Maths Solve the Following Question.(4 Marks)

Let O-ABC be a tetrahedron. Then o (OA, BC), (OB, CA) and (OC, AB) are the pair of opposite edges.

Take $O$ as the origin of reference and let $\bar{a} \bar{b}$ and

$\begin{aligned} & \overline{\mathrm{OA}}=\bar{a}, \overline{\mathrm{OB}}=\bar{b}, \overline{\mathrm{OC}}=\bar{c} \\ & \overline{\mathrm{AB}}=\bar{b}-\bar{a}, \overline{\mathrm{BC}}=\bar{c}-\bar{b} \text { and } \overline{\mathrm{CA}}=\bar{a}-\bar{c}\end{aligned}$

Now, suppose the pairs $(O A, B C)$ and $(O B, C A)$ are

perpendicular to each other.

Then $\overline{\mathrm{OA}} \cdot \overline{\mathrm{BC}}=0$, i.e. $\bar{a} \cdot(\bar{c}-\bar{b})=0$

$\therefore \bar{a} \cdot \bar{c}-\bar{a} \cdot \bar{b}=0$

... (1)

and $\overline{\mathrm{OB}} \cdot \overline{\mathrm{CA}}=0$, i.e. $\bar{b} \cdot(\bar{a}-\bar{c})=0$

$\begin{aligned} & \therefore \bar{b} \cdot \bar{a}-\bar{b} \cdot \bar{c}=0 \\ & \therefore \bar{a} \cdot \bar{b}-\bar{b} \cdot \bar{c}=0\end{aligned}$

$\ldots(2)$

Adding (1) and (2), we get

$\begin{aligned} & \bar{a} \cdot \bar{c}-\bar{b} \cdot \bar{c}=0 \\ \therefore & \bar{c} \cdot \bar{b}-\bar{c} \cdot \bar{a}=0\end{aligned}$

$\begin{aligned} & \text { i.e. } \bar{c} \cdot(\bar{b}-\bar{a})=0 \\ & \therefore \overline{\mathrm{OC}} \cdot \overline{\mathrm{AB}}=0\end{aligned}$

∴ the third pair (OC, AB) is perpendicular.

points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ arid $\mathrm{D}$ respectively.

$\therefore \overline{\mathrm{AB}}=\bar{b}-\bar{a}, \overline{\mathrm{AC}}=\overline{\mathrm{c}}-\bar{a}, \overline{\mathrm{AD}}=\bar{d}-\bar{a}$

The points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are coplanar.

$\therefore$ the vectors $\overline{\mathrm{AB}}, \overline{\mathrm{AC}}, \overline{\mathrm{AD}}$ are woplanar.

$\begin{aligned} & \therefore[\overline{\mathrm{AB}} \overline{\mathrm{AC}} \overline{\mathrm{AD}}]=0 \\ & \therefore[\bar{b}-\bar{a} \bar{c}-\bar{a} \bar{d}-\bar{a}]=0 \\ & \therefore(\bar{b}-\bar{a}) \cdot[(\bar{c}-\bar{a}) \times(\bar{d}-\bar{a})]=0 \\ & \therefore(\bar{b}-\bar{a}) \cdot(\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{a} \times \bar{d}+\bar{a} \times \bar{a})=0\end{aligned}$

where $\bar{a} \times \bar{a}=\overline{0}$

$\therefore(\bar{b}-\bar{a}) \cdot(\bar{c} \times \bar{d}-\bar{c} \times \bar{a}-\bar{a} \times \bar{d})=0$

$\begin{array}{r}\therefore \bar{b} \cdot(\bar{c} \times \bar{d})-\bar{b} \cdot(\bar{c} \times \bar{a})-\bar{b} \cdot(\bar{a} \times \bar{d})-\bar{a} \cdot(\bar{c} \times \bar{d})+ \\ \bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{a} \times \bar{d})=0 \quad \ldots(1)\end{array}$

Now, $\bar{a} \cdot(\bar{c} \times \bar{a})=0, \bar{a} \cdot(\bar{a} \times \bar{d})=0$,

$\begin{aligned} & -\bar{b} \cdot(\bar{a} \times \bar{d})=\bar{b} \cdot(\bar{d} \times \bar{a})=[\bar{b} \bar{d} \bar{a}]=[\bar{a} \bar{b} \bar{d}], \\ & -\bar{a} \cdot(\bar{c} \times \bar{d})=\bar{a} \cdot(\bar{d} \times \bar{c})=[\bar{a} \bar{d} \bar{c}]=[\bar{c} \bar{a} \bar{d}]\end{aligned}$

Also, $\bar{b} \cdot(\bar{c} \times \bar{d})=[\bar{b} \bar{c} \bar{d}]$,

$-\bar{b} \cdot(\bar{c} \times \bar{a})=-\bar{a} \cdot(\bar{b} \times \bar{c})=-[\bar{a} \bar{b} \bar{c}]$

$\therefore$ from (1)

$\begin{aligned} & {[\bar{b} \bar{c} \bar{d}]-[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{d}]+[\bar{c} \bar{a} \bar{d}]+0+0=0} \\ & \therefore[\bar{a} \bar{b} \bar{d}]+[\bar{b} \bar{c} \bar{d}]+[\bar{c} \bar{a} \bar{d}]=[\bar{a} \bar{b} \bar{c}] .\end{aligned}$

