Handbook of mathematics for engineers and scienteists part 25 ppsx

If the straight lines are given by canonical equations, then the condition that they are parallel can be written as l1 l2 = m1 m2 = n1 Remark.. If parallel lines have a common point i.e.

i.e., if their direction vectors R1 and R2 are collinear If the straight lines are given by canonical equations, then the condition that they are parallel can be written as

l1

l2 =

m1

m2 =

n1

Remark. If parallel lines have a common point (i.e., r1= r2in parametric equations), then they coincide.

Example 2 Let us show that the lines

x– 1

y– 3

1 =

z

x– 3

4 =

y+ 1

2 =

z

4

are parallel to each other.

Indeed, condition (4.6.3.2) is satisfied,

2

4 =

1

2 =

2

4,

and hence the lines are parallel.

4.6.3-3 Conditions for two lines to be perpendicular

Two straight lines given by vector parametric equation r = r1+ tR1and r = r2+ tR2 are perpendicular if

If the lines are given by canonical equations, then the condition that they are perpendicular can be written as

l1l2+ m1m2+ n1n2 =0, (4.6.3.3a) which coincides with formula (4.6.3.3) written in coordinate form

Example 3 Let us show that the lines

x– 1

y– 3

1 =

z

x– 2

1 =

y+ 1

2 =

z

– 2

are perpendicular.

Indeed, condition (4.6.3.3a) is satisfied,

2 ⋅ 1 + 1 ⋅ 2 + 2 ⋅ (– 2 ) = 0 , and hence the lines are perpendicular.

4.6.3-4 Theorem on the arrangement of two lines in space

THEOREM ON THE ARRANGEMENT OF TWO LINES IN SPACE Two lines in space can:

a) be skew;

b) lie in the same plane and not meet each other, i.e., be parallel;

c) meet at a point;

d) coincide

A general characteristic of all four cases is the determinant of the matrix

2– x1 y2– y1 z2– z1

)

whose entries are taken from the canonical equations

x – x1

l1 =

y – y1

m1 =

z – z1

x – x2

l2 =

y – y2

m2 =

z – z2

n2

of the lines

In cases a–d of the theorem, for the matrix (4.6.3.4) we have, respectively:

a) the determinant is nonzero;

b) the last two rows are proportional to each other but are not proportional to the first row; c) the last two rows are not proportional, and the first row is their linear combination; d) all rows are proportional.

4.6.3-5 Angles between planes

Consider two planes given by the general equations

A1x + B1y + C1z + D1=0,

A2x + B2y + C2z + D2=0, or

r⋅N1+ D1=0,

r⋅N2+ D2=0, (4.6.3.5)

where N1 = (A1, B1, C1) and N2 = (A2, B2, C2) are the normals to the planes and r is the

position vector of the point (x, y, z).

N

N

1 2

φ

Figure 4.49 Angles between planes.

The angle between two planes (see Fig 4.49) is defined as any of the two adjacent dihedral angles formed by the planes (if the planes are parallel, then the angle between them is by definition equal to0or π) One of these dihedral angles is equal to the angle ϕ

between the normal vectors N1 = (A1, B1, C1) and N2 = (A2, B2, C2) to the planes, which can be determined by the formula

cos ϕ = A1A2+ B1B2+ C1C2

A2

1+ B12+ C12

A2

2+ B22+ C22

= N1⋅N2

|N1| |N2| . (4.6.3.6)

If the planes are given by vector parametric equations

r = r1+ R1t+ R2s or r = r1+ R1t+ R2s, (4.6.3.7) then the angle between the planes is given by the formula

cos ϕ = (R1×R2)⋅(R1×R2)

|R1×R2| |R

1×R2| . (4.6.3.8)

4.6.3-6 Conditions for two planes to be parallel.

Two planes given by the general equations (4.6.3.5) in coordinate form are parallel if and

only if the following condition for the planes to be parallel is satisfied:

A1

A2 =

B1

B2 =

C1

C2 ≠ D1

in this case, the planes do not coincide For planes given by the general equations (4.6.3.5)

in vector form, the condition becomes

N2= λN1 or N2×N1=0; (4.6.3.10) i.e., the planes are parallel if their normals are parallel

Example 4 Let us show that the planes x – y + z =0 and 2x– 2y+ 2z+ 5 = 0 are parallel.

Since condition (4.6.3.9) is satisfied,

1

2 =

– 1

– 2 = 12,

we see that the planes are parallel to each other.

4.6.3-7 Conditions for planes to coincide

Two planes coincide if they are parallel and have a common point

Two planes given by the general equations (4.6.3.5) coincide if and only if the following

condition for the planes to coincide is satisfied:

A1

A2 =

B1

B2 =

C1

C2 =

D1

Remark Sometimes the case in which the planes coincide is treated as a special case of parallel straight lines and is not distinguished as an exceptional case.

