Applications of Vectors

COLLINEAR AND COPLANAR VECTORS

METHODS TO PROVE COLLINEARITY

$\bullet$ Two vectors $\bar{p}$and $\bar{q}$are collinear if there exists k$\in$R such that $\bar{p}=k\bar{q}$.

$\bullet$ If $\bar{p}\times \bar{q}=\bar{o}then\bar{p},\bar{q}$are collinear.

$\bullet$ Three points A($\bar{a}$), B($\bar{b}$), C($\bar{c}$)are collinear if there exists k$\in$R such that $\overline{AB}=k(\overline{BC})$ that is $\bar{b}-\bar{a}=k(\bar{c}-\bar{b})$.

$\bullet$ If $(\bar{b}-\bar{a})\times (\bar{c}-\bar{b})=\bar{o}$then A, B, C are collinear.

$\bullet$ A($\bar{a}$), B($\bar{b}$), C($\bar{c}$) are collinear if there exists scalars l, m, n, (not all zero) such that $l\bar{a}+m\bar{b}+n\bar{c}=\bar{o}$where l + m + n = 0.

All the above methods are equivalent and any of them can be utilized to prove the collinearity. (of 2 vectors or 3 points)

Illustration 1: Let $\vec{a},\vec{b},\vec{c}$ be three non–zero vectors such that any two of them are non–collinear. If $\frac{a}{\sqrt{a^2+b^2+k^{2^{\prime }}}}\frac{b}{\sqrt{a^2+b^2+k^{2^{\prime }}}}\frac{k}{\sqrt{a^2+b^2+k^2}}$ is collinear with $\vec{c}$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$, , then prove that $\vec{a}+2\vec{b}+6\vec{c}=0$.

Key concept: Two vectors $\bar{p}$and $\bar{q}$are collinear if there exists k$\in$R such that $\bar{p}=k\bar{q}$.

Solution: It is given that $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$

$\Rightarrow$$\vec{a}+2\vec{b}=\lambda \vec{c}$for some scalar $\lambda$;;;;;;;;;;;;;;;…(i);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Also;$\vec{b}+3\vec{c}$ is collinear with $\vec{a}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$ $\vec{b}+3\vec{c}=\mu \vec{a}$ for some scalar $\mu$;;;;;;;;;;;;;;;;;;…(ii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;from (i) and (ii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$(1 + 2$\mu$$\mu$) $\vec{b}$ + (3 – $\mu$$\lambda$) $\vec{c}$ = 0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$1 + 2$\mu$ = 0;and;3 – $\mu$$\lambda$ = 0 {$\because$ $\vec{b}$and $\vec{c}$ are non–collinear vectors}

$\Rightarrow$ $\mu$ = – 1/2 and $\lambda$ = – 6

Substituting the values of $\lambda$ and $\mu$ in (i) & (ii), we get

$\vec{a}+2\vec{b}+6\vec{c}=0$

METHODS TO PROVE COPLANARITY

$\bullet$ Three vectors $\bar{p}\bar{q},\bar{r}$are coplanar if there exists l, m$\in$R such that $\bar{r}=l\bar{p}+m\bar{q}$ i.e., one can be expressed as a linear combination of the other two.

$\bullet$ If $[\bar{p}\bar{q}\bar{r}]=0,then\bar{p},\bar{q}and\bar{r}$ are coplanar (necessary and sufficient condition).

$\bullet$ Four points A$(\bar{a})$, B$(\bar{b})$, C$(\bar{c})$ and D$(\bar{d})$lie in the same plane if there exist l, m$\in$R such that $\overline{AB}=l(\overline{BC})+m(\overline{CD})$ i.e. $(\bar{b}-\bar{a})=l(\bar{c}-\bar{b})+m(\bar{d}-\bar{c})$.

$\bullet$ If $[\bar{b}-\bar{a}\bar{c}-\bar{b}\bar{d}-\bar{c}]$= 0 then A, B, C, D are coplanar.

$\bullet$ A, B, C, D are coplanar if there exists scalars k, l, m, n (not all zero), such that $k\bar{a}+l\bar{b}+m\bar{c}+n\bar{d}=\bar{o}$where k + l + m + n = o.

Again all the above methods are equivalent. Choose the best amongst them depending on convenience.

