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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.