4.6.3-8 Conditions for two planes to be perpendicular

Planes are perpendicular if their normals are perpendicular

Two planes determined by the general equations (4.6.3.5) are perpendicular if and only

if the following condition for the planes to be perpendicular is satisfied:

A1A2+ B1B2+ C1C2=0 or N1⋅N2=0, (4.6.3.12)

where N1= (A1, B1, C1) and N2= (A2, B2, C2) are the normals to the planes

Example 5 Let us show that the planes x – y + z =0and x – y –2z+ 5 = 0 are perpendicular.

Since condition (4.6.3.12) is satisfied,

1 ⋅ 1 + (– 1 ) ⋅ (– 1 ) + 1 ⋅ (– 2 ) = 0 ,

we see that the planes are perpendicular to each other.

4.6.3-9 Intersection of three planes.

The point of intersection of three planes given by the equations

A1x + B1y + C1z + D1=0, A2x + B2y + C2z + D2=0, A3x + B3y + C3z + D3 =0, has the following coordinates:

x0= –1

Δ







D1 B1 C1

D2 B2 C2

D3 B3 C3





, y0= –

1

Δ







A1 D1 C1

A2 D2 C2

A3 D3 C3





, z0= –

1

Δ







A1 B1 D1

A2 B2 D2

A3 B3 D3





, (4.6.3.13)

whereΔ is given by the formula

Δ =





A1 B1 C1

A2 B2 C2

A3 B3 C3





Remark Three planes are concurrent at a single point if Δ ≠ 0 If Δ = 0 and at least one of the second-order minors is nonzero, then all planes are parallel to a single line If all minors are zero, then the planes are concurrent in a single line.

4.6.3-10 Intersection of four planes

If four planes given by the equations

A1x + B1y + C1z + D1=0, A2x + B2y + C2z + D2=0,

A3x + B3y + C3z + D3=0, A4x + B4y + C4z + D4=0,

are concurrent at a single point, then









x1 y1 z1 1

x2 y2 z2 1

x3 y3 z3 1









To find the points of intersection, it suffices to find the point of intersection of any three

of them (see Paragraph 4.6.3-9) The remaining equation follows from the three other equations

4.6.3-11 Angle between straight line and plane

Consider a plane given by the general equation

Ax + By + Cz + D =0, or r⋅N + D =0 (4.6.3.16) and a line given by the canonical equation

x – x1

l = y – y1

m = z – z1

n , or (r – r1)×R =0, (4.6.3.17)

where N = (A, B, C) is the normal to the plane, r and r1are the respective position vectors

of the points (x, y, z) and (x1, y1, z1), and R = (l, m, n) is the direction vector of the line.

The angle between the line and the plane (see Fig 4.50) is defined as the complementary

angle θ of the angle ϕ between the direction vector R of the line and the normal N to the

plane For this angle, one has the formula

A2+ B2+ C2√

l2+ m2+ n2 =

N⋅R

|N| |R| . (4.6.3.18)

N R

φ θ

M0

Figure 4.50 Angle between straight line and plane.

4.6.3-12 Conditions for line and plane to be parallel

A plane given by the general equation (4.6.3.16) and a line given by the canonical equa-tion (4.6.3.17) are parallel if

Al + Bm + Cn =0,

Ax1+ By1+ Cz1+ D≠ 0, or

N⋅R =0,

N⋅r1+ D≠ 0; (4.6.3.19) i.e., a line is parallel to a plane if the direction vector of the line is perpendicular to the normal to the plane Conditions (4.6.3.19) include the condition under which the line is not contained in the plane

4.6.3-13 Condition for line to be entirely contained in plane

A straight line given by the canonical equation (4.6.3.17) is entirely contained in a plane given by the general equation (4.6.3.16) if

Al + Bm + Cn =0,

Ax1+ By1+ Cz1+ D =0, or

N⋅R =0,

N⋅r1+ D =0 (4.6.3.20)

Remark Sometimes the case in which a line is entirely contained in a plane is treated as a special case of parallel straight lines and is not distinguished as an exception.

4.6.3-14 Condition for line and plane to be perpendicular

A line given by the canonical equation (4.6.3.17) and a plane given by the general equa-tion (4.6.3.16) are perpendicular if the line is collinear to the normal to the plane (is a normal itself), i.e., if

A

l = B

m = C

n, or N = λR, or N×R =0 (4.6.3.21)

4.6.3-15 Intersection of line and plane

Consider a plane given by the general equation (4.6.3.16) and a straight line given by the parametric equation

x = x1+ lt, y = y1+ mt, z = z1+ nt, or r = r1+ tR.

The coordinates of the point M0(x0, y0, z0) of intersection of the line with the plane (see Fig 4.50), if the point exists at all, are determined by the formulas

x0 = x1+ lt0, y0= y1+ mt0, z0 = z1+ nt0, or r = r1+ t0R, (4.6.3.22)

where the parameter t0is determined from the relation

t0= –Ax1+ By1+ Cz1+ D

N⋅r1+ D

N⋅R . (4.6.3.23)

Remark To obtain formulas (4.6.3.22) and (4.6.3.23), one should rewrite the equation of the straight line

in parametric form and replace x, y, and z in equation (4.6.3.16) of the plane by their expressions via t From the resulting expression, one finds the parameter t0and then the coordinates x0, y0, and z0 themselves.

