Handbook of mathematics for engineers and scienteists part 24 pot

Plane passing through a point and parallel to two straight lines.. Equation of plane passing through point and parallel to two straight lines.. If the straight line passing through point

R R

R

R

M x y z( , , )

M1

1 1

2

2

Figure 4.40 Plane passing through a point and parallel to two straight lines.

4.6.1-9 Equation of plane passing through point and parallel to two straight lines

The equation of the plane passing through a point M1(x1, y1, z1) and parallel to two straight

lines with direction vectors R1= (l1, m1, n1) and R2= (l2, m2, n2) (see Fig 4.40) is







x – x1 y – y1 z – z1

l1 m1 n1

l2 m2 n2





=0, or



(r – r1)R1R2

where r and r1are the position vectors of the points M (x, y, z) and M1(x1, y1, z1), respec-tively

Example 7 Let us find the equation of the plane passing through the point M1 ( 0 , 1 , 0 ) and parallel to the

straight lines with direction vectors R1 = ( 1 , 0 , 1) and R2 = ( 0 , 1 , 2 ).

According to (4.6.1.13), the desired equation is







x– 0 y– 1 z– 0





=0, whence

–x –2y+ z +2 = 0

4.6.1-10 Plane passing through two points and perpendicular to given plane

The plane (see Fig 4.41) passing through two points M1(x1, y1, z1) and M2(x2, y2, z2) and

perpendicular to the plane given by the equation Ax + By + Cz + D =0is determined by the equation







x – x1 y – y1 z – z1

x2– x1 y2– y1 z2– z1





=0, or



(r – r1)(r2– r1)N

where r, r1, and r2 are the position vectors of the points M (x, y, z), M1(x1, y1, z1), and

M2(x2, y2, z2), respectively.

Remark. If the straight line passing through points M1(x1, y1, z1) and M2(x2, y2, z2) is perpendicular to the original plane, then the desired plane is undetermined and equations (4.6.1.14) become identities.

N

M M

M x y z( , , )

2

1

Figure 4.41 Plane passing through two points and perpendicular to given plane.

Example 8 Let us find an equation of the plane passing through the points M1 ( 0 , 1 , 2) and M2 ( 2 , 2 , 3 )

and perpendicular to the plane x – y + z +5 = 0

According to (4.6.1.14), the desired equation is







x– 0 y– 1 z– 2





=0, whence

2x– y –3z + 7 = 0

4.6.1-11 Plane passing through point and perpendicular to two planes

The plane (see Fig 4.42) passing through a point M1(x1, y1, z1) and perpendicular to two

(nonparallel) planes A1x + B1y + C1z + D1=0and A2x + B2y + C2z + D2=0is given by the equation







x – x1 y – y1 z – z1

A1 B1 C1

A2 B2 C2





=0, or



(r – r1)N1N2

where N1 = (A1, B1, C1) and N2 = (A2, B2, C2) are the normals to the given planes and r and r1are the position vectors of the points M (x, y, z) and M1(x1, y1, z1), respectively

Figure 4.42 Plane passing through a point and perpendicular to two planes.

Remark 1. Equations (4.6.1.15) mean that the vectors −−−→ M1M, N1, and N2are coplanar.

Remark 2 If the original planes are parallel, then the desired plane is undetermined In this case, equations (4.6.1.15) become identities.

Example 9 Let us find an equation of the plane passing through the point M1 ( 0 , 1 , 2 ) and perpendicular

to the planes x – y + z +5 = 0and –x + y + z –1 = 0

According to (4.6.1.15), the desired equation is







x– 0 y– 1 z– 2





=0, whence

x + y –1 = 0

4.6.1-12 Equation of plane passing through line of intersection of planes

The planes passing through the line of intersection of the planes A1x + B1y + C1z + D1=0

and A2x + B2y + C2z + D2 =0are given by the equation

α (A1x + B1y + C1z + D1) + β(A2x + B2y + C2z + D2) =0, (4.6.1.16)

which is called the equation of a pencil of planes Here α and β are arbitrary parameters Let α≠ 0 Set β/α = λ; then equation (4.6.1.16) becomes

A1x + B1y + C1z + D1+ λ(A2x + B2y + C2z + D2) =0 (4.6.1.17)

By varying the parameter λ from – ∞ to +∞, we obtain all the planes in the pencil For

λ= 1, we obtain equations of the planes that bisect the angles between the given planes provided that the equations of the latter are given in normalized form

Remark. The passage from equation (4.6.1.16) to equation (4.6.1.17) excludes the case α =0

Equa-tion (4.6.1.17) does not define the plane A2x + B2y + C2z + D2= 0; i.e., equation (4.6.1.17) for various λ

defines all the planes in the pencil but one (the second of the two given planes).

