Three Dimensional Plane
In three dimensional Geometry, it is not a new geometry though it is the refined or extension form of the two dimension geometry. In 3-dimensional geometry. Three axes i.e. x-axis, y-axis and z-axis are perpendicular to each other is considered.
Let \[X'OX',Y'OY\] & \[Z'OZ\] be three mutually perpendicular lines which be intersect at 0. It is called origin.
\[X'OX\xrightarrow{{}}x-axis\]
\[Y'OY\xrightarrow{{}}y-axis\]
\[Z'OZ\xrightarrow{{}}z-axis\]
Plane XOY is called xy plane
YOZ is called yz plane
and ZOX is called zx plane
In 3-D, there are 8 quadrents
Equation of x-axis be y= 0 & z =0
Equation of y-axis be x = 0 & z = 0
and equation of z-axis be x=0 & y=0
Note: In 3-D, a straight line is represented by two equations where as a plane is represented by single equation in at most three variables.
- Some basic formula which are used in 3-dimension. The distance between points \[A({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] and \[B({{x}_{1}},\,{{y}_{2}},\,{{z}_{3}})\]be
\[AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}+{{({{z}_{2}}-{{z}_{1}})}^{2}}}\]
e.g. Let two points are A (2, 3, 1) & B = (- 5, 2-1)
\[\therefore \,\,\,\,AB=\sqrt{{{(-5-2)}^{2}}+{{(2-3)}^{2}}+{{(-1-1)}^{2}}}\]
\[=\sqrt{49+5}=\sqrt{54}\]
- Section Formula: The coordinate of the point P dividing the line joining \[A({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] & \[({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}})\] in the ratio m:n internally are
\[P=\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\frac{m{{z}_{2}}+n{{z}_{1}}}{m+n} \right)\]
The co-ordinate of the point P dividing the line joining \[A({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] and \[({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}})\] in the ratio m:n externally are
\[P=\left( \frac{m{{x}_{2}}-n{{x}_{1}}}{m-n},\frac{m{{y}_{2}}-n{{y}_{1}}}{m-n},\frac{m{{z}_{2}}-n{{z}_{1}}}{m-n} \right)\]
Midpoint of AB be
\[P=\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2},\frac{{{z}_{1}}+{{z}_{2}}}{2} \right).\]
e.g. Find the co-ordinate of the point which divides the line segment joining the point (-2, 3, 5) & (1, - 4, 6) in the ratio (i) 2: 3 internally (ii) 2:3 externally.
Sol. Here, Let A= (-2, 3, 5) & B= (1, -4, 6)
and m:n =2:3 internally
Let P divides AB in the ratio m: n internally
\[\therefore \,\,\,P=\left( \frac{m{{x}_{2}}+n{{x}_{1}}}{m+n},\frac{m{{y}_{2}}+n{{y}_{1}}}{m+n},\frac{m{{z}_{2}}+m{{z}_{1}}}{m+n} \right)\]
\[=\left( \frac{2.1+3(-2)}{2+3},\frac{2(-4)+3.3}{2+3},\frac{2\times 6-3\times 5}{2+3} \right)\]
\[=\left( \frac{-4}{5},\frac{1}{5},\frac{27}{5} \right)\]
When P divides AB in the ratio m : n externally
\[\therefore \,\,\,P=\left( \frac{2.1-3(-2)}{2-3},\frac{2(-4)-3.3}{2-3},\frac{2\times 6-3\times 5}{2-3} \right)\]
\[P=(-8,+17,3).\]
- Controid of triangle: The co-ordinate of the centroid of the triangle ABC, whose vertices are
\[A({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}}),\] \[B({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}}),\] & \[C({{x}_{2}},\,{{y}_{2}},\,{{z}_{3}}),\] are
\[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3},\frac{{{z}_{1}}+{{z}_{2}}+{{z}_{3}}}{3} \right)\]
- Centroid of the tetrahedran: If \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}}),\] \[({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}}),\] \[({{x}_{3}},\,{{y}_{3}},\,{{z}_{3}})\,a({{x}_{4}},\,{{y}_{4}},\,{{z}_{4}})\] be the vertices of the tetrahedran, then its centroid G is given by
\[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}}{4},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}+{{y}_{4}}}{4},\frac{z1+z2+z3+z4}{4} \right)\]
A point R with x-coordinate 4 lies on the segment joining the points \[\mathbf{P(2,-3,4)}\] & \[\mathbf{Q(8,0,1,0)}\mathbf{.}\]
Find the co-ordinate of the point R.
Sol: Let \[P(2,-3,4)\] and \[Q(8,0,1,0).\]
Let R divides PQ in the ratio l: 1 internally
\[\therefore \,\,\,\,R=\left( \frac{8\lambda +2}{\lambda +1},\frac{0.\lambda +(-3)}{\lambda +1},\frac{10\lambda +1\times 4}{\lambda +1} \right)\]………… (1)
Here x co-ordinate =4
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