Introduction to Three Dimensional Geometry — Class 11 solved problems
6 problems from this chapter, each solved step by step.
- Find the equation of set of points P such that PA²+PB²=2 k², where A and B are the points (3,4,5) and (-1,3,-7), respectively.
x²-2x + y²-7y + z²+2z + (109)/(2) - k² = 0
- The centroid of a triangle ABC is at the point (1,1,1). If the coordinates of A and B are (3,-5,7) and (-1,7,-6), respectively, find the coordinates of the poin
The coordinates of point C are (1,1,2).
- Show that the points A (1, 2, 3), B (-1, -2, -1), C (2, 3, 2) and D(4,7,6) are the vertices of a parallelogram ABCD, but it is not a rectangle.
The points A, B, C, D form a parallelogram because AB = - CD and BC = - DA. It is not a rectangle because the dot product of adjacent sides AB · BC = -38 ≠ 0, i
- Are the points A (3,6,9), B (10,20,30) and C(25,-41,5), the vertices of a right angled triangle?
No, the points A (3,6,9), B (10,20,30) and C (25,-41,5) do not form the vertices of a right-angled triangle.
- Find the equation of the set of the points P such that its distances from the points A(3,4,-5) and B(-2,1,4) are equal.
The equation of the set of points P is 10x + 6y - 18z - 29 = 0.
- Show that the points P(-2,3,5), Q(1,2,3) and R(7,0,-1) are collinear.
The points P, Q, and R are collinear because QR = 2 PQ.
More Class 11 chapters
- Sets19 solved
- Relations and Functions26 solved
- Trigonometric Functions33 solved
- Complex Numbers and Quadratic Equations53 solved
- Linear Inequalities9 solved
- Permutations and Combinations49 solved
- Binomial Theorem17 solved
- Sequences and Series34 solved
- Straight Lines19 solved
- Conic Sections68 solved
- Limits and Derivatives32 solved
- Statistics15 solved
- Probability8 solved