4.6.2 Line in Space

4.6.2-1 Parametric equation of straight line

The parametric equation of the line that passes through a point M1(x1, y1, z1) and is parallel

to a direction vector R = (l, m, n) (see Fig 4.43) is

where r = −−→ OMand r1= −−−→ M1M As the parameter t varies from – ∞ to +∞, the point M with

position vector r = (x, y, z) determined by formula (4.6.2.1) runs over the entire straight

line in question It is convenient to use the parametric equation (4.6.2.1) if one needs to find the point of intersection of a straight line with a plane

The numbers l, m, and n characterize the direction of the straight line in space; they are

called the direction coefficients of the straight line For a unit vector R = R0, the coefficients

l , m, n are the cosines of the angles α, β, and γ formed by this straight line (the direction of

the vector R0) with the coordinate axes OX, OY , and OZ These cosines can be expressed

via the coordinates of the direction vector R as

l2+ m2+ n2 , cos β =

m

√

l2+ m2+ n2 , cos γ =

n

√

l2+ m2+ n2 (4.6.2.2)

Example 1 Let us find the equation of the straight line that passes through the point M1 ( 2 , – 3 , 1 ) and is

parallel to the direction vector R = (1 , 2 , – 3 ).

According to (4.6.2.1), the desired equation is

x= 2+ t, y= – 3 + 2t , z= 1 – 3t

tR

R

r

1

1

M M x y z( , , )

X

Z

Figure 4.43 Straight line passing through a point and parallel to direction vector.

4.6.2-2 Canonical equation of straight line

The equation

x – x1

l = y – y1

m = z – z1

is called the canonical equation of the straight line passing through the point M1(x1, y1, z1)

with position vector r1= (x1, y1, z1) and parallel to the direction vector R = (l, m, n).

Remark 1 One can obtain the canonical equation (4.6.2.3) from the parametric equations (4.6.2.1) by

eliminating the parameter t.

Remark 2. In the canonical equation, all coefficients l, m, and n cannot be zero simultaneously, since

that the corresponding numerator is also zero.

Example 2 The equations (x –1)/1= (y –3)/4= (z –3)/0 determine the straight line passing through

the point M1 ( 1 , 3 , 3) and perpendicular to the axis OZ This means that the line lies in the plane z =3 , and

hence z –3 = 0 for all points of the line.

Example 3 Let us find the equation of the straight line passing through the point M1( 2 , – 3 , 1 ) and parallel

to the direction vector R = (1 , 2 , – 3 ).

According to (4.6.2.3), the desired equation is

x– 2

y+ 3

z– 1

– 3 .

4.6.2-3 General equation of straight line

The general equation of a straight line in space defines it as the line of intersection of two

planes (see Fig 4.44) and is given analytically by a system of two linear equations

A1x + B1y + C1z + D1=0,

A2x + B2y + C2z + D2=0, or

r⋅N1+ D1=0,

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).

The direction vector R is equal to the cross product of the normals N1and N2; i.e.,

Figure 4.44 Straight line as intersection of two planes.

and its coordinates l, m, and n can be obtained by the formulas

l=B1 C1

B2 C2



 , m =C1 A1

C2 A2



 , n =A1 B1

A2 B2



Remark 1 Simultaneous equations of the form (4.6.2.4) define a straight line if and only if the coefficients

A1, B1, and C1in one of them are not proportional to the respective coefficients A2, B2, and C2 in the other.

Remark 2. For D1= D2 = 0 (and only in this case), the line passes through the origin.

Example 4 Let us reduce the equation of the straight line

x+ 2y– z +1 = 0 , x – y + z +3 = 0

to canonical form.

We choose one of the coordinates arbitrarily; say, x =0 Then

2y– z +1 = 0 , –y + z +3 = 0 ,

and hence y = –4, z = –7 Thus the desired line contains the point M (0 , – 4 , – 7 ) We find the cross product of the

vectors N1= ( 1 , 2 , – 1) and N2= ( 1 , – 1 , 1) and, according to (4.6.2.5), obtain the direction vector R = (1 , – 2 , – 3 )

of the desired line Therefore, with (4.6.2.3) taken into account, the equation of the line becomes

x

y+ 4

– 2 =

z+ 7

– 3 .