$\begin{aligned} & =\frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{c} \bar{a} \bar{b}]}+\frac{[\bar{b} \bar{a} \bar{c}]}{[\bar{c} \bar{a} \bar{b}]} \\ & =\frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}-\frac{[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]} \\ & =0=\text { RHS. }\end{aligned}$

Take origin O as one vertex of the cube and OA, OB and OC as the positive directions of the X-axis, the Y-axis and the Z-axis respectively. Here, the sides of the cube are OA = OB = OC = a ∴ the coordinates of all the vertices of the cube will be O = (0, 0, 0) A = (a, 0, 0) B = (0, a, 0) C = (0, 0, a) N = (a, a, 0) L = (0, a, a) M = (a, 0, a) P = (a, a, a) ON, OL, OM are the three diagonals which meet at the vertex O

$\begin{aligned} & \overline{\mathrm{ON}}=\bar{a} \hat{i}+\bar{a} \hat{j}, \quad \overline{\mathrm{OL}}=\bar{a}+a \hat{j} \\ & \overline{\mathrm{OM}}=\bar{a} \hat{i}+a \hat{k} .\end{aligned}$

$[\overline{\mathrm{ON}} \overline{\mathrm{OL}} \overline{\mathrm{OM}}]=\left|\begin{array}{lll}a & a & 0 \\ 0 & a & a \\ a & 0 & a\end{array}\right|$

$\begin{aligned} & =a\left(a^2-0\right)-a\left(0-a^2\right)+0 \\ & =a^3+a^3=2 a^3\end{aligned}$

$\therefore$ required volume $=[\overline{\mathrm{ON}} \overline{\mathrm{OL}} \overline{\mathrm{OM}}]$

$=2 a^3$ cubic units.

Let $\mathrm{V}$ be the volume of the parallelopiped formed by $\bar{p}, \bar{q}, \bar{r}$.

Then $V=[\bar{p} \bar{q} \bar{r}]$

$=\left|\begin{array}{lll}1 & a & 1 \\ 0 & 1 & a \\ a & 0 & 1\end{array}\right|$

$\begin{aligned} & =1(1-0)-a\left(0-a^2\right)+1(0-a) \\ & =1+a^3-a\end{aligned}$

$\begin{aligned} \therefore \frac{d V}{d a} & =\frac{d}{d a}\left(1+a^3-a\right) \\ & =0+3 a^2-1=3 a^2-1\end{aligned}$

and $\begin{aligned} \frac{d^2 V}{d a^2} & =\frac{d}{d a}\left(3 a^2-1\right) \\ & =3 \times 2 a-0=6 a\end{aligned}$

For maximum and minimum $V, \frac{d V}{d a}=0$

$\begin{aligned} & \therefore 3 a^2-1=0 \\ & \therefore a^2=\frac{1}{3} \quad \therefore a= \pm \frac{1}{\sqrt{3}}\end{aligned}$

Now, $\left(\frac{d^2 V}{d a^2}\right)_{\text {at } a=\frac{1}{\sqrt{3}}}=6\left(\frac{1}{\sqrt{3}}\right)=2 \sqrt{3}>0$

$\therefore V$ is minimum when $a=\frac{1}{\sqrt{3}}$

Also, $\left(\frac{d^2 V}{d a^2}\right)_{\text {at } a=-\frac{1}{\sqrt{3}}}=6\left(-\frac{1}{\sqrt{3}}\right)=-2 \sqrt{3}<0$

$\therefore V$ is maximum when $a=-\frac{1}{\sqrt{3}}$

Hence, $a=\frac{1}{\sqrt{3}}$

$\mathrm{B}(0, b, 0), \mathrm{C}(0,0, c)$ are

$\begin{aligned} & \bar{p}=a \hat{i}, \bar{q}=b \hat{j}, \bar{r}=c \hat{k} \\ & \overline{\mathrm{AB}}=\bar{q}-\bar{p}=b \hat{j}-a \hat{i}=-a \hat{i}+b \hat{j}\end{aligned}$

$\overline{\mathrm{BC}}=\bar{r}-\bar{q}=c \hat{k}-b \hat{j}=-b \hat{j}+c \hat{k}$

$\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ -a & b & 0 \\ 0 & -b & c\end{array}\right|$

$\begin{aligned} & =(b c-0) \hat{i}-(-a c-0) \hat{j}+(a b-0) \hat{k} \\ & =b c \hat{i}+a c \hat{j}+a b \hat{k}\end{aligned}$

$\begin{aligned}|\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}| & =\sqrt{(b c)^2+(a c)^2+(a b)^2} \\ & =\sqrt{b^2 c^2+a^2 c^2+a^2 b^2}\end{aligned}$

$\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}$ is perpendicular to the plane containing $\mathrm{A}, \mathrm{B}, \mathrm{C}$.

$\therefore$ the required unit vector

$=\frac{\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}}{|\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}|}=\frac{b c \hat{i}+c a \hat{j}+a b \hat{k}}{\sqrt{b^2 c^2+c^2 a^2+a^2 b^2}}$

Area of $\triangle \mathrm{ABC}=\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{BC}}|$

$=\frac{1}{2} \sqrt{b^2 c^2+a^2 c^2+a^2 b^2}$ sq units.

