ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 19 -Worked Out Examples 2

Friday, 8 August 2014

CHAPTER 19 -Worked Out Examples 2

Example: 21

If $a$ , $b$ , $c$ are the lengths of the sides of $\Delta ABC$ opposite to the angles $A$ , $B$ and $C$ 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.

Friday, 8 August 2014

CHAPTER 19 -Worked Out Examples 2

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