Category : 12th Class
Three Dimensional Geometry
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\[\Rightarrow \] \[{{\ell }^{2}}+{{m}^{2}}+{{n}^{2}}=1\]
Consider P(a, b, c) be any point in the space at length r from the origin to the axis. Here a, b and c are said to be direction ratios.
\[\therefore OP=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] (By distance formula)
\[|r|=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\]
Note: Direction cosine is proportional to the direction ratio.
Let a, b, c be d.r.,s of the line OP and its d.c.'s be \[\ell \],m and n respectively.
Then \[\frac{\ell }{a}=\frac{m}{b}=\frac{n}{c}=K\] (say)
Convesion from direction ratios (d.r.'s) to direction cosines (d.c.'s)
\[\ell =\pm \frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]
\[m=\pm \frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\] and \[n=\pm \frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]
If \[\vec{r}=a\hat{i}+b\hat{j}+c\hat{k}\]
Then a, b and c be the d.r.'s of r and d.c.'s of r is given by
\[\ell =\frac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\],\[m=\frac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]and \[n=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\]
\[{{\ell }_{1}},{{m}_{1}},{{n}_{1}}\]and \[{{\ell }_{2}},{{m}_{2}},{{n}_{2}}\]then
\[\cos \theta ={{\ell }_{1}}{{\ell }_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}\]
and\[\sin \theta =\sqrt{{{({{m}_{2}}{{\ell }_{1}}-{{m}_{1}}{{\ell }_{2}})}^{2}}-{{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{\ell }_{2}}{{n}_{1}}-{{\ell }_{1}}{{n}_{2}})}^{2}}}\]
If \[\vec{a}={{a}_{1}}\vec{i}+{{b}_{1}}\vec{j}+{{c}_{1}}\vec{k}\]
and\[\vec{b}+{{a}_{2}}\vec{i}+{{b}_{2}}\vec{j}+{{c}_{2}}\vec{k}\]be the two vectors, then their direction ratios be \[{{a}_{1}},{{b}_{1}},{{c}_{1}}\] and \[{{a}_{2}},{{b}_{2}},{{c}_{2}}\] respectively. Let \[\theta \] is the .angle between these two vectors. Then \[\cos \theta =\]\[\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}.\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}.\]
Solved Examples
Sol. Here \[P\equiv (1,1,2)\]
\[Q\equiv (3,4,1)\]
Given \[d.r.'s\equiv (2,3,6)\]
The first step is to find the \[d.c.'s\]
\[\ell =\frac{2}{\sqrt{{{2}^{2}}+{{3}^{2}}+{{6}^{2}}}}=\frac{2}{7}\],\[m=\frac{3}{7}\]and\[n=\frac{6}{7}.\]
\[\therefore \]Projection of the line joining the points (1, 1, 2) and (3, 4, 1) on the line whose \[d.c.'s\]are\[\left( \frac{2}{7},\frac{3}{7},\frac{6}{7} \right)\], is
\[=(3-1)\times \frac{2}{7}+(4-1)\times \frac{3}{7}+(1-2)\times \frac{6}{7}\]\[=2\times \frac{2}{7}+3\times \frac{3}{7}+\frac{-6}{7}\]\[=\frac{4+9-6}{7}=\frac{7}{7}=1\]
Note:
(i) Equation of xy plane is \[z=0\]
(ii) Equation of yz plane is \[x=0\]
(iii) Equation of zx plane is \[y=0\]
(iv) Equation of the any plane parallel to xy plane is \[z=c\]
(v) Equation of the any plane parallel to yz plane \[x=a\] &
(vi) Equation of any plane parallel to zx plane is \[y=b\] etc.