Illustration 2: Prove that if cos a $\ne$ 1, cos$\beta$ $\ne$ 1 and cos $\gamma$ $\ne$ 1, then the vectors $\vec{a}=\hat{i}\cos \alpha +\hat{j}+\hat{k},$$\vec{b}=\hat{i}+\hat{j}\cos \beta +\hat{k},\vec{c}=\hat{i}+\hat{j}+\hat{k}\cos \gamma$ can never be coplanar. Solution:;;;;;;;;;;;;;;;;;;Suppose that $\vec{a},\vec{b},\vec{c}$ are coplanar.;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;$\left|\begin{array}{c};\cos \alpha ;11;\\ 1\cos \beta ;1;\\ ;11\cos \gamma ;;\end{array}\right|=0$( R₂ $\to$ R₂ – R₁ and R₃ $\to$ R₃ – R₁ );;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;or;;;;;;;;$\left|\begin{array}{c};;\cos \alpha ;11;\\ ;1-\cos \alpha ;\cos \beta -10;\\ ;;1-\cos \alpha ;0\cos \gamma -1;\end{array}\right|=0$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;or;;;;;;;;cos$\alpha$ (cos$\beta$ – 1)(cos$\gamma$ – 1) – (1 – cos$\alpha$)(cos$\gamma$ – 1) – (1 – cos$\alpha$)(cos$\beta$ – 1) = 0;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;dividing through out by (1 – cos$\alpha$) (1 – cos$\beta$)(1 – cos$\gamma$); we get;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\frac{\cos \alpha }{1-\cos \alpha }+\frac{1}{1-\cos \beta }+\frac{1}{1-\cos \gamma }=0$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;or;;;;;;;;–1 + $\frac{1}{1-\cos \alpha }+\frac{1}{1-\cos \beta }+\frac{1}{1-\cos \gamma }=0$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;$\frac{1}{1-\cos \alpha }+\frac{1}{1-\cos \beta }+\frac{1}{1-\cos \gamma }=0$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;$\cos e{c}^2\frac{\alpha }{2}+\cos e{c}^2\frac{\beta }{2}+\cos e{c}^2\frac{\gamma }{2}=2$, which is not possible;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;as $\cos e{c}^2\frac{\alpha }{2}\ge 1,\cos e{c}^2\frac{\beta }{2}\ge 1,\cos e{c}^2\frac{\gamma }{2}\ge 1$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Hence they can not be coplanar;;VECTOR EQUATION OF A STRAIGHT LINE Following are the two most useful forms of the equation of a line.

(i);;;;;;;;Line passing through a given point A;;;;;;;;;;;$(\bar{a})$and parallel to a vector $(\bar{b})$: $r=\bar{a}+\lambda \bar{b}$
(ii);;;;;;Line passing through two given points A$(\bar{a})$and B$(\bar{b})$: $\bar{r}=\bar{a}+\lambda (\bar{b}-\bar{a})$For each particular value of $\lambda$, we get a particular point on the line. Each of the above equations can be written easily in Cartesian form also.For example, in case (i), writing$\bar{r}=x\hat{i}+y\hat{j}+z\hat{k},\bar{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}and\bar{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$,

