IDZ Ryabushko 3.1 Option 12

No. 1 Given four points A1(4;4;10); A2(7;10;2); A3(2;8;4); A4(9;6;9). Make up the equations: a) plane A1 A2 A3; b) straight A1A2; c) straight line A4M, perpendicular to the plane A1A2A3; d) straight line A3N parallel to straight line A1A2; e) a plane passing through point A4, perpendicular to straight line A1A2. Calculate: e) the sine of the angle between straight line A1A4 and plane A1A2A3; g) cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.

a) Find the vectors AB1 and AB2: AB1 = (7-4; 10-4; 2-10) = (3;6;-8) AB2 = (2-4; 8-4; 4-10) = (- 2;4;-6) Then the vector product of AB1 and AB2 gives the normal vector to the plane: n = AB1 x AB2 = (36;18;18) Thus, the equation of the plane A1A2A3 has the form: 36(x-4)+18( y-4)+18(z-10)=0

b) The direction vector of straight line A1A2 is equal to: d = (7-4; 10-4; 2-10) = (3;6;-8) Point A1 has coordinates (4;4;10), so the equation of straight line A1A2 has the form : x=4+3t y=4+6t z=10-8t

c) The direction vector of straight line A4M must be perpendicular to the normal vector of the plane A1A2A3, therefore, it must be collinear to the vector product of this vector and the vector directed from point A4 to an arbitrary point M on this straight line. Take, for example, point M(9;0;0): AM = (9-9; 0-6; 0-9) = (0;-6;-9) d = n x AM = (-54;162; -54) Point A4 has coordinates (9;6;9), therefore the equation of the desired line A4M has the form: x=9-6t y=6+18t z=9-6t

d) The direction vector of straight line A3N must be collinear to the direction vector of straight line A1A2, therefore, it is equal to: d = (3;6;-8) Point A3 has coordinates (2;8;4), so the equation of straight line A3N has the form: x= 2+3t y=8+6t z=4-8t

e) The equation of the desired plane has the form: Ax+By+Cz+D=0 Since the plane passes through the point A4(9;6;9), its coordinates satisfy the equation of the plane: 36(x-9)+18(y- 6)+18(z-9)=0 Let us expand the left side of this equation into the scalar product of the normal vector and the vector with coordinates (x-9; y-6; z-9): 36x-288+18y-108+18z-162 =0 Let's simplify: 36x+18y+18z=558 Thus, the equation of the plane passing through point A4 and perpendicular to straight line A1A2 has the form: 36x+18y+18z-558=0

e) Find the vector product of vectors A1A4 and A1A2: n = (26;34;-14) Then the sine of the angle between straight line A1A4 and plane A1A2A3 is equal to the modulus of the projection of vector A1A4 onto the normal vector of the plane, divided by the product of the lengths of these vectors: sinα = |n * A1A4| / (|n| * |A1A4|) |n * A1A4| = |(265)+(34(-2))+((-14)*6)| = 22√29 |n| = √(26²+34²+(-14)²) = 42 |A1A4| = √(5²+2²+(-1)²) = √30 Thus, the sine of the angle α between straight line A1A4 and plane A1A2A3 is equal to: sinα = (22√29) / (42 * √30)

g) The cosine of the angle between the plane A1A2A3 and the coordinate plane Oxy is equal to the projection of the normal vector of the plane A1A2A3 onto the Ox axis, divided by the length of the normal vector: cosα = |n|ₓ / |n| where |n|ₓ is the projection of the normal vector onto the Ox axis. The normal vector of the plane A1A2A3 is equal to (36;18;18), so its projection onto the Ox axis is 36. The length of the normal vector is √(36²+18²+18²) = 6√13. Thus, the cosine of the angle α between the plane A1A2A3 and the coordinate plane Oxy is equal to: cosα = 36 / (6√13)

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IDZ Ryabushko 3.1 Option 12 is a set of tasks in mathematics, geometry and algebra. The tasks give the coordinates of points and require you to create equations of straight lines and planes, calculate the angles between them, and also solve other problems. In task No. 1 you need to create equations for a plane and straight lines, as well as calculate the sine and cosine of angles. In task No. 2 you need to create an equation of a plane passing through two given points and perpendicular to a given plane. In task No. 3 you need to create an equation of a line passing through the origin of coordinates and parallel to a given line.

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