Analytic geometry. Textbook

~

~{ ~| k

~ ~e1 ~e2 = 4 3 1 = 15~{ 10~| + 30k;

2 9 2

~

~{ ~| k

~ ~e1 ~e2 = 4 3 2 = 3~{ + 4~| + 12k;

8 3 3

Answer: a) d = 7; b) d = 13.

150. a) Parametric equations of the straight lines have the form

:

172 Chapter 5. Answers and solutions

Remark. It is essential that the parameters are denoted by di erent letters since the intersection point can be obtained from di erent parameters on each of the lines. The given lines are not parallel since their direction vectors are not collinear. Therefore there are possible only two cases: the straight lines do not intersect (the system of equations is inconsistent), the lines intersect (they have a single common point and the system is de nite).

Solving the system we get that t = 2, = 3, the lines intersect and

the point M(5; 2; 1) and has the direction vector ~e = f4; 3; 2g. The desired perpendicular lies in the plane passing through this line and the origin O.

Therefore we can take ~n = f1; 2; 1g.

The direction vector ~e1 of the desired perpendicular is to be orthogonal

~ ~n ~e1 = 1 2 1 = 7~{ 2~| + 11k;

4 3 2

so we can take ~e1 = f7; 2; 11g, and the perpendicular has the equation x7 = y2 = 11z :

152. Let us denote the desired straight line by `0, its canonical equation has the form

x 4 = y = z + 1:

Let `1 and `2 denote the given straight lines. Applying the condition that two lines lie in the same plane to the pairs `1, `0 and `2, `0 we get:

Solving this system we obtain a nontrivial solution = 13, = 37, = 58.

Applying the condition that this line and each of the given lines lie in the

174 Chapter 5. Answers and solutions

Taking any solution of this indeterminate system we get a point of this line. For instance we can take x0 = 0, y0 = 8, z0 = 9.

Remark. The system written above determines the desired straight line

as the intersection if two planes.

Answer: x = y + 8 = z + 9. 8 7 1

154. The direction vector ~e of the common perpendicular is to be or-

of the given straight lines. Therefore the vectors ~e and ~e1 ~e2 are collinear,

In order to nd the values x0, y0, z0 we use the condition that this line lies in the same planes with each of the given lines. These conditions take the

We can take any solution of this indeterminate system of linear equations, for instance, x0 = 1, y0 = 0, z0 = 3, so the equation of the perpendicular

takes the form

x 1 = y = z + 3: 2 1 4

c)x 112 = y 13+ 5 = z 17 3.

156.The parametric equations of the straight line have the form

x = 4 + t; y = 2 4t; z = 9 + 3t:

We put the right sides of these equations in the equation of the sphere:

(4 + t)2 + ( 2 4t)2 + (9 + 3t)2 = 49:

After simpli cation we get:

159.x 3y + 4z 5 = 0

160.The straight line passing through the given points is de ned by

176 Chapter 5. Answers and solutions

We put the right sides of these equations in the equation of the plane and solve the resulting equation:

6(2 + 4t) 8(2 2t) + 2( 4 + 6t) 4 = 0;

From the parametric equations we nd the coordinates of the intersection

161. Let Ax + By + Cz + D = 0 be the equation of the desired plane.

Equating the distance of this plane to the given value we get

The given straight line passes through the point M0(0; 0; 6) and has the direction vector ~e = f1; 1; 1g. Therefore 6C + D = 0 (the plane passes

through the point M0) and A + B + C = 0 (the normal vector of the plane and the direction vector ~e are orthogonal). Let C = 1. Then D = 6 and we get the relations A + B = 1, A2 + B2 = 5. Solving this system of equations we get two solutions A = 1, B = 2 and A = 2, B = 1.

Squaring the both sides and simplifying the obtained relations we get the equation

500 2 + 428 + 29 2 = 0:

If = 0 then 500 2 = 0, = 0, which is impossible. Dividing the both sides of the equation under consideration by 2 we get the quadratic equation with

the unknown :

500 2 + 428 + 29 = 0:

Solving this equation we get that = 12 or = 25089 . In the rst case we take = 1, = 2, in the second case we take = 29, = 250.

Substituting these values in the equation of the bundle and simplifying the obtained equations we get the answer.

Answer: 3x 4y 5 = 0, 387x 164y 24z 421 = 0.