$\therefore|\bar{a}|=|\bar{b}|=|\bar{c}|=1$.

Also, $\bar{a} \cdot \bar{a}=\bar{b} \cdot \bar{b}=\bar{c} \cdot \bar{c}=1$

Now, $\bar{a}+\bar{b}+\bar{c}=\overline{0}$

$\ldots$ (1)

Taking scalar product of both sides with $\bar{a}$, we get

$\bar{a} \cdot(\bar{a}+\bar{b}+\bar{c})=\bar{a} \cdot \overline{0}$

$\begin{aligned} & \therefore \bar{a} \cdot \bar{a}+\bar{a} \cdot \bar{b}+\bar{a} \cdot \bar{c}=0 \\ & \therefore \bar{a} \cdot \bar{b}+\bar{a} \cdot \bar{c}=-\bar{a} \cdot \bar{a}=-1\end{aligned}$

$\ldots(2)$

Similarly taking scalar product of both sides of (1) with

$\bar{b}$ and $\bar{c}$, we get

$\ldots(3)$

$\bar{c} \cdot \bar{a}+\bar{c} \cdot \bar{b}=-1$

$\ldots(4)$

Adding (2), (3), (4) and using the fact that scalar product commutative, we get

$\begin{aligned} & 2(\bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a})=-3 \\ & \therefore \bar{a} \cdot \bar{b}+\bar{b} \cdot \bar{c}+\bar{c} \cdot \bar{a}=-\frac{3}{2}\end{aligned}$

Let $\bar{r}=x \hat{i}+y \hat{j}+z \hat{k}$ be the required vector.

Then $\bar{r} \cdot \bar{a}=-1, \bar{r} \cdot \bar{b}=6, \bar{r} \cdot \bar{c}=5$

$\begin{aligned} \therefore & (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-5 \hat{k})=-1 \\ & (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+7 \hat{j})=6 \text { and } \\ & (x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k})=5\end{aligned}$

∴ 3x – 5z= -1 … (1) ∴ 2x + 7y = 6 … (2) ∴ x + y + z = 5 … (3) From (3), z = 5 – x – y Substituting this value of z in (1), we get ∴ 3x – 5(5 – x – y)= -1 ∴ 8x + 5y = 24 … (4) Multiplying (2) by 4 and subtracting from (4), we get 8x + 5y – 4(2x + 7y) = 24 – 6 × 4 ∴ -23y = 0 ∴ y = 0 Substituting y = 0 in (2), we get ∴ 2x = 6 ∴ x = 3 Substituting x = 3 in (1), we get ∴ 3(3) – 5z = -1 ∴ 5z = -10 ∴ z = 2

$\therefore \bar{r}=3 \hat{i}+0 \cdot \hat{j}+2 \hat{k}=3 \hat{i}+2 \hat{k}$

Hence, the required vector is $3 \hat{i}+2 \hat{k}$

Let $\mathrm{A}, \mathrm{B}, \mathrm{D}, \mathrm{E}, \mathrm{P}$ have position vectors $\bar{a}_t \bar{b}_t \bar{d}_t \bar{e}_t \bar{p}$ respectively w.r.t. O,

∵ AD : DB = 2 : 1. ∴ D divides AB internally in the ratio 2 : 1. Using section formula for internal division, we get

$\begin{aligned} & \bar{d}=\frac{2 \bar{b}+\bar{a}}{2+1} \\ & \therefore 3 \bar{d}=2 \bar{b}+\bar{a}\end{aligned}$

$\ldots(1)$

Since $\mathrm{E}$ is the midpoint of $\mathrm{OB}, \bar{e}=\overline{\mathrm{OE}}=\frac{1}{2} \overline{\mathrm{OB}}=\frac{\bar{b}}{2}$

$\therefore \bar{b}=2 \bar{e}$

$\ldots(2)$

$\therefore$ from (1),

$\begin{aligned} 3 \bar{d} & =2(2 \bar{e})+\bar{a} \\ & =4 \bar{e}+\bar{a}\end{aligned}$

$\therefore \frac{3 \vec{d}+2 \cdot \overline{0}}{3+2}=\frac{4 e+\bar{a}}{4+1}$

$\ldots$... By (2)]

$\therefore \frac{3 \vec{d}+2 \cdot \overline{0}}{3+2}=\frac{4 e+\bar{a}}{4+1}$

LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.

Choose any point P on the angle bisector of ∠AOB. Draw PM parallel to OB.
∴ ∠OPM = ∠POM
= ∠POB
Hence, OM = MP

$\therefore O M$ and MP is the same scalar multiple of unit vectors $\hat{a}$ and $\hat{b}$ along these directions,

where $\hat{a}=\frac{\bar{a}}{|\bar{a}|}$ and $\hat{b}=\frac{\vec{b}}{|\bar{b}|}$

$\begin{aligned} & \therefore \overline{\mathrm{OM}}=\hat{\lambda} \text { and } \overline{\mathrm{MP}}=\lambda \hat{b} \\ & \therefore \overline{\mathrm{OP}}=\overline{\mathrm{OM}}+\overline{\mathrm{MP}}=\lambda \hat{a}+\lambda \hat{b}\end{aligned}$

$=\lambda(\hat{a}+\hat{b})$

Hence, the vector along angle bisector of $\angle A O B$ is given

by

$\bar{d}=\overline{\mathrm{OP}}=\lambda\left(\frac{\bar{a}}{|\bar{a}|}+\frac{\bar{b}}{|\bar{b}|}\right)$

Slope of tangent at $P(2,4)=\left(\frac{d y}{d x}\right)_{\text {at } \mathrm{P}(2,4)}=2 \times 2=4$

$\therefore$ the equation of tangent at $P$ is

$\begin{aligned} & y-4=4(-2) \\ & \therefore y=4 x-4\end{aligned}$

∴ y = 4x is equation of line parallel to the tangent at P and passing through the origin O.

