ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 18 -Worked Out Examples

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.

Friday, 8 August 2014

CHAPTER 18 -Worked Out Examples

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