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Friday, 8 August 2014

CHAPTER 18 -Worked Out Examples

     Example: 16  

Find the component of a vector \vec b perpendicular to the vector \vec a.
Solution: 16

We need to find \vec r, the component of \vec b perpendicular to \vec a
We have
\overrightarrow {PQ}  = \left( {\dfrac{{\vec b \cdot \vec a}}{{{{\left| {\vec a} \right|}^2}}}} \right)\vec a
 \Rightarrow  \,\,\,\,\vec r = \vec b - \overrightarrow {PQ}
 = \vec b - \left( {\dfrac{{\vec a \cdot \vec b}}{{{{\left| {\vec a} \right|}^2}}}} \right)\vec a
     Example: 17
Let \vec a,\;\vec b and \vec c be three mutually perpendicular vectors of equal magnitude. Prove that the vector \vec a + \vec b + \vec c is equally inclined to each of the three vectors.
Solution: 17

Let {\theta _a} represent the angle between \vec a and \vec a + \vec b + \vec c. We have,
\cos {\theta _a} = \dfrac{{\vec a \cdot (\vec a + \vec b + \vec c)}}{{\left| {\vec a} \right|\;\left| {\vec a + \vec b + \vec c} \right|}}
 = \dfrac{{\vec a \cdot \vec a + \vec a \cdot \vec b + \vec a \cdot \vec c}}{{\left| {\vec a} \right|\;\sqrt {{{\left| {\vec a + \vec b + \vec c} \right|}^2}} }}\ldots(1)
Let the magnitude of \vec a,\;\;\vec b and \vec c be \lambda . Also since the three vectors are mutually perpendicular, we have \vec a \cdot \vec b = \vec a \cdot \vec c = 0. Now,
{\left| {\vec a + \vec b + \vec c} \right|^2} = (\vec a + \vec b + \vec c) \cdot (\vec a + \vec b + \vec c)
 = \vec a \cdot \vec a + \vec b \cdot \vec b + \vec c \cdot \vec c
 = 3{\lambda ^2}
Using these facts in (1), we have
\cos {\theta _a} = \dfrac{{{\lambda ^2}}}{{\lambda  \cdot \sqrt 3 \,\lambda }} = \dfrac{1}{{\sqrt 3 }}
It is easy to see that \cos {\theta _b} and \cos {\theta _c} will have the same value. Thus, \vec a + \vec b + \vec c is equally inclined to \vec a,\;\;\vec b and \vec c.
     Example: 18      

Find the angle between the two diagonals of a cube.
Solution: 18

We now have,
\overrightarrow {OP}  \equiv a\hat i + a\hat j + a\hat k \Rightarrow \left| {\overrightarrow {OP} } \right| = \sqrt 3 a
\overrightarrow {AB}  = \overrightarrow {OB}  - \overrightarrow {OA}
 \equiv  - a\hat i + a\hat j + a\hat k \Rightarrow \left| {\overrightarrow {AB} } \right| = \sqrt 3 \,a
Let \theta  denote the angle between OP and AB. Thus,
\cos \theta  = \dfrac{{\overrightarrow {OP}  \cdot \overrightarrow {AB} }}{{\left| {\overrightarrow {OP} } \right|\;\left| {\overrightarrow {AB} } \right|}}
 = \dfrac{{(a\hat i + a\hat j + a\hat k) \cdot ( - a\hat i + a\hat j + a\hat k)}}{{(\sqrt 3 a)\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(\sqrt 3 \,a)}}
 = \dfrac{{ - {a^2} + {a^2} + {a^2}}}{{3\;{a^2}}}
 = \dfrac{1}{3}
 \Rightarrow  \,\,\,\,\theta  = {\cos ^{ - 1}}\dfrac{1}{3}
This is the angle between any two diagonals of (any) cube.
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