$4 x=y, z=0 \therefore \frac{x}{1}=\frac{y}{4}, z=0$

∴ the direction ratios of this line are 1, 4, 0 ∴ its direction cosines are

$\pm \frac{1}{\sqrt{1^2+4^2+0^2}}, \pm \frac{4}{\sqrt{1^2+4^2+0^2}}, 0$

i.e. $\pm \frac{1}{\sqrt{17}}, \pm \frac{4}{\sqrt{17}}, 0$

$\therefore$ unit vectors parallel to tangent line at $P(2,4)$ is

$\pm \frac{1}{\sqrt{17}}(\hat{i}+4 \hat{j})$

Since $\bar{a}$ lies in YZ-plane, it is perpendicular to $X$-axis

$\therefore \alpha=90^{\circ}$

It is given that $\beta=60^{\circ}$

$\begin{aligned} & \because \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \therefore \cos ^2 90^{\circ}+\cos ^2 60^{\circ}+\cos ^2 \gamma=1 \\ & \therefore 0+\left(\frac{1}{2}\right)^2+\cos ^2 \gamma=1 \\ & \therefore \cos ^2 \gamma=1-\frac{1}{4}=\frac{3}{4} \\ & \therefore \cos \gamma= \pm \frac{\sqrt{3}}{2}\end{aligned}$

Unit vector along a is given by

$\vec{a}=(\cos \alpha) \hat{i}+(\cos \beta) \hat{j}+(\cos \gamma) \hat{k}=0-\hat{i}+\frac{1}{2} \hat{j}+\frac{\sqrt{3}}{2} \hat{k}$

$=\frac{1}{2} \hat{j} \pm \frac{\sqrt{3}}{2} \hat{k}$

$\therefore \bar{a}=|\bar{a}| \hat{a}=4\left(\frac{1}{2} \hat{j} \pm \frac{\sqrt{3}}{2} \hat{k}\right) \quad \ldots[\because|\bar{a}|=4]$

$\therefore \bar{a}=2 \hat{j} \pm 2 \sqrt{3} \hat{k}$

$
\begin{aligned}
& \quad=(3+10) \hat{i}-(-6-15) \hat{j}+(4-3) \hat{k} \\
& \quad=13 \hat{i}+21 \hat{j}+\hat{k} \\
& \therefore|\overline{P Q} \times \overline{P R}|=\sqrt{169+441+1}=\sqrt{611}
\end{aligned}
$
If $\hat{n}$ is a unit vector perpendicular to $\overline{P Q}$ and $\overline{P R}$, then
$
\hat{n}=\frac{\overline{P Q} \times \overline{P R}}{|\overline{P Q} \times \overline{P R}|}=\frac{13 \hat{i}+21 \hat{j}+\hat{k}}{\sqrt{611}}
$
If $\theta$ is the angle between $\overline{P Q}$ and $\overline{P R}$ then $\sin \theta=\left|\frac{\overline{P Q} \times \overline{P R}}{\overline{|\overline{P Q}|}|\overline{\mid P R}|}\right|=\frac{\sqrt{611}}{\sqrt{30} \sqrt{22}}$

Now equation (2) gives $t=-\frac{5}{11}$
Put $t=-\frac{5}{11}$ in equation (1) and equation (3), we get $q=9$ and $p=-4$.
The negative sign of $t$ suggests that the point $\mathrm{C}$ divides the line segment AB externally in the ratio $5: 11$.

Let M be the foot of the perpendicular drawn from B to the line joining O and A. Let M = (x, y, z) OM has direction ratios x – 0, y – 0, z – 0 = x, y, z OA has direction ratios 1 – 0, 2 – 0, 3 – 0 = 1, 2, 3 But O, M, A are collinear.

$\therefore \frac{x}{1}=\frac{y}{2}=\frac{z}{3}=k \ldots($ Let $)$

∴ x = k, y = 2k, z = 3k ∴ M = (k, 2k, 3k) ∵ BM has direction ratios k – 4, 2k – 5, 3k – 6 BM is perpendicular to OA ∴ (l)(k – 4) + 2(2k – 5) + 3(3k – 6) ∴ = k – 4 + 4k – 10 + 9k – 18 = 0 ∴ 14k = 32

$\begin{aligned} & \therefore \mathrm{k}=\frac{16}{7} \\ & \therefore M=(k, 2 k, 3 k)=\left(\frac{16}{7}, \frac{32}{7}, \frac{48}{7}\right)\end{aligned}$

From (1), $\mathrm{n}=-\left(\frac{a l+b m}{c}\right) \ldots . .(3)$

Substituting this value of n in equation (2), we get (fm + gl)∙[latex]-\left(\frac{a l+b m}{c}\right)[/latex] + hlm = 0

$\begin{aligned} & \left.\therefore-\left(a f l m+b f m^2+a g\right)^2+b g l m\right)+c h l m=0 \\ & \left.\therefore a g\right|^2+(a f+b g-c h) l m+b f m^2=0 \ldots(4)\end{aligned}$

Note that both l and m cannot be zero, because if l = m = 0, then from (3), we get

$n=0$, which is not possible as $1^2+m^2+n^2=1$.