Example 6 Let us find the point of intersection of the line x/2= (y –1)/1= (z +1)/2 with the plane

x+ 2y+ 3z– 29 = 0

We use formula (4.6.3.23) to find the value of the parameter t0 :

t0= – 1 ⋅ 0 + 2 ⋅ 1 + 3 ⋅ (– 1 ) – 29

1 ⋅ 2 + 2 ⋅ 1 + 3 ⋅ 2 = –

– 30

10 =3.

Then, according to (4.6.3.22), we finally obtain the coordinates of the point of intersection in the form

x0= 0 – 2 ⋅ 3 = 6 , y0= 1 – 1 ⋅ 3 = 4 , z0= – 1 – 2 ⋅ 3 = 5

4.6.3-16 Distance from point to plane

The deviation of a point from a plane is defined as the number δ equal to the length of the

perpendicular drawn from this point to the plane and taken with sign + if the point and the origin lie on opposite sides of the plane and with sign – if they lie on the same side of the plane Obviously, the deviation is zero for the points lying on the plane

To obtain the deviation of a point M1(x1, y1, z1) from a given plane, one should

re-place the current Cartesian coordinates (x, y, z) on the left-hand side in the normal equa-tion (4.6.1.7) of this plane by the coordinates of the point M1:

δ = x1cos α + y1cos β + z1cos γ – p = r1⋅N0– p, (4.6.3.24)

where N0 = (cos α, cos β, cos γ) is a unit vector and r1 is the position vector of the point

M1(x1, y1, z1) If the plane is given by the parametric equation (4.6.1.5), then the deviation

of the point M1from the plane is equal to

δ = [(r1– r0)R1R2]

|R1×R2| . (4.6.3.25)

The distance from a point to a plane is defined as the nonnegative number d equal to

the absolute value of the deviation; i.e.,

d=|δ|=|x0cos α + y0cos β + z0cos γ – p| (4.6.3.26) Let us write out some more representations of the distance for the cases in which the plane

is given by the general equation (4.6.1.3) and the parametric equation (4.6.1.5):

d= |Ax0+ By0+ Cz0+ D|

√

A2+ B2+ C2 =

[(r1– r0)R1R2]

|R1×R2| . (4.6.3.27)

Example 7 Let us find the distance from the point M (5 , 1 , – 1) to the plane x –2y– 2z+ 1 = 0 Using formula (4.6.3.27), we obtain the desired distance

d= |1 ⋅ 5 + (– 2 ) ⋅ 1 + (– 2 ) ⋅ (– 1 ) + 1|

1 2 + (– 2 ) 2 + (– 2 ) 2 = 6

3 =2.

4.6.3-17 Distance between two parallel planes.

We consider two parallel planes given by the general equations Ax + By + Cz + D1=0and

Ax + By + Cz + D2 =0 The distance between them is

d= |D1– D2|

√

A2+ B2+ C2 . (4.6.3.28)

4.6.3-18 Distance from point to line

The distance from a point M0(x0, y0, z0) to a line given by the canonical equation (4.6.2.3)

is determined by the formula

d= R×(r0– r1)|

|R =

y1 – y0 z1 – z0



z1 – z0 x1 – x0



x1 – x0 y1 – y0



2

√

Note that the last formulas are significantly simplified if R is the unit vector (l2+m2+n2=1)

Remark. The numerator of the fraction (4.6.3.29) is the area of the triangle spanned by the vectors r0– r1

and R, while the denominator of this fraction is the length of the base of the triangle Hence the fraction itself

is the altitude d of this triangle.

Example 8 Let us find the distance from the point M0 ( 3 , 0 , 4) to the line x/1= (y –1)/2= z/2

We use formula (4.6.3.29) to obtain the desired distance

d=  2 2

1 – 0 0 – 4 2

+  2 1

0 – 4 0 – 3 2

+  1 2

0 – 3 1 – 0 2

√

√

153

3 .

4.6.3-19 Distance between lines

Consider two nonparallel lines (see Fig 4.51) given in the canonical form

x – x1

l1 =

y – y1

m1 =

z – z1

n1 , or (r – r1)×R1=0

and

x – x2

l2 =

y – y2

m2 =

z – z2

n2 , or (r – r2)×R2 =0

R

R

1

2

M

M

1 2

Figure 4.51 Distance between lines.

in parametric form and replace x, y, and z in equation (4.6.3.16) of the... opposite sides of the plane and with sign – if they lie on the same side of the plane Obviously, the deviation is zero for the points lying on the plane

To obtain the deviation of a point... the plane, r and r1are the respective position vectors

of the points (x, y, z) and (x1, y1, z1), and R = (l,