0, 2).

0 < t < 1: P lies between A and B; AP : PB = 1 : 1.

The blue point uses the same fraction along each coordinate. At t = 0 or 1 it coincides with an endpoint. Outside this interval, directed displacement and ordinary positive length ratios must be distinguished.

02 / Name parallelogram vertices in boundary order

For ABCD, AB and DC have the same direction.

A = (1, 0, 2), B = (4, 1, 0), D = (0, 3, 1). Find C so that ABCD is a parallelogram.Worked example

AB = b − a = (3, 1, −2)

Walk from A to B.

DC = AB, so c = d + b − a

Translate the same displacement from D.

C = (3, 4, −1)

Add componentwise.

BC = (−1, 3, −1) = AD

The other pair agrees too.

01 · Directed opposite sides

For parallelogram ABCD, is AB equal to CD or DC?

Hint

Follow the arrows around the boundary.

Worked solution

AB = DC = −CD.

02 · Missing vertex

A = (0, 1, 2), B = (2, 0, 3), D = (−1, 4, 0). Find C in parallelogram ABCD.

Hint

Use c = b + d − a.

Worked solution

C = (1, 3, 1).

03 / Three vertices alone allow three parallelograms

You need to know which given point is opposite the missing point.

Given positions a, b and c, list the possible fourth positions.Worked example

a + b − c

C is opposite the new point.

a + c − b

B is opposite the new point.

b + c − a

A is opposite the new point.

03 · Why order matters

Why is “find the fourth vertex from A, B, C” incomplete unless an order or opposite pair is given?

Hint

Any of the three points could be between the other two along the boundary.

Worked solution

There are generally three possible parallelograms. The intended boundary order selects one.

04 / Parallelogram diagonals share a midpoint

a + c = b + d is the position-vector test.

(a + c)/2 = (b + d)/2

The diagonals AC and BD bisect one another.

Check the first parallelogram.Worked example

a + c = (4, 4, 1)

Add A and C.

b + d = (4, 4, 1)

Add B and D.

Both diagonal midpoints are (2, 2, 1/2)

The independent midpoint check agrees.

04 · Recover a vertex from diagonals

A = (1, 2, 3), C = (5, 0, 1), B = (2, −1, 4). Find D in parallelogram ABCD.

Hint

d = a + c − b.

Worked solution

D = (4, 3, 0).

05 / Convert a ratio into a fraction of the whole segment

If AP : PB = m : n internally, t = m/(m + n).

p = a + [m/(m + n)](b − a) = (na + mb)/(m + n)

The coefficient of b uses the distance ratio from A.

A = (−1, 2, 0), B = (3, −2, 4), and AP : PB = 1 : 3.Worked example

AP is 1/(1 + 3) = 1/4 of AB

The denominator is the whole ratio.

p = a + (1/4)(4, −4, 4)

Move one quarter of the displacement.

P = (0, 1, 1)

Check AP = (1, −1, 1) and PB = (3, −3, 3).

Watch: use the same fraction in all coordinates

Pause, replay or seek freely. The notes explain the same idea and stay in view.

05 · Reverse the ratio

For those A and B, find P when AP : PB = 3 : 1 internally.

Hint

Use t = 3/4.

Worked solution

P = (2, −1, 3).

06 · Unequal parts

A = (2, 0, −1), B = (7, 5, 4). Find P dividing AB internally in ratio AP : PB = 2 : 3.

Hint

Use t = 2/5.

Worked solution

P = (4, 2, 1).

06 / Verify collinearity and the ratio together

All three displacement components must share the same scale.

Is P = (0, 1, 2) on the line through A = (−1, 2, 0) and B = (3, −2, 4)?Worked example

AP = (1, −1, 2)

Compare with AB = (4, −4, 4).

The first two components suggest t = 1/4

But the third would then be 1.

P is not on the line

Matching only two coordinates is insufficient.

07 · Recover t

For the model A and B, P = (2, −1, 3). Find t and the internal ratio.

Hint

AP = (3, −3, 3) = tAB.

Worked solution

t = 3/4 and AP : PB = 3 : 1.

08 · A zero component

A = (1, 0, 2), B = (1, 4, 6). What must the x coordinate of every point on AB be?

Hint

The x displacement is zero.

Worked solution

x = 1. Use another nonzero component to recover t.

07 / An external point uses t outside the interval from zero to one

State the order of points and distinguish lengths from directed vectors.

For the model A and B, let t = 2.Worked example

P = a + 2(b − a) = (7, −6, 8)

Continue beyond B.

AP = 2AB and BP = AB

These directed vectors have the same orientation.

AP : PB = 2 : 1 as positive lengths

But the vector PB = −AB, so vector signs differ.

09 · Before A

Find P for t = −1 using A = (−1, 2, 0), B = (3, −2, 4).

Hint

Move one whole AB in the opposite direction.

Worked solution

P = (−5, 6, −4). The order is P, A, B and the positive lengths AP : PB are 1 : 2.

10 · Why the internal formula fails

If P is beyond B and AP : PB = 2 : 1, why is t not 2/3?

Hint

For an external point, AB is the difference AP − PB.

Worked solution

t = AP/AB = 2/(2 − 1) = 2. The sum denominator applies to internal division.

08 / Reflection in a line uses a perpendicular midpoint

Find the foot H on the line, then use p′ = 2h − p.

H = (P + P′)/2; HP′ = −HP

H must lie on the reflecting line and HP must be perpendicular to it.

Reflect P = (3, 1, 2) in the line through O with direction (1, 2, 2), given that H = (1, 2, 2) is the perpendicular foot.Worked example

p′ = 2h − p

H is the midpoint of the original and reflected points.

P′ = (−1, 3, 2)

Calculate all coordinates.

HP = (2, −1, 0); HP′ = (−2, 1, 0)

The two perpendicular offsets are opposites.

11 · Reflection in the x-axis

Reflect (2, −3, 4) in the x-axis.

Hint

The perpendicular foot is (2, 0, 0).

Worked solution

The image is (2, 3, −4). Both components perpendicular to the axis change sign.

12 · A point on the line

What happens when the original point already lies on the reflecting line?

Hint

Its foot is the point itself.

Worked solution

It is unchanged: 2p − p = p.

10 / Reflection in a line differs from reflection in a plane

Name the geometric object precisely.

Compare images of P = (2, −3, 4).Worked example

In the xy-plane: (2, −3, −4)

Only the perpendicular z coordinate reverses.

In the x-axis: (2, 3, −4)

Both y and z reverse.

In the origin: (−2, 3, −4)

All three coordinates reverse.

15 · Midpoint alone is insufficient

If H is any point on a line, does 2h − p always reflect P in that line?

Hint

Which extra condition is required?

Worked solution

No. H must be the perpendicular foot. Otherwise the construction gives the point reflection in H, not the required line reflection.

11 / Use the geometry to choose the vector equation

A correct-looking weighted average still needs the correct ratio and order.

  • Name parallelogram vertices in boundary order.
  • Check diagonals using a + c = b + d.
  • For internal ratios, divide by the sum of the parts.
  • For external points, use the directed line parameter carefully.
  • Check all three components for collinearity.
  • For a line reflection, find the perpendicular foot first.
  • Distinguish reflection in a line, plane and point.

16 · Final ratio

A = (0, 2, 1), B = (6, −4, 7). Find the point one third of the way from B towards A.

Hint

Start from B and use (a − b)/3.

Worked solution

P = (4, −2, 5). It is two thirds of the way from A towards B.

Section 1 of 11 · Write a position as a start plus a fraction of a displacement