IDZ Ryabushko 3.1 Option 9

No. 1. Four points are given: A1(7;5;3), A2(9;4;4), A3(4;5;7), A4(7;9;6). It is necessary to create equations:

a) Planes A1A2A3 Solution: Vectors AB1(2;-1;1) and AB2(-3;1;3) lying on the plane A1A2A3 can be obtained as the difference of the corresponding coordinates of points (A2-A1) and (A3-A1) . The normal to the plane A1A2A3 is the vector product of these vectors: N = AB1 x AB2. We get: AB1 = A2 - A1 = (9-7; 4-5; 4-3) = (2; -1; 1) AB2 = A3 - A1 = (4-7; 5-5; 7-3) = (-3; 0; 4) N = AB1 x AB2 = (-1 - 4; -2 - 12; -5 - 3) = (-5; -14; -8) Thus, the equation of the plane A1A2A3 has the form: -5x - 14y - 8z + d = 0 To find the unknown coefficient d, substitute the coordinates of any point, for example, A1: -57 - 145 - 8*3 + d = 0 d = 131 Equation of plane A1A2A3: -5x - 14y - 8z + 131 = 0

b) Straight A1A2 Solution: The direction vector of straight A1A2 is equal to the difference in the coordinates of these points: (A2-A1) = (2; -1; 1). The coordinates of point A1 are already known. Equation of a straight line in vector form: x = 7 + 2t y = 5 - t z = 3 + t

c) Line A4M perpendicular to the plane A1A2A3 Solution: A vector drawn from point A4 to an arbitrary point M(x,y,z) lying on the desired line must be perpendicular to the normal vector to the plane A1A2A3. The normal to the plane A1A2A3 has already been found in point (a) and is equal to N = (-5; -14; -8). Let the vector AM = (x-7; y-9; z-6). The condition for the perpendicularity of vectors N and AM is that their scalar product is equal to zero: (-5)(x-7) + (-14)(y-9) + (-8)*(z-6) = 0 Simplifying, we get: 5x + 14y + 8z - 131 = 0 Thus, the equation of straight line A4M: x = 5t + 7 y = 14t + 9 z = -8t+6

d) Line A3N parallel to line A1A2 Solution: The direction vector of line A1A2 is equal to (2; -1; 1). To find the direction vector of straight line A3N, you can use the property of parallel lines: vectors corresponding to the directions of straight lines are collinear. Thus, the direction vector of straight line A3N will also be equal to (2; -1; 1). The coordinates of point A3 are already known. Equation of a straight line in vector form: x = 4 + 2t y = 5 - t z = 7 + t

e) A plane passing through point A4, perpendicular to line A1A2 Solution: A vector drawn from point A4 to an arbitrary point on the desired plane must be perpendicular to the direction vector of line A1A2. Therefore, the normal to the desired plane will be collinear to the vector A1A2. The normal can be obtained by the vector product of the vectors AB1 and AB2 found in part (a). The normal to the plane will have the form: N = AB1 x AB2 = (-5; -14; -8) The plane equation has the form: -5x - 14y - 8z + d = 0 To find the unknown coefficient d, substitute the coordinates of point A4: - 57 - 149 - 8*6 + d = 0 d = 233 Thus, the equation of the plane passing through point A4 and perpendicular to line A1A2: -5x - 14y - 8z + 233 = 0

f) Sine of the angle between straight line A1A4 and plane A1A2A3 Solution: Find the projection of vector A1A4 onto the normal to plane A1A2A3. The normal to the plane was found in point (a) and is equal to N = (-5; -14; -8). Vector A1A4 can be found as the difference between the coordinates of points A1 and A4: (A4-A1) = (0; 4; 3). The scalar product of vectors N and A1A4 is equal to the product of their lengths multiplied by the cosine of the angle between them. Thus, the sine of this angle is: sin(angle) = |N x A1A4|/|N||A1A4| where |N| and |A1A4| - lengths of vectors N and A1A4, respectively. Let's calculate the numerator: N x A1A4 = (-5; -14; -8) x (0; 4; 3) = (-68; 15; 20) |N x A1A4| = sqrt((-68)^2 + 15^2 + 20^2) = sqrt(4819) Calculate the denominator: |N| = sqrt((-5)^2 + (-14)^2 + (-8)^2) = sqrt(325) |A1A4| = sqrt(0^2 + 4^2 + 3^2) = 5 Thus, the sine of the angle between straight line A1A4 and plane A1A2A3 is equal to: sin(angle) = sqrt(4819)/(5*sqrt(325))

g) Cosine of the angle between the coordinate plane Oxy and the plane A1A2A3 Solution: The coordinate plane Oxy can be given by the equation z = 0. The angle between the coordinate plane and the plane A1A2A3 can be found as the angle between them

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