Let us take m # 0.

Dividing equation (4) by $\mathrm{m}^2$, we get

$a g\left(\frac{l}{m}\right)^2+(a f+b g-c h)\left(\frac{l}{m}\right)+b f=0 \ldots(5)$

This is quadratic equation in $\left(\frac{l}{m}\right)$.

If $I_1, m_1, n_1$ and $I_2, m_2, n_2$ are the direction cosines of the two lines given by the equation

(1) and $(2)$, then $\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$ are the roots of the equation (5).

From the quadratic equation (5), we get

product of roots $=\frac{l_1}{m_1} \cdot \frac{l_2}{m_2}=\frac{b f}{a g}$

$\therefore \frac{l_1 l_2}{m_1 m_2}=\frac{(f / a)}{(g / b)}$

$\therefore \frac{l_1 l_2}{(f / a)}=\frac{m_1 m_2}{(g / b)}$

Similarly, we can show that,

$\frac{l_1 l_2}{(f / a)}=\frac{n_1 n_2}{(h / c)}$

$\therefore \frac{l_1 l_2}{(f / a)}=\frac{m_1 m_2}{(g / b)}=\frac{n_1 n_2}{(h / c)}=\lambda \quad \ldots$ (Say)

$\therefore l_1 l_2=\lambda\left(\frac{f}{a}\right), m_1 m_2=\lambda\left(\frac{g}{b}\right), n_1 n_2=\lambda\left(\frac{h}{c}\right)$

Now, the lines are perpendicular if

$l_1 l_2+m_1 m_2+n_1 n_2=0$

i.e. if $\lambda\left(\frac{f}{a}\right)+\lambda\left(\frac{g}{b}\right)+\lambda\left(\frac{h}{c}\right)=0$

i.e. if $\frac{f}{a}+\frac{g}{b}+\frac{h}{c}=0$.

Let $\angle X O P$ and $\angle X O Q$ be in standard position and $m \angle X O P=-\alpha, m \angle X O Q=\beta$.

Take a point A on ray OP and a point B on ray OQ such that OA = OB = 1. Since cos (-α) = cos α and sin (-α) = -sin α,

$\mathrm{A}$ is $(\cos (-\alpha), \sin (-\alpha))$

i.e. $\left(\cos \alpha_1-\sin \alpha\right)$

$B$ is $\left(\cos \beta_1 \sin \beta\right)$

$\therefore \overline{\mathrm{OA}}=(\cos \alpha) \bar{i}-(\sin \alpha) \cdot \bar{j}+0 \cdot \bar{k}$

$\overline{\mathrm{OB}}=(\cos \beta) \cdot \bar{i}+(\sin \beta) \cdot \bar{j}+0 \cdot \bar{k}$

$\therefore \overline{\mathrm{OA}} \times \overline{\mathrm{OB}}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \cos \alpha & -\sin \alpha & 0 \\ \cos \beta & \sin \beta & 0\end{array}\right|$

$=(\cos \alpha \sin \beta+\sin \alpha \cos \beta) \bar{k}$

... (1)

The angle between $\overline{\mathrm{OA}}$ and $\overline{\mathrm{OB}}$ is $\alpha+\beta$.

Also $\overline{\mathrm{OA}}, \overline{\mathrm{OB}}$ lie in the $\mathrm{XY}$-plane.

$\therefore$ the unit vector perpendicular to $\mathrm{OA}$ and $\mathrm{OB}$ is $\bar{k}$.

$\therefore \overline{\mathrm{OA}} \times \overline{\mathrm{OB}}=[\mathrm{OA} \cdot \mathrm{OB} \sin (\alpha+\beta)] \bar{k}$

$=\sin (\alpha+\beta) \cdot \bar{k}_{\ldots(2)}$

∴ from (1) and (2), sin (α + β) = sin α cos β + cos α sin β.

Let $\bar{b}=x \hat{i}+y \hat{j}+z \hat{k}$

Then $\bar{a} \cdot \bar{b}=3$ gives

$(\hat{i}+\hat{j}+\hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=3$

$\therefore(1)(x)+(1)(y)+(1)(z)=3$

Also, $x+y+z=3$

... (1)

Also, $\bar{c}=\bar{a} \times \bar{b}$

$\begin{aligned} \therefore \hat{j}-\hat{k} & =\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z\end{array}\right| \\ & =(z-y) \hat{i}-(z-x) \hat{j}+(y-x) \hat{k} \\ & =(z-y) \hat{i}+(x-z) \hat{j}+(y-x) \hat{k}\end{aligned}$

By equality of vectors, z – y = 0 ….(2) x – z = 1 ……(3) y – x = -1 ……(4) From (2), y = z. From (3), x = 1 + z Substituting these values of x and y in (1), we get

$\begin{aligned} & 1+z+z+z=3 \therefore z=\frac{2}{3} \\ & \therefore y=z=\frac{2}{3} \\ & \therefore x=1+z=1+\frac{2}{3}=\frac{5}{3}\end{aligned}$

