المستوى الذي يوازي المستوى 2x − 4y + 6z = 10 ويمر بالنقطة (1, 1, 1) هو:
- 1 x − 2y + 3z = 2
- 2 x − 2y + 3z = 0
- 3 2x − 4y + 6z = 4
- 4 x + 2y + 3z = 6
This unit focuses on the algebra of planes in three-dimensional space. It begins by introducing the vector equation of a plane, which requires a point on the plane and a normal vector. It then derives the standard form (a(x-x1)+b(y-y1)+c(z-z1)=0) and the general form (ax+by+cz+d=0) from the vector form. Students learn to find the equation of a plane passing through three non-collinear points by using the cross product to find a normal vector, and also to find the equation of a plane containing two intersecting lines. The unit covers finding the intersection point of a line and a plane, and the angle between two planes using the dot product of their normal vectors. Conditions for parallel and perpendicular planes are given, along with methods to find the line of intersection of two planes. The distance from a point to a plane is derived, both in vector and Cartesian forms, and the distance between two parallel planes is found by taking a point on one plane and calculating its distance to the other. Finally, the equation of a plane in terms of its intercepts with the coordinate axes is presented. The unit includes worked examples, 'Try to solve' exercises, and a set of exercises at the end, including multiple-choice questions and multi-requirement problems. By the end, a student should be able to write equations of planes in various forms, determine relationships between planes, compute distances, and solve related geometric problems.
Multiple choice questions on this lesson in Pure Mathematics for 3rd secondary, First Term, with practice exercises for revision and exam preparation.
المستوى الذي يوازي المستوى 2x − 4y + 6z = 10 ويمر بالنقطة (1, 1, 1) هو:
المستوى المار بالنقاط (1, 0, 0) و (0, 2, 0) و (0, 0, 3) تكون معادلته العامة:
المستوى 5x − 2y + 3z = 15 يقطع محور z عند النقطة:
خط تقاطع المستويين x + y + z = 6 و x − y + z = 2 يمر بالنقطة:
إذا كان المستوى 3x + ky + 6z = 12 عمودياً على المستوى x + 2y − z = 5 فإن قيمة k تساوي:
المستوى المار بنقطة تقاطع المستقيمين x = y = z و x − 1 = y − 2 = z − 3 وعمودي على المتجه (1, 1, 1) هو:
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