we get $\frac{x-a_1}{b_1}=\frac{y-a_2}{b_2}=\frac{z-a_3}{b_3}=\lambda$$\Rightarrow$ x = a₁ + $\lambda$b₁, y = a₂ + $\lambda$b₂, z = a₃ + $\lambda$b₃. Illustration 3:;;;;;;;;;;;Given vectors $\overrightarrow{CB}=\vec{a},\overrightarrow{CA}=\vec{b}and\overrightarrow{CO}=\vec{x}$ where O is the centre of circle circumscribed about $\Delta$ABC, then find vector $\overrightarrow{c}\times \overrightarrow{a}$.Solution:;;;;;;;;;;;;;;;;;;Here, $\overrightarrow{OC}=\overrightarrow{x},\overrightarrow{CA}=\overrightarrow{b,}\overrightarrow{CB}=\overrightarrow{a}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$ $\overrightarrow{OA}=\overrightarrow{b}-\overrightarrow{x}$ and $\overrightarrow{OB}=\overrightarrow{a}-\overrightarrow{x}$ ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;Now;;;;;;;;;;;$\overrightarrow{O{A}^2}=\overrightarrow{O{B}^2}=\overrightarrow{O{C}^2}$;[as radii of circle];;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;;;;$\overrightarrow{{(x)}^2}=\overrightarrow{(a}-\overrightarrow{x{)}^2}=\overrightarrow{(b}-\overrightarrow{x{)}^2}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;or;;;;;;;;$\overrightarrow{x}\cdot \overrightarrow{x}=\overrightarrow{(a}-\overrightarrow{x)}\cdot \overrightarrow{(a}-\overrightarrow{x)}=(\vec{b}-\vec{x})\cdot (\vec{b}-\overrightarrow{x})$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;$\overrightarrow{x}\cdot \overrightarrow{x}=\overrightarrow{a}\cdot \overrightarrow{x}-2\overrightarrow{a}\cdot \overrightarrow{x}+\overrightarrow{x}\cdot \overrightarrow{x}=\vec{b}\cdot \overrightarrow{b}-2\overrightarrow{b}\cdot \overrightarrow{x}+\overrightarrow{x}\cdot \overrightarrow{x}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;$\overrightarrow{a}\cdot \overrightarrow{x}=\frac{\overrightarrow{a}}{2}$ and $\overrightarrow{b}\cdot \overrightarrow{x}=\frac{\overrightarrow{b}}{2}$;;;;;;;;;;;…(i) ;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
Now if we take $\overrightarrow{x}=\lambda \vec{a}+\overrightarrow{b}$ then from (i),;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\lambda a^2+\mu \overrightarrow{a}\cdot \overrightarrow{b}=\frac{a^2}{2}$;;;;;;;;;;;;;;;;;;;…(ii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;and ;;;$\lambda \overrightarrow{a}\cdot \overrightarrow{b}+\mu b^2=\frac{b^2}{2}$;;;;;;;;;;…(iii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\therefore$ Solving (ii) and (iii);;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\lambda =\frac{a^2{b}^2-b^2(\overrightarrow{a}\cdot \vec{b})}{2(a^2{b}^2)-{(\overrightarrow{a}\cdot \overrightarrow{b})}^2}$ and $\mu =\frac{a^2{b}^2-a^2(\overrightarrow{a}\cdot \vec{b})}{2(a^2{b}^2)-{(\overrightarrow{a}\cdot \overrightarrow{b})}^2}$;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;$\Rightarrow$;;;;;;;;;;;$\overrightarrow{x}=\frac{1}{2}\frac{a^2{b}^2-b^2(\overrightarrow{a}\cdot \vec{b})}{(a^2{b}^2)-{(\overrightarrow{a}\cdot \overrightarrow{b})}^2}\overrightarrow{a}+\frac{1}{2}\frac{a^2{b}^2-a^2(\overrightarrow{a}\cdot \vec{b})}{(a^2{b}^2)-{(\overrightarrow{a}\cdot \overrightarrow{b})}^2}.\overrightarrow{b}$
SHORTEST DISTANCE BETWEEN TWO LINES;Two lines in space can be parallel, intersecting or neither (called skew lines). Let $\bar{r}={\bar{a}}_1+\lambda {\bar{b}}_{1}and\bar{r}={\bar{a}}_2+\mu {\bar{b}}_2$be two lines.(i);;;;;;;;They intersect if $({\bar{b}}_1\times {\bar{b}}_2)\cdot ({\bar{a}}_2-{\bar{a}}_1)=0$.(ii);;;;;;;They are parallel if ${\bar{b}}_{1}and{\bar{b}}_2$are collinear. Parallel lines are of the form $\bar{r}={\bar{a}}_1+\lambda \bar{b}andr={\bar{a}}_2+\lambda \bar{b}.$ Perpendicular distance between them is constant and is equal to;;;$\left|\frac{b\times ({\bar{a}}_2-{\bar{a}}_1)}{|\bar{b}|}\right|$.(iii);;;;;;For skew lines, shortest distance between them (along common perpendicular) is given by $\left|\frac{({\bar{b}}_1\times {\bar{b}}_2)\cdot ({\bar{a}}_2-{\bar{a}}_1)}{|{\bar{b}}_1\times {\bar{b}}_2|}\right|$.;EQUATION OF A PLANE IN VECTOR FORM; Following are the four useful ways of specifying a plane.;(i);;;;;;;;A plane at a perpendicular distance d from the origin and normal to a given direction $(\hat{n})$ has the equation $(\hat{r}-d\hat{n})\cdot d\hat{n}=0$;;;;;;;;;;;or $\bar{r}\cdot \hat{n}=d$ ($\hat{n}$ is a unit vector).;
(ii);;;;;;;A plane passing through the point A$(\bar{a})$and normal to $\hat{n}$ has the equation $(\bar{r}-\bar{a})\cdot \hat{n}=0$.