$\begin{aligned} & \therefore \bar{b}=\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k} \\ & \text { i.e. } \bar{b}=\frac{1}{3}(5 \hat{i}+2 \hat{j}+2 \hat{k}) .\end{aligned}$

and $\bar{a} \times(\bar{b}+\bar{c})=\left|\begin{array}{rrr}\vec{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 4 & -3 & 1\end{array}\right|$

$\begin{aligned} & =(-6+12) \hat{i}-(3-15) \hat{j}+(-4+10) \hat{k} \\ & =6 \hat{i}+12 \hat{j}+6 \hat{k}\end{aligned}$

$\ldots(1)$

Also, $\bar{a} \times \bar{b}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 4 & -3 & 1\end{array}\right|$

$\begin{aligned} & =(-2+9) \hat{i}-(1-12) \hat{j}+(-3+8) \hat{k} \\ & =7 \hat{i}+11 \hat{j}+5 \hat{k}\end{aligned}$

and $\bar{a} \times \bar{c}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 3 \\ 1 & -1 & 2\end{array}\right|$

$\begin{aligned} & =(-4+3) \hat{i}-(2-3) \hat{j}+(-1+2) \hat{k} \\ & =-\hat{i}+\hat{j}+\hat{k}\end{aligned}$

$\begin{aligned} \therefore \bar{a} \times \bar{b}+\bar{a} \times \bar{c} & =(7 \hat{i}+11 \hat{j}+5 \hat{k})+(-\hat{i}+\hat{j}+\hat{k}) \\ & =6 \hat{i}+12 \hat{j}+6 \hat{k}\end{aligned}$

$\ldots(2)$

From (1) and (2), we get

$\bar{a} \times(\bar{b}+\bar{c})=\bar{a} \times \bar{b}+\bar{a} \times \bar{c}$

$\begin{aligned} & \therefore-25 \ln -15 n^2-2 n|-30|^2-18 \ln =0 \\ & \therefore-\left.30\right|^2-45 \ln -15 n^2=0 \\ & \left.\therefore 2\right|^2+3 \ln +n^2=0 \\ & \left.\therefore 2\right|^2+2 \ln +\ln +n^2=0\end{aligned}$

∴ 2l(l + n) + n(l + n) = 0
∴ (l + n)(2l + n) = 0
∴ l + n = 0 or 2l + n = 0
l = -n or n = -2l
Now, m = -(5l + 3n), therefore, if l = -n,
m = -(-5n + 3n) = 2n

$\begin{aligned} & \therefore-1=\frac{m}{2}=n \\ & \therefore \frac{l}{-1}=\frac{m}{2}=\frac{n}{1}\end{aligned}$

∴ the direction ratios of the first line are

$\begin{aligned} & a_1=-1, b_1=2, c_1=1 \\ & \text { If } n=-2 l_1 m=-(51-6 l)-1\end{aligned}$

$\begin{aligned} & \therefore 1=m=\frac{n}{-2} \\ & \therefore \frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\end{aligned}$

$\begin{aligned} & \therefore 1=m=\frac{n}{-2} \\ & \therefore \frac{l}{1}=\frac{m}{1}=\frac{n}{-2}\end{aligned}$

$\therefore$ the direction ratios of the second line are

$a_2=-1, b_2=1, c_2=-2$

Let θ be the angle between the lines.

Then $\cos \theta=\left|\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1{ }^2+c_1^2} \cdot \sqrt{a_2{ }^2+b_2{ }^2+c_2^2}}\right|$

$=\left|\frac{(-1)(1)+2(1)+1(-2)}{\sqrt{(-1)^2+2^2+1^2} \cdot \sqrt{1^2+1^2+(-2)^2}}\right|$

$=\left|\frac{-1+2-2}{\sqrt{6} \cdot \sqrt{6}}\right|=\left|\frac{-1}{6}\right|=\frac{1}{6}$

$\therefore \theta=\cos ^{-1}\left(\frac{1}{6}\right)$

Let the triangle be denoted by ABC,

where $\overline{\mathrm{AB}}=\bar{p}, \overline{\mathrm{AC}}=\bar{q}$ and $\overline{\mathrm{BC}}=\bar{r}$

$\because \bar{p}, \bar{r}, \bar{r}$ are unit vectors.

∴ l(AB) = l(BC) = l(CA) = 1 ∴ the triangle is equilateral ∴ ∠A = ∠B = ∠C = 60°

Using the formula for angle between two vectors,

$\begin{aligned} & \cos A=\frac{\bar{p} \cdot \bar{q}}{|\bar{p}||\bar{q}|} \\ & \therefore \cos 60^{\circ}=\frac{\bar{p} \cdot \bar{q}}{1 \times 1}\end{aligned}$

$\begin{aligned} & \therefore \frac{1}{2}=\bar{p} \cdot \bar{q} \\ & \therefore \bar{p} \cdot \bar{q}=\frac{1}{2} .\end{aligned}$

2. Using $\cos B=\frac{\bar{p} \cdot \bar{r}}{|\bar{p} \| \bar{r}|}$, we get $\bar{p} \cdot \bar{r}=\frac{1}{2}$

$\cos 60^{\circ}=\frac{\bar{p} \cdot \bar{r}}{1 \times 1} \quad \therefore \frac{1}{2}=\vec{p} \cdot \vec{r}$

$\therefore \bar{p} \cdot \bar{r}=\frac{1}{2}$.

