CBSE Class 11 Maths Notes Chapter 12 Introduction to Three Dimensional Geometry

The coordinate system in three-dimensional geometry is a fundamental concept that helps us locate points in space. Unlike the familiar two-dimensional Cartesian coordinate system, which consists of two perpendicular axes (x and y), the three-dimensional coordinate system adds a third axis (usually labeled z) perpendicular to the x-y plane. Each point in three-dimensional space can be uniquely identified by its coordinates, typically represented as (x, y, z), where x represents the horizontal position, y represents the vertical position, and z represents the position along the third dimension.

Distance between Two Points

i) Considering two points A(x1, y1, z1) and B(x2, y2, z2), forming a right-angled triangle ΔACB, we apply the Pythagoras theorem. This yields AB^2 = AC^2 + BC^2. Similarly, for the triangle ΔBCH, BC^2 = CH^2 + BH^2. Combining these equations, we find AB^2 = AC^2 + CH^2 + BH^2. Substituting the coordinates of BH = x2 - x1, CH = y2 - y1, and AC = z2 - z1, we arrive at AB = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2), representing the distance between two points in three-dimensional space. ii) The distance of any point (x1, y1, z1) from the origin is given by √(x1^2 + y1^2 + z1^2). iii) According to the rule, the sum of two collinear points equals the third collinear point, applicable only when dealing with three collinear points A, B, and C. For instance, AB + BC = AC. The application of the distance formula helps determine collinearity among points.

Section Formula

i) The Section Formula enables us to determine the internal division ratio of a line segment by a point. ii) If point C divides the line segment AB internally, the ratio m:n in which AB is divided internally can be found using the section formula. iii) Consider two points A(x1, y1, z1) and B(x2, y2, z2), with C(x, y, z) dividing the line segment AB internally in the ratio m:n. iv) Perpendiculars are drawn from A, B, and C on the XY plane, such that AH||CD||BE. Additionally, line HG is drawn through point C, parallel to FE. v) Quadrilaterals CDFH and DEGC are parallelograms based on the figure. vi) ΔACH and ΔBCG are right-angled triangles with vertically opposite angles, thus similar. Hence, mn = AC/BC = AH/BG = AF - HF/GE - BE = AF - CD/CD - BE. vii) Writing the corresponding coordinates, we derive mn = (z - z1)/(z2 - z) = (mz2 - mz)/(nz - nz1) = z(m + n)/(mz2 + nz1). viii) Similarly, for other coordinates, x = (mx2 + nx1)/(m + n) and y = (my2 + ny1)/(m + n). ix) Therefore, the required coordinates of the point dividing internally are (mx2 + nx1)/(m + n), (my2 + ny1)/(m + n), (mz2 + nz1)/(m + n). x) For a midpoint, where m = 1 and n = 1, the coordinates of the point will be ((x2 + x1)/2, (y2 + y1)/2, (z2 + z1)/2).

Rectangular Coordinate System