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Saturday, 9 August 2014

CHAPTER 6 - Worked Out Examples

     Example: 10     

Find the equation of the plane passing through the points P(1,1,0),Q(1,2,1) and R( - 2,2, - 1).
Solution: 10

Let S(x,y,z) be any arbitrary point in the plane whose equation we wish to determine:
Since \overrightarrow {PQ}  \times \overrightarrow {PR}  will be perpendicular to this plane, we must have
\overrightarrow {PS}  \cdot \left( {\overrightarrow {PQ}  \times \overrightarrow {PR} } \right) = 0
 \Rightarrow  \,\,\,\,\left\{ {\left( {x - 1} \right)\hat i + \left( {y - 1} \right)\hat j + z\hat k} \right\} \cdot \left| {\begin{array}{*{20}{c}}  {\hat i}\,\,\,\,{\hat j}\,\,\,\,\,{\hat k}\\  0\,\,\,\,1\,\,\,\,1\\  { - 3}\,\,\,\,\,1\,\,\,\,{ - 1}  \end{array}} \right| = 0
 \Rightarrow  \,\,\,\, \left\{ {\left( {x - 1} \right)\hat i + \left( {y - 1} \right)\hat j + z\hat k} \right\} \cdot \left( { - 2\hat i - 3\hat j + 3\hat k} \right) = 0
 \Rightarrow  \,\,\,\, 2\left( {x - 1} \right) + 3\left( {y - 1} \right) - 3z = 0
 \Rightarrow  \,\,\,\, 2x + 3y - 3z = 5
We could have proceeded alternatively as follows: using the result of the last example, any arbitrary plane through P(1,1,0) will be of the form
a\left( {x - 1} \right) + b\left( {y - 1} \right) + cz = 0
 \Rightarrow  \,\,\,\,{\lambda _1}\left( {x - 1} \right) + {\lambda _2}\left( {y - 1} \right) + z = 0 \,\,\,\, ;{\lambda _1} = \dfrac{a}{c},\,{\lambda _2} = \dfrac{b}{c}
If this passes through Q(1,2,1) and R( - 2,1, - 1), we have
{\lambda _2} + 1 = 0
and
 - 3{\lambda _1} + {\lambda _2} - 1 = 0
 \Rightarrow  \,\,\,\, {\lambda _1} =  - \dfrac{2}{3},\,\,{\lambda _2} =  - 1
Thus the equation of the plane is
 - \dfrac{2}{3}\left( {x - 1} \right) - \left( {y - 1} \right) + z = 0
 \Rightarrow  \,\,\,\, 2\left( {x - 1} \right) + 3\left( {y - 1} \right) - 3z = 0
 \Rightarrow  \,\,\,\, 2x + 3y - 3z = 5
     Example: 11     

Find the equation of the plane intercepting lengths ab and c on the x-, y- and z-axis respectively.
Solution: 11

The plane passes through the points A(a,0,0)B(0,b,0) and C(0,0,c). For any variable point S(x,y,z) in this plane, we have (as discussed in the previous section),
\overrightarrow {AS}  \cdot \left( {\overrightarrow {AB}  \times \overrightarrow {AC} } \right) = 0
 \Rightarrow  \,\,\,\, \left\{ {\left( {x - a} \right)\hat i + y\hat j + z\hat k} \right\} \cdot \left| {\begin{array}{*{20}{c}}  {\hat i}\,\,\,\,{\hat j}\,\,\,\,{\hat k}\\  { - a}\,\,\,\,b\,\,\,\,0\\  { - a}\,\,\,\,0\,\,\,\,c  \end{array}} \right| = 0
 \Rightarrow  \,\,\,\, \left\{ {\left( {x - a} \right)\hat i + y\hat j + z\hat k} \right\} \cdot \left( {bc\hat i + ac\hat j + ab\hat k} \right) = 0
 \Rightarrow  \,\,\,\, \left( {bc} \right)\left( {x - a} \right) + acy + abz = 0
 \Rightarrow  \,\,\,\, bcx + acy + abz = abc
\Rightarrow\,\,\,\,\,{\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} =1 }
This general equation has the same form as the equation of the line in intercept form; which further proves the analogy between the formulae in two and in three dimensions.
     Example: 12      

A plane is at a distance p from the origin and the direction cosines of the (outward) normal to it are lmn. Find its equation.
Solution: 12

The unit vector \hat n normal to the plane is
\hat n = l\hat i + m\hat j + n\hat k
For any point \vec r\left( {x\hat i + y\hat j + z\hat k} \right) in the plane, we have
\vec r \cdot \hat n = p
 \Rightarrow  \,\,\,\,lx + my + nz = p
This is the required equation; it is called the normal form of the plane’s equation. As an exercise, convert the general equation of the plane
ax + by + cz + d = 0
into normal form.
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