$\begin{aligned} & \bar{p}=-\hat{j}-2 \hat{k}, \bar{q}=3 \hat{i}+\hat{j}+4 \hat{k} \\ & \bar{r}=5 \hat{i}+7 \hat{j}+\hat{k}\end{aligned}$

$\begin{aligned} & \therefore \overline{\mathrm{PQ}}=\bar{q}-\bar{p}=(3 \hat{i}+\hat{j}+4 \hat{k})-(-\hat{j}-2 \hat{k})=3 \hat{i}+2 \hat{j}+6 \hat{k} \\ & \text { and } \overline{\mathrm{PR}}=\bar{r}-\bar{p}=(5 \hat{i}+7 \hat{j}+\hat{k})-(-\hat{j}-2 \hat{k})=5 \hat{i}+8 \hat{j}+3 \hat{k} \\ & \begin{aligned} & \therefore \overline{\mathrm{PQ}} \cdot \overline{\mathrm{PR}}=(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(5 \hat{i}+8 \hat{j}+3 \hat{k}) \\ &=(3)(5)+(2)(8)+(6)(3)\end{aligned}\end{aligned}$

$=15+16+18=49$

$\begin{aligned} & |\overline{\mathrm{PQ}}|=\sqrt{3^2+2^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7 \\ & |\overline{\mathrm{PR}}|=\sqrt{5^2+8^2+3^2}=\sqrt{25+64+9}=\sqrt{98}=7 \sqrt{2}\end{aligned}$

Using the formula for angle between two vectors,

$\cos \mathrm{P}=\frac{\overline{\mathrm{PQ}} \cdot \overline{\mathrm{PR}}}{|\overline{\mathrm{PQ}}| \overline{\mathrm{PR}} \mid}=\frac{49}{7 \times 7 \sqrt{2}}=\frac{1}{\sqrt{2}}=\cos 45^{\circ}$

∴ P = 45°

$=-\frac{3}{2}(6 \hat{i}-4 \hat{j}-19 \hat{k})$

$\therefore \bar{a}=-\frac{3}{2} \bar{b}$

i.e. $\bar{a}$ is a non-zero scalar multiple of $\bar{b}$

Hence, $\bar{a}$ is parallel to $\bar{b}$.

2. $\begin{aligned} & \bar{a} \cdot \bar{b}=(2 \hat{i}+3 \hat{j}-\hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k}) \\ & =(2)(5)+(3)(-2)+(-1)(4)\end{aligned}$ = 10 – 6 – 4 = 0

Since, $\bar{a}, \bar{b}$ are non-zero vectors and $\bar{a} \cdot \bar{b}=0, \bar{a}$ is orthogonal to $\bar{b}$.

3. $\begin{aligned} & \overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=\left(-\frac{3}{5} \hat{\mathrm{i}}+\frac{1}{2} \hat{\mathrm{j}}+\frac{1}{3} \hat{\mathrm{k}}\right) \cdot(5 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & =\left(-\frac{3}{5}\right)(5)+\left(\frac{1}{2}\right)(4)+\left(\frac{1}{3}\right)(3)\end{aligned}$

= -3 + 2 + 1 = 0

Since $\bar{a} \cdot \bar{b}$ are non-zero vectors and $\bar{a} \cdot \bar{b}=0$

$\bar{a}$ is orthogonal to $\bar{b}$

4. $\begin{aligned} & \bar{a} \cdot \bar{b}=(4 \hat{i}-\hat{j}+6 \hat{k}) \cdot(5 \hat{i}-2 \hat{j}+4 \hat{k}) \\ & =(4)(5)+(-1)(-2)+(6)(4) \\ & =20+2+24 \\ & =46 \neq 0\end{aligned}$

$\therefore \bar{a}$ is not orthogonal to $\bar{b}$.

It is clear that $\bar{a}$ is not a scalar multiple of $\bar{b}$.

$\therefore \bar{a}$ is not parallel to $\bar{b}$.

Hence, $\bar{a}$ is neither parallel nor orthogonal to $\bar{b}$.

$\begin{aligned} & \text { Then } \bar{p}=5 \hat{i}-7 \hat{j}+0 \cdot \hat{k}, \bar{l}=\vec{i}+5 \hat{j}+3 \hat{k}, \\ & \bar{m}=4 \hat{i}-6 \hat{j}+3 \hat{k}, \bar{n}=6 \hat{i}-4 \hat{j}+2 \hat{k} .\end{aligned}$

Let $G(\bar{g})$ be the centroid of the tetrahedron.

Then by centroid formula

$\bar{g}=\frac{\bar{p}+\bar{l}+\bar{m}+\bar{n}}{4}$

$\begin{aligned} & =\frac{1}{4}[(5 \hat{i}-7 \hat{j}+0 \cdot \hat{k})+(\hat{i}+5 \hat{j}+3 \hat{k}) \\ & +(4 \hat{i}-6 \hat{j}+3 \hat{k})+(6 \hat{i}-4 \hat{j}+2 \hat{k})] \\ & \end{aligned}$

$\begin{aligned} & =\frac{1}{4}(16 \hat{i}-12 \hat{j}+8 \hat{k}) \\ & =4 \hat{i}-3 \hat{j}+2 \hat{k}\end{aligned}$

Hence, the centroid of the tetrahedron is G = (4, -3, 2).