(iii);;;;;;Parameteric equation of the plane passing through A$(\bar{a})$and parallel to the plane of vectors $(\bar{b})and(\bar{c})$is given by $\bar{r}=\bar{a}+\lambda \bar{b}+\mu \bar{c}$$\Rightarrow$. $\overline{r}.\left(\overline{b}\times \overline{c}\right)=\left[\overline{a}\overline{b}\overline{c}\right]$

(iv) Parameteric equation of the plane passing through A$(\bar{a})$, B$(\bar{b}),$ C$(\bar{c})$(A, B, C non-collinear) is given by $\bar{r}=(1-\lambda -\mu )\bar{a}+\lambda \bar{b}+\mu \bar{c}$ $\Rightarrow$ $\bar{r}\cdot [\bar{b}\times \bar{c}+\bar{c}\times \bar{a}+\bar{a}\times \bar{b}]=[\bar{a}\bar{b}\bar{c}]$.

In Cartesian form, the equation of the plane assumes the form Ax + By + Cz = D. The vector normal to this plane is $A\hat{i}+B\hat{j}+C\hat{k}$and the perpendicular distance of the plane from the origin is $\frac{|D|}{\sqrt{A^2+B^2+C^2}}$.

Angle Between a Line and a Plane:

The angle between a line and a plane is the complement of the angle between the line and the normal to the plane.

Angle Between Two Planes:

It is equal to the angle between their normal unit vectors ${\hat{n}}_{1}and{\hat{n}}_2$. i.e. cos$\theta$ = ${\hat{n}}_1\cdot {\hat{n}}_2$

SOME MISCELLANEOUS RESULTS

(i) Volume of the tetrahedron ABCD = $\frac{1}{6}[\overline{AB}\overline{AC}\overline{AD}].$

(ii) Area of the quadrilateral with diagonals

$\overline{d_1}and\overline{d_2}=\frac{1}{2}\left|\overline{d_1}\times \overline{d_2}\right|.$

RECIPROCAL SYSTEM OF VECTORS

If $\bar{a},\bar{b},\bar{c}$ are three non-coplanar vectors, then a system of vectors $\bar{a},\bar{b},\bar{c}$defined by ${\bar{a}}^{\prime }=\frac{\bar{b}\times \bar{c}}{[\bar{a}\bar{b}\bar{c}]},{\bar{b}}^{\prime }=\frac{\bar{c}\times \bar{a}}{[\bar{a}\bar{b}\bar{c}]},{\bar{c}}^{\prime }=\frac{\bar{a}\times \bar{b}}{[\bar{a}\bar{b}\bar{c}]}$is called the reciprocal system of vectors because $\bar{a}\cdot {\bar{a}}^{\prime }=\bar{b}\cdot {\bar{b}}^{\prime }=\bar{c}\cdot {\bar{c}}^{\prime }=1$.

Further $[{\bar{a}}^{\prime }{\bar{b}}^{\prime }{\bar{c}}^{\prime }]=\frac{1}{[abc]}$

The scalar product of any vector of one system with a vector of other system which does not correspond to it is zero i.e.

$\bar{a}\cdot {\bar{b}}^{\prime }=\bar{a}\cdot {\bar{c}}^{\prime }=\bar{b}\cdot {\bar{c}}^{\prime }=\bar{c}\cdot {\bar{a}}^{\prime }=\cdots =0$

If ${\bar{a}}^{\prime },{\bar{b}}^{\prime },{\bar{c}}^{\prime }$is a reciprocal system to $\bar{a},\bar{b},\bar{c}$ then $\bar{a},\bar{b},\bar{c}$ is also reciprocal system to ${\bar{a}}^{\prime },{\bar{b}}^{\prime },{\bar{c}}^{\prime }$.

Illustration 4: Show that the points $i-j+3k$and $3\hat{i}+3\hat{j}+3\hat{k}$ are equi-distant from the plane r × (5i + 2j – 7k) + 9 = 0 and are on the opposite sides of it.

Solution: The given plane is $r\cdot (5\hat{i}+2\hat{j}-7\hat{k})$ = -9

Length of the perpendicular from $\hat{i}-\hat{j}+3\hat{k}$to it is

$\frac{-9-(i-j+3k)\cdot (5i+2j-7k)}{\sqrt{25+4+49}}=\frac{-9-5+2+21}{\sqrt{78}}=\frac{9}{\sqrt{78}}$

Length of the perpendicular from $3\hat{i}+3\hat{j}+3\hat{k}$

= $\frac{-9-(3i+3j+3k)\cdot (5i+2j-7k)}{\sqrt{78}}$

= $\frac{-9-15-6+21}{\sqrt{78}}=\frac{-9}{\sqrt{78}}$

Thus the length of the two perpendiculars are equal in magnitude but opposite in sign. Hence they are located on opposite sides of the plane.