Analytic geometry. Textbook

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We check the property 1): M1M2 = f5; 2; 4g,

since there are two equal rows.

The normal vector of the plane is orthogonal to the both vectors ~e1 and ~e2,

~ ~e1 ~e2 = 5 2 4 = 8~{ + 22~| k:

3 1 2

Let us take ~n = f8; 22; 1g, equation of the plane with this normal vector

passing through the point M1 (we could take the point M2) has the form

8(x 7) 22(y + 1) + (z 2) = 0;

or 8x 22y + z 48 = 0.

173.The given straight lines pass respectively through the points

M1(0; 2; 1) and M2(1; 3; 1) and have the common direction vector ~e = f7; 3; 5g. The normal vector of the desired plane is orthogonal to the both

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vectors M1M2 = f1; 5; 3g and ~e,

Let us take the vector f17; 13; 16g as the normal vector of the desired plane. This plane passes through the point M1 (we also can take the point

M2) and has the equation

17x 13(y + 2) 16(z 1) = 0

or 17x 13y 16z 10 = 0.

174. The normal vector of the desired plane is orthogonal to the direction vectors ~e1 = f6; 2; 3g and ~e2 = f5; 4; 2g of the given straight lines,

~ ~e1 ~e2 = 6 2 3 = 16~{ 27~| + 14k;

5 4 2

taking this vector as the normal one we get the equation of the plane

16(x 4) 27(y + 3) + 14(z 1) = 0;

or 16x 27y + 14 159 = 0.

175. Let ~e1 = f2; 1; 3g and ~e2 = f4; 7; 2g be the direction vectors of the given straight lines. The normal vector of the desired plane is orthogonal to the vectors ~e1 and ~e2,

~ ~e1 ~e2 = 2 1 3 = 23~{ 16~| + 10k:

4 7 2

Equation of the plane passing through the point of the rst straight line (3; 4; 2) and having the normal vector f23; 16; 10g has the form

23(x 3) 16(y + 4) + 10(z 2) = 0;

or 23x 16y + 10z 153 = 0.

176. The direction vector f3; 1; 4g of the given straight line and the normal vector f1; 1; 1g are orthogonal, therefore, the given line and plane are parallel. The desired plane passes through the point ( 5; 2; 0) and has the normal vector f1; 1; 1g. Equation of this plane has the form

(x + 5) + (y 2) z = 0

or x + y z + 3 = 0.

177. The desired plane belongs to the bundle de ned by the given plane and the plane xOz which is de ned be the equation y = 0. Therefore this

bundle can be de ned by the equation

(x 3y + 2z 5) + y = 0

or

x + ( 3 )y + 2 z 5 = 0; j j + j j =6 0:

From the orthogonality condition we get the equation

3( 3 ) + 4 = 0

or 14 3 = 0.Taking the values = 13, = 3 we get the answer 3x + 5y + 6z 15 = 0.

178. The normal vector ~n1 of the desired plane is orthogonal to the normal vector ~n = f1; 4; 3g of the given plane and to the direction vector

~e = f5; 1; 2g of the given straight line, therefore the vectors ~n1 and ~n ~e are collinear,

~ ~n ~e = 1 4 3 = 11~{ 17~| 19k:

5 1 2

The desired plane passing through the point (2; 3; 1) of the given straight line and having the normal vector f11; 17; 19g is de ned by the equation

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

We may take M1(2; 3; 1), M2(1; 0; 5) or M1(1; 0; 5), M2(2; 3; 1). The equation takes the form

x = 7 + 5t; y = 4 + t; z = 5 + 4t; t 2 R;

and substitute these expressions in the equation of the plane:

lines lie in one plane using the condition obtained above with the points M1(1; 7; 5) and M2(6; 1; 0):

The lines lie on the same plane and are not parallel. Therefore these lines intersect at a single point.

Parametric equations of the lines have the following form:

88

In order to nd the common point of the lines we solve the following system

We obtain the single solution t = 2, = 3, and x = 3, y = 5,

z = 3, the straight lines intersect at the point M( 3; 5; 3).

182. We may take the normal vector of the plane ~n = f4; 1; 6g as the direction vector of the desired straight line, so the equations take the form

x 3 = y 2 = z + 1: 4 1 6

183. The normal vectors of the given planes: ~n1 = f2; 1; 1g, ~n2 = f1; 3; 1g we may take the vector ~e = ~n1 ~n2 as the direction vector of the given straight line and as the direction vector of the desired straight line,

and now we write the equations of the desired straight line:

x 2 = y 1 = z + 1: 4 1 7

10 15

=( 255 + 10) = 245;

~

~{ ~| k

p

j~e1 ~e2j = ( 15)2 + ( 10)2 + 302

pp

187. The given planes are not parallel since their normal vectors ~n1 =