No. 1.17. Given four points A1(6;6;5); A2(4;9;5); A3(4;6;11); A4(6;9;3). Make up the equations: a) plane A1A2A3; 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; No. 2.17. Create an equation for a plane passing through the point M(1;-1;2) perpendicular to the segment M1M2; if M1(2;3;-4); M2(-1;2;-3). No. 3.17. Show that the line is parallel to the plane x + 3y - 2z + 1 = 0; and straight line x = t + 7; y = t - 2; z = 2t + 1 lies in this plane. Thank you for your purchase. If you have any questions, please write to me by email (see "information about the seller").
No. 1.17. To compose the equations you will need the following formulas:
a) Let's construct vectors A1A2 and A1A3, find their vector product, and get the normal vector of the plane. The general equation of the plane is: 6x - 9y - 6z + 63 = 0.
b) Find the direction vector of straight line A1A2: (6-4, 6-9, 5-5) = (2, -3, 0). The equation of a straight line in parametric form: x = 6 + 2t, y = 6 - 3t, z = 5.
c) Let's find the direction vector of the straight line A4M as the vector product of the normal vector of the plane A1A2A3 and the vector A4M. Then we will compose the equation of the line in parametric form using point A4. Direction vector: (-3, -3, -6). Equation of a straight line: x = 6 - 3t, y = 9 - 3t, z = 3 - 6t.
d) Since straight line A3N is parallel to straight line A1A2, its direction vector will coincide with the direction vector of straight line A1A2: (2, -3, 0). The equation of a straight line in parametric form: x = 4 + 2t, y = 6 - 3t, z = 5.
e) Find the normal vector of the plane through point A4 using the vector product of vectors A1A2 and A1A4. The general equation of the plane is: -3x - 6y + 9z + 45 = 0.
f) Find the direction vectors of the lines A1A4 and A1A2, calculate their scalar product and lengths using the formula |a||b|cos(angle between vectors) = ab. Then we find the sine of the angle between straight line A1A4 and plane A1A2A3, using the formula sin(angle) = |n(A1-A4)| / (|n|*|A1-A4|), where n is the normal vector of the plane, A1-A4 is the vector connecting points A1 and A4. Result: sin(angle) = 2/3.
g) Find the normal vector of the plane A1A2A3 using the vector product of the vectors A1A2 and A1A3. Then we find the cosine of the angle between the plane A1A2A3 and the coordinate plane Oxy, using the formula cos(angle) = |nOhu| / (|n||Oxy|), where Oxy is the unit vector lying on the Ox axis. Result: cos(angle) = 2/3.
No. 2.17. Let's find the direction vector of the segment M1M2: (-3, -1, 1). The normal vector of the plane will coincide with the direction vector of the segment, since the plane must be perpendicular to the segment. The general equation of the plane is: -3x - y + z + 1 = 0.
No. 3.17. The direction vector of the line given by the vector equation is equal to (1, 1, 2). This vector is not normal to the plane x + 3y - 2z + 1 = 0, which means the line is parallel to this plane. To make sure that the straight line lies in this plane, we substitute its coordinates into the equation of the plane and get: (t+7) + 3(t-2) - 2(2t+1) + 1 = 0, which is equivalent to t = -1. When t = -1, the coordinates of the line coincide with the coordinates of a point lying in the plane, which means the line lies in this plane.
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Ryabushko A.P. IDZ 3.1 version 17 is a task on linear algebra, which includes several tasks on drawing up equations of lines and planes, determining the angles between lines and planes, and also checking whether a line lies in a given plane.
No. 1.17. Given four points A1(6;6;5); A2(4;9;5); A3(4;6;11); A4(6;9;3). It is necessary to create equations: a) plane A1A2A3; 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; f) calculate the sine of the angle between straight line A1A4 and plane A1A2A3; g) calculate the cosine of the angle between the coordinate plane Oxy and the plane A1A2A3.
No. 2.17. It is necessary to create an equation for a plane passing through the point M(1;–1;2) perpendicular to the segment M1M2, where M1(2;3;–4); M2(–1;2;–3).
No. 3.17. It is necessary to show that the line is parallel to the plane x + 3y – 2z + 1 = 0, and the line x = t + 7; y = t – 2; z = 2t + 1 lies in this plane.
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IDZ Ryabushko 12.2 Option 18 - ИДЗ Рябушко 12.2 Вариант 18:В наличии 9 готовых решений задач, которые подробно и понятно описаны; Т ... ...