4.6.2-4 Equation of line in projections

The equation of a line in projections can be obtained by eliminating first z and then y from

the general equations (4.6.2.4):

Each of two equations (4.6.2.7) defines a plane projecting the straight line onto the planes

OXY and OXZ (see Fig 4.45).

Remark 1. For straight lines parallel to the plane OYZ, this form of the equations cannot be used; one

should take the projections onto some other pair of coordinate planes.

Remark 2 Equations (4.6.2.7) can be represented in the canonical form

x– 0

y – a

k = z – b

O

y z

Figure 4.45 Straight line with equation in projections.

4.6.2-5 Equation of straight line passing through two points

The canonical equation of the straight line (see Fig 4.46) passing through two points

M1(x1, y1, z1) and M2(x2, y2, z2) is

x – x1

x2– x1 =

y – y1

y2– y1 =

z – z1

z2– z1, or (r – r1)×(r2– r1) =0, (4.6.2.9)

where r, r1, and r2 are the position vectors of the points M (x, y, z), M1(x1, y1, z1), and

M2(x2, y2, z2), respectively.

The parametric equations of the straight line passing through two points M1(x1, y1, z1)

and M2(x2, y2, z2) in the rectangular Cartesian coordinate system OXY Z can be written as

x = x1(1– t) + x2t,

y = y1(1– t) + y2t,

z = z1(1– t) + z2t,

or r = (1– t)r1+ tr2 (4.6.2.10)

Remark. Eliminating the parameter t from equations (4.6.2.10), we obtain equations (4.6.2.9).

r

r

r

2 1

2

1

M

X

Z

Figure 4.46 Straight line passing through two

points.

N

M0

Figure 4.47 Straight line passing through point

and perpendicular to plane.

4.6.2-6 Equation of straight line passing through point and perpendicular to plane

The equation of the straight line passing through a point M0(x0, y0, z0) and perpendicular

to the plane given by the equation Ax + By + Cz + D =0, or r⋅N + D =0(see Fig 4.47), is

x – x0

A = y – y0

B = z – z0

where N = (A, B, C) is the normal to the plane.

4.6.3 Mutual Arrangement of Points, Lines, and Planes

4.6.3-1 Angles between lines in space

Consider two straight lines determined by vector parametric equations r = r1+ tR1 and

r = r2+ tR2 The angle ϕ between these lines (see Fig 4.48) can be obtained from the

formulas

cos ϕ = R1⋅R2

|R1| |R2| , sin ϕ =

|R1×R2|

|R1| |R2| .

If the lines are given by the canonical equations

x – x1

l1 =

y – y1

m1 =

z – z1

n1 and

x – x2

l2 =

y – y2

m2 =

z – z2

n2 ,

then the angle ϕ between the lines can be found from the formulas

cos ϕ = l1l2+ m1m2+ n1n2

l2

1+ m21+ n21

l2

2+ m22+ n22

,

sin ϕ =

m1 n1

m2 n2



2+n1 l1

n2 l2



2+l1 m1

l2 m2



2

l2

1+ m21+ n21

l2

2+ m22+ n22

,

(4.6.3.1)

which coincide with formulas (4.6.3.1) written in coordinate form

R

R

2

1

φ

Figure 4.48 Angles between lines in space.

Example 1 Let us find the angle between the lines

x

y– 2

z+ 1

x

y– 2

z+ 1

Using the first formula in (4.6.3.1), we obtain

cos ϕ = √ 1 ⋅ 0 + 3 ⋅ 2 + 4 ⋅ 2

and hence ϕ≈ 0 3672 rad.

4.6.3-2 Conditions for two lines to be parallel

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

R2= λR1 or R2×R1 =0,

,

(4.6.3.1)

which coincide with formulas (4.6.3.1) written in coordinate form

R

R

2

1...

Using the first formula in (4.6.3.1), we obtain

cos ϕ = √ 1 ⋅ + ⋅ + ⋅ 2

and hence ϕ≈...

4.6.3-2 Conditions for two lines to be parallel

Two straight lines given by vector parametric equations r = r1+ tR1and r = r2+