(i) Vector form: If \[\vec{n}\] be a unit vector normal to a given plane & p be the length of perpendicular from the origin to the plane, then the equation of the plane is written as \[\vec{r}.\vec{n}=p\]
(ii) Cartesian form: If\[\ell \], m, n be the direction cosines of the normal to given plane and P is distance from the origin to the plane, then the equation of the plane is \[\ell x+my+nz=p\]
\[=\frac{|(a-{{r}_{1}})\times b|}{|b|}\]
Let the two lines AB and PQ are respectively, \[{{r}_{1}}={{a}_{1}}+\lambda {{b}_{1}}\]and \[{{r}_{2}}={{a}_{2}}+\mu {{b}_{2}}.\]
Then the shortest distance between the lines \[{{r}_{1}}={{a}_{1}}+\lambda {{b}_{1}}\]and \[{{r}_{2}}={{a}_{2}}+\mu {{b}_{2}}.\]is given by
\[d=\frac{|({{a}_{2}}-{{a}_{1}}).({{b}_{1}}\times {{b}_{2}})}{|{{b}_{1}}\times {{b}_{2}}|}\]
Note: The shortest distance between two parallel lines\[r={{a}_{1}}+lb\]&\[r={{a}_{2}}+\mu b\]are\[d=\frac{({{a}_{2}}-{{a}_{1}})\times b|}{|b|}\]
\[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\] & \[\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}}\]
Let \[\theta \] be the angle between them, then \[\cos \theta =\frac{{{a}_{1}}.{{a}_{2}}+{{b}_{1}}.{{b}_{2}}+{{c}_{1}}.{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}\times \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\]
\[\frac{x-a}{\ell }=\frac{y-b}{m}=\frac{z-c}{n}=r\]be the given point. Then
\[\therefore \] \[P\equiv (\ell r+a,mr+b,nr+c)\]
\[\therefore \] Length of the perpendicular is AP
\[=\sqrt{{{(\ell r+a-\alpha )}^{2}}+{{(mr+b-\beta )}^{2}}+{{(nr+c-\gamma )}^{2}}}\](By distance formula)
So, equation of the line AP be
where \[r=(\alpha -a)\ell +(\beta -b)m+(\gamma -c)n\]
The vector equation of the straight line passing through given point with the position vector \[\vec{a}\] & parallel to a given vector \[\vec{b}\]is given as follows:
\[\vec{r}=\vec{a}+\lambda \,\vec{b}\]
where \[\lambda =\] scalar quantity.
Note: If \[\vec{r}\] is the position vector of the any point P(x, y, z) on the line, then \[\vec{r}=x\vec{i}+y\vec{j}+z\vec{k}\]
\[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\]
If \[\ell ,\] m & n be the direction cosines of the line, then the equation of the straight line is
\[\frac{x-{{x}_{1}}}{\ell }=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\]
Solved Examples
Sol. \[\overrightarrow{OA}=2\hat{i}-\hat{j}+4\hat{k}\]
\[\vec{b}=\hat{i}-2j-\hat{k}\]
\[\therefore \] Equation of the straight line be
\[\vec{r}=(2\vec{i}-\vec{j}+4\vec{k})+\lambda (\vec{i}+2\vec{j}-\vec{k})\]
\[\vec{r}=(2+\lambda )\vec{i}-(1-2\lambda )\vec{j}+(4-\lambda )\vec{k}\] (1)
Equation (1) is the eqaution of straight line in vector form. Now to find the equation of line in certesian form
Let \[\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\] (2)
Comparing (1) & (2), we have
\[x=2+\lambda \] \[\Rightarrow x-2=\lambda \]
