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Friday, 8 August 2014

CHAPTER 19 -Worked Out Examples 2

    Example: 21     

If abc are the lengths of the sides of \Delta ABC opposite to the angles ABand C respectively, prove using vector methods that
a(1 + \cos A) + b(1 + \cos B) + c(1 + \cos C) = (a + b + c)(\cos A + \cos B + \cos C)
Solution: 21

We have, by the triangle law,
\vec a + \vec b + \vec c = \vec 0
 \Rightarrow  \,\,\,\, \vec a =  - (\vec b + \vec c)
 \Rightarrow  \,\,\,\, {a^2} = ab\cos C + ac\cos B\left( \begin{array}{l}  \vec a \cdot \vec b =  - ab\cos C\\  \,\,\,\,\,\vec a \cdot \vec c =  - ac\cos B  \end{array} \right)
 \Rightarrow  \,\,\,\,a = b\cos C + c\cos B\ldots(1)
Similarly,
b = c\cos A + a\cos C\ldots(2)
c = a\cos B + b\cos A\ldots(3)
Adding (1) (2)  and (3) we have
a + b + c = a(\cos B + \cos C) + b(\cos C + \cos A) + c(\cos A + \cos B)
Adding
a\cos A + b\cos B + c\cos C on both sides, we have
a(1 + \cos A) + b(1 + \cos B) + c(1 + \cos C) = (a + b + c)(\cos A + \cos B + \cos C)
     Example: 22    

Find three-dimensional vectors {\vec v_1},\;\;{\vec v_2} and {\vec v_3} satisfying the relations
\begin{array}{l}  {{\vec v}_1} \cdot {{\vec v}_1} = 4 \,\,\,\, {{\vec v}_1} \cdot {{\vec v}_2} =  - 2 \,\,\,\,{{\vec v}_1} \cdot {{\vec v}_3} = 6\\  {{\vec v}_2} \cdot {{\vec v}_2} = 2 \,\,\,\, {{\vec v}_2} \cdot {{\vec v}_3} =  - 5 \,\,\,\, {{\vec v}_3} \cdot {{\vec v}_3} = 29  \end{array}
Solution: 22

A reference frame for the vectors has not been specified; therefore, it is up to us to choose a reference frame and then use it consistently and evaluate the required vectors in that reference frame.
Assume {\vec v_1} to be along the x-direction, i.e.
{\vec v_1} = 2\hat i
Let {\vec v_2} = a\hat i + b\hat j + c\hat k
{\vec v_3} = p\hat i + q\hat j + r\hat k
Now we step by step use all the given relations to determine the unknown constraints:
{\vec v_1} \cdot {\vec v_2} =  - 2 \,\,\,\,  \Rightarrow  \,\,\,\, 2a =  - 2
 \Rightarrow  \,\,\,\, a =  - 1\ldots(1)
{\vec v_2} \cdot {\vec v_2} = 2 \,\,\,\,  \Rightarrow  \,\,\,\, {a^2} + {b^2} + {c^2} = 2
 \Rightarrow  \,\,\,\, {b^2} + {c^2} = 1\ldots(2) from (1)
{\vec v_1} \cdot {\vec v_3} = 6 \,\,\,\,  \Rightarrow   2p = 6
 \Rightarrow  \,\,\,\, p = 3\ldots(3)
{\vec v_2} \cdot {\vec v_3} =  - 5 \,\,\,\,  \Rightarrow  \,\,\,\, ap + bq + cr =  - 5
 \Rightarrow  \,\,\,\, bq + cr =  - 2\ldots(4) Using (1) and (3)
{\vec v_3} \cdot {\vec v_3} = 29 \,\,\,\,  \Rightarrow  \,\,\,\, {p^2} + {q^2} + {r^2} = 29
 \Rightarrow  \,\,\,\, {q^2} + {r^2} = 20\ldots(5) Using (3)
Notice that (2)(4)  and (5)  are three equations in four unknowns. To get over this problem (it is not a problem actually! There will be an infinite set of vectors satisfying the given constraints. We have to find any one of them), when we chose {\vec v_1} to be along the x-axis, we could also have adjusted the co-ordinate frame, so that {\vec v_1} and {\vec v_2} lie in the x - z plane. This can always be done; since it is upto us to choose the frame of reference, we chose it so that the x - z plane co-insides with the plane of {\vec v_1} and {\vec v_2}.
How does this help? Now we’ll have one unknown less, since the y-component of {\vec v_2} is zero, i.e., b = 0.
Thus, (2) (4)  and (5)  reduce to
{c^2} = 1,  \,\,\,\,\,\,cr = 0 - 2,  \,\,{q^2} + {r^2} = 20
 \Rightarrow  \,\,\,\, c =  \pm \;1,  \,\,\,\,\,\,\,\,r =  \mp \;2,  \,\,q =  \pm 4
Thus, the three dimensional vectors that satisfy the given constraints can be
{\vec v_1} = 2\hat i  \,\,\,\,\,\,\,\,\,{\vec v_2} =  - \hat i + \hat k \,\,\,\,\,\,\, {\vec v_3} = 3\hat i \pm 4\hat j - 2\hat k
OR
{\vec v_1} = 2\hat i  \,\,\,\,\,\,\,\,\,{\vec v_2} =  - \hat i - \hat k \,\,\,\,\,\,\, {\vec v_3} = 3\hat i \pm 4\hat j + 2\hat k
To emphasize once again, we were required to find vectors satisfying the given constraints. This meant that absolute positions of the vectors were not important; what mattered was their relative sizes and orientation; and thus the coordinate axes was our choice. We selected it in a way which made the calculations most convenient.
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