163. We write out parametric equations of the given lines, put the right sides of this equations in the equation of the plane and solve the resulting equation with the unknown parameter.

a) x = 12 + 4t, y = 9 + 3t, z = 1 + t,

3(12 + 4t) + 5(9 + 3t) (9 + 3t) 2 = 0;

after simpli cation we get the equation 26t + 78 = 0, t = 3. From the parametric equations we obtain the coordinates of the desired point: x = 0, y = 0, z = 2.

b) x = 1 + 2t, y = 3 + 4t, z = 3t,

3( 1 + 2t) 3(3 + 4t) + 2 3t 5 = 0;

after simpli cation we get the equation 0 t 17 = 0. It has no solutions. It means that the straight line is parallel to the plane and does not lie on the plane.

c) x = 13 + 8t, y = 1 + 2t, z = 4 + 3t,

(13 + 8t) + 2(1 + 2t) 4(4 + 3t) + 1 = 0;

178 Chapter 5. Answers and solutions

after simpli cation we get the equation 0 t = 0, the set of equations equals the entire set of real numbers, the straight line lies on the plane.

d) x = 7 + 5t, y = 4 + t, z = 5 + 4t,

3(7 + 5t) (4 + t) + 2(5 + 4t) 5 = 0,

after simpli cation we get the equation t + 1 = 0, t = 1. From the

parametric equations we obtain the coordinates of the desired point: x = 2, y = 3, z = 1.

Answer: a) (0; 0; 2); b) no intersection points (the line and the plane

are parallel); c) the straight line lies on the plane; d) (2; 3; 1).

164. To nd the intersection point M1 of the rst straight line with the plane we write out parametric equations of this line, put the right sides of these equation in the equation of the plane and solve the arising equations

Using the second approach we nd the intersection point M2(3; 1; 2) of the second straight line with the plane. The direction vector of the desired

! 5 7 1

straight line and the vector M1M2 = 3; 3; 3 are collinear. Wee take into account that the desired straight line passes through the point M2 and choose f5; 7; 1g as the direction vector. The equation of the line takes the form

This line is parallel to the given plane, hence the direction vector f ; ; g of this line and the normal vector f3; 1; 2g of the plane are orthogonal. We obtain the equation 3 + 2 = 0: The given straight line passes through the the point (1; 3; 0) and has the direction vector f4; 2; 1g. This line and the desired line intersect, therefore they lie in the same plane and we obtain the relation

17 28 12 = 0:

Now we have to nd a non-trivial solution of the homogeneous system of

linear equations

(

3 + 2 = 0;

17 28 12 = 0:

Solving this system in the matrix form we get

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169.a) the line lies on the plane; b) the single intersection point (1; 0; 2);

c)no common points.

170.The given line passes through the point M0(4; 3; 0) and has the direction vector ~e = f5; 2; 1g. Let us denote M1(3; 1; 2). The normal

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vector ~n of the desired plane is orthogonal to the vectors M0M1 and ~e.

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Therefore the vectors M0M1 ~e and ~n are collinear,

Equation of the plane passing through the point M0 and having the normal vector ~n = f8; 9; 22g has the form

8(x 4) 9(y + 3) 22z = 0;

or 8x 9y 22z 59 = 0.

171. The desired plane passes through the point M0(2; 3; 1) lying in the given straight line, the normal vector of this plane is orthogonal to the direction vector ~e = f5; 1; 2g of this line and to the normal vector

~n = f1; 4; 3g of the given plane,

~ ~e ~n = 5 1 2 = 11~{ + 17~| + 19k:

1 4 3

Taking the vector f11; 17; 19g as the normal vector of the desired plane we obtain its equation

11(x 2) 17(y 3) 19(z + 1) = 0

or 11x 17y 19z + 10 = 0.

172. The given lines pass respectively through the points M1(3; 1; 2),

M2(8; 1; 6) and have the direction vectors ~e1 = f5; 2; 4g, ~e2 = f3; 1; 2g. These lines intersect if and only if the following two conditions are valid:

a) the lines lie in the same plane, in the vector form this condition can

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be written as the equality M1M2~e1~e2 = 0;

b) the lines are not parallel, it means that the direction vectors ~e1 and ~e2 are not collinear, this property is obviously valid.