\[y=-(1-2\lambda )\] \[\Rightarrow \left( \frac{y+1}{2} \right)=\lambda \]
\[z=4-\lambda \] \[\Rightarrow (z-4)=-\lambda \]
\[\therefore \] Equation of straight line will be
\[\frac{x-2}{1}=\frac{(y+1)}{2}=\frac{(z-4)}{-1}\]
\[\vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})\]
\[\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}\]
Vector form: The angle between two lines
\[{{r}_{1}}={{a}_{1}}+\lambda \,{{b}_{1}}\] & \[{{r}_{2}}={{a}_{2}}+\mu {{b}_{2}}\]is written as
\[\cos \theta =\frac{{{b}_{1}}.{{b}_{2}}}{|{{b}_{1}}|.|{{b}_{2}}|}\]
Solved Example
Sol. Equation of the straight line passes through A & B be
\[\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}\]
\[\Rightarrow \frac{x-3}{5-3}=\frac{y-4}{1-4}=\frac{z-1}{6-1}\] \[\Rightarrow \frac{x-3}{2}=\frac{y-4}{-3}=\frac{z-1}{5}\]
Let point \[P\equiv (a,b,0)\]
According to question, straight line crosses the XY-plane at P. It means the co-ordinate of point satisfies the given straight line
\[\therefore \,\,\,\frac{a-3}{2}=\frac{b-4}{-3}=\frac{0-1}{5}\]
\[\Rightarrow \frac{a-3}{2}=\frac{b-4}{-3}=\frac{0-1}{5}\] ……..(i)
On solving equ (i), we get
\[\frac{a-3}{2}=\frac{-1}{5}\Rightarrow 5a-15=-2\] \[\Rightarrow 5a=13\Rightarrow a\frac{13}{5}\]
Similarly, \[\frac{b-4}{-3}=\frac{-1}{5}\] \[\Rightarrow 5b-20=3\Rightarrow b=\frac{23}{5}\]
Hence the required point, \[P\equiv (a,b,0)\equiv \left( \frac{13}{5},\frac{23}{5},0 \right)\]
Sol. The vector equation of a plane be
\[\vec{r}.\vec{n}=p\]
\[\because \vec{n}=\frac{{\vec{a}}}{|\vec{a}|}=\frac{3\vec{i}+5\vec{j}-6\vec{k}}{\sqrt{{{3}^{2}}+{{(5)}^{2}}+{{(-6)}^{2}}}}=\frac{3\vec{i}+5\vec{j}-6\vec{k}}{\sqrt{70}}\]
\[\Rightarrow \vec{r}.\frac{(3\vec{i}+5\vec{j}-6\vec{k}}{\sqrt{70}}=7\]
\[\Rightarrow (x\vec{i}+y\vec{j}+z\vec{k}).\frac{(3\vec{i}+5\vec{j}-6\vec{k})}{\sqrt{70}}=7\]
\[\Rightarrow 3x+5y-6z=7\sqrt{70}\]
This is the required equation of the plane.
\[2x+y+z=7\]
Sol. Vector equation of the straight line through the points (5, 1, 6) and (3, 4, 1) be
\[\Rightarrow \vec{r}=\vec{a}+\lambda (\vec{b}-\vec{a})\]
\[\Rightarrow \vec{r}=5\vec{i}+\vec{j}+6\vec{k}+\lambda (-2\vec{i}+3\vec{j}-5\vec{k})\]
\[\Rightarrow \vec{r}=(5-2\lambda )\vec{i}+(1+3\lambda )\vec{j}+(6-5\lambda )\vec{k}\] ….. (1)
Let \[P\equiv (a,b,c)\] be any point where the straight line crosses the plane \[2x+y+z=7.\] ……(2)
So, the point P must satisfies the equation (2)
\[\therefore 2a+b+c=7\]
On putting the values in equation (2), we get
\[\Rightarrow 2(5-2\lambda )+1+3\lambda +6-5\lambda =7\]
\[\Rightarrow 10-4\lambda +1+3\lambda +6-5\lambda =7\]
\[\Rightarrow -6\lambda +17=7\] \[\Rightarrow \lambda =\frac{10}{6}=\frac{5}{3}\]
Thus,
\[a=5-2\times \frac{5}{3}=\frac{5}{3}\]
\[b=1+3\times \frac{5}{3}=6\]
\[c=6-5\times \frac{5}{3}=\frac{-7}{3}\]
Hence, the required point be \[\left( \frac{5}{3},6\frac{-7}{3} \right)\].
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