JEE Mathematics Introduction to Three Dimensional Geometry Notes Set 01

Coordinates of a Point in Space

Axes, planes and octants. Three mutually perpendicular axes OX, OY, OZ meet at the origin and divide space into 8 octants. A point is written \((x,y,z)\), where x, y, z are its perpendicular distances (with sign) from the \(yz\), \(zx\) and \(xy\) planes. Points on the x-axis have the form \((x,0,0)\), and the coordinate planes are \(x=0\), \(y=0\), \(z=0\). The signs \((+,+,+)\) mark the first octant and \((-,-,-)\) the eighth.

Distances from planes, axes and the origin. The distance of \(P(x,y,z)\) from the yz, zx and xy planes is \(|x|,|y|,|z|\). Its distance from the x, y, z axes is \(\sqrt{y^2+z^2}\), \(\sqrt{z^2+x^2}\), \(\sqrt{x^2+y^2}\), and from the origin \(\sqrt{x^2+y^2+z^2}\).

Distance and Section Formulae

Distance. For \(P(x_1,y_1,z_1)\) and \(Q(x_2,y_2,z_2)\), \(PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}\).

Section formula. The point dividing PQ in the ratio \(m:n\) internally is \(\left(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n},\frac{mz_2+nz_1}{m+n}\right)\). For external division replace \(n\) by \(-n\). The mid-point is the average of the coordinates, and the centroid of a triangle is the average of its three vertices.

Triangles in Space

Area and collinearity. The area of triangle ABC is \(\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|\), or \(\sqrt{\Delta_{xy}^2+\Delta_{yz}^2+\Delta_{zx}^2}\) where each term is the area of the projection on a coordinate plane. Three points are collinear when this area is zero, that is, when \(\overrightarrow{AC}\) is a multiple of \(\overrightarrow{AB}\).

Centres. Centroid \(G\) is the average of the vertices. For a right triangle the orthocentre is the right-angle vertex and the circumcentre is the mid-point of the hypotenuse. In an equilateral triangle all centres coincide. In any triangle \(H\), \(G\), \(S\) are collinear with \(HG:GS=2:1\).

Worked Examples

Example 1. Find the point on the y-axis at distance \(5\sqrt2\) from \((3,-2,5)\). Put \((0,y,0)\): \(9+(y+2)^2+25=50\), so \(y^2+4y-12=0\) and \(y=2\) or \(-6\).

Example 2. Three vertices of a parallelogram ABCD are \(A(3,-1,2)\), \(B(1,2,-4)\), \(C(-1,1,2)\). Since the diagonals bisect each other, \(D=A+C-B=(1,-2,8)\).

Example 3. Show that \((2,3,4)\), \((-1,2,-3)\), \((-4,1,-10)\) are collinear. The vectors from the first point are \((-3,-1,-7)\) and \((-6,-2,-14)\), and the second is twice the first.

Quick Tips

To find the locus of points equidistant from two points, equate the squared distances and cancel the quadratic terms; the result is a plane. Always use signs carefully when working with octants and external division.