Let $\bar{a}_i, \bar{b}, \bar{c}, \bar{g}$ be the position vectors of the points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{G}$ respectively w.r.t. O.

$\begin{aligned} & \text { Then } \bar{a}=a \hat{i}+2 \hat{j}+3 \hat{k}, \bar{b}=\hat{i}+b \hat{j}+2 \hat{k} \\ & \bar{c}=2 \hat{i}+\hat{j}+c \hat{k}, \bar{g}=\hat{i}+2 \hat{j}-\hat{k}\end{aligned}$

By formula of centroid of a tetrahedron,

$\bar{g}=\frac{\overline{0}+\bar{a}+\bar{b}+\bar{c}}{4}$

$\begin{aligned} & \therefore 4 \bar{g}=\bar{a}+\bar{b}+\bar{c} \\ & \therefore 4(\hat{i}+2 \hat{j}-\hat{k})=(a \hat{i}+2 \hat{j}+3 \hat{k})+(\hat{i}+b \hat{j}+2 \hat{k})+ \\ & \quad(2 \hat{i}+\hat{j}+c \hat{k})\end{aligned}$

$\begin{aligned} & \therefore 4 \hat{i}+8 \hat{j}-4 \hat{k}=(a+1+2) \hat{i}+(2+b+1) \hat{j}+(3+2+c) \hat{k} \\ & \therefore 4 \hat{i}+8 \hat{j}-4 \hat{k}=(a+3) \hat{i}+(b+3) \hat{j}+(c+5) \hat{k}\end{aligned}$

By equality of vectors a + 3 = 4, b + 3 = 8, c + 5= -4 ∴ a = 1, b = 5, c = -9 ∴ P = (a, b, c) = (1, 5, -9)

Distance of $P$ from origion $=\sqrt{1^2+5^2+(-9)^2}$

$\begin{aligned} & =\sqrt{1+25+81} \\ & =\sqrt{107}\end{aligned}$

Let $A, B, D, E, P$ have position vectors $\bar{a}_r \bar{b}, \bar{d}, \bar{e}, \bar{p}$ respectively w.r.t. $O$.

∵ AD : DB = 2 : 1. ∴ D divides AB internally in the ratio 2 : 1. Using section formula for internal division, we get

$\begin{aligned} & \bar{d}=\frac{2 \bar{b}+\bar{a}}{2+1} \\ & \therefore 3 \bar{d}=2 \bar{b}+\bar{a}\end{aligned}$

$\ldots(1)$

Since $\mathrm{E}$ is the midpoint of $\mathrm{OB}, \bar{e}=\overline{\mathrm{OE}}=\frac{1}{2} \overline{\mathrm{OB}}=\frac{\bar{b}}{2}$

$\begin{aligned} & \therefore \bar{b}=2 \bar{e} \\ & \therefore \operatorname{from}(1)\end{aligned}$

$\ldots(2)$

$3 \vec{d}=2(2 \bar{e})+\bar{a}$

... [By (2)]

$\begin{aligned} & =4 \bar{e}+\bar{a} \\ \therefore & \frac{3 \bar{d}+2 \cdot \overline{0}}{3+2}=\frac{4 \bar{e}+\bar{a}}{4+1}\end{aligned}$

LHS is the position vector of the point which divides OD internally in the ratio 3 : 2.
RHS is the position vector of the point which divides AE internally in the ratio 4 : 1.
But OD and AE intersect at P
∴ P divides OD internally in the ratio 3 : 2.
Hence, OP : PD = 3 : 2.

$\bar{a}=\hat{i}-\hat{j}+\hat{k}, \bar{b}=-\hat{i}+\hat{j}+\hat{k}, \bar{c}=\hat{i}+\hat{j}+\hat{k}, \bar{d}=2 \hat{i}-3 \hat{j}+4 \hat{k}$

$\therefore \overline{\mathrm{AB}}=\bar{b}-\bar{a}=(-\hat{i}+\hat{j}+\hat{k})-(\vec{i}-\vec{j}+\hat{k})$

$=-2 \hat{i}+2 \hat{j}$

$\overline{\mathrm{AC}}=\bar{c}-\bar{a}=(\hat{i}+\hat{j}+\hat{k})-(\hat{i}-\hat{j}+\hat{k})=2 \hat{j}$

and $\overline{\mathrm{AD}}=\bar{d}-\bar{a}=(2 \hat{i}-3 \hat{j}+4 \hat{k})-(\hat{i}-\hat{j}+\hat{k})$

$=\hat{i}-2 \hat{j}+3 \hat{k}$

If $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are coplanar, then there exist scalars $x, y$ such that

$\overline{\mathrm{AB}}=x \cdot \overline{\mathrm{AC}}+y \cdot \overline{\mathrm{AD}}$

$\begin{aligned} & \therefore-2 \hat{i}+2 \hat{j}=x(2 \hat{j})+y(\hat{i}-2 \hat{j}+3 \hat{k}) \\ & \therefore-2 \hat{i}+2 \hat{j}=y \hat{i}+(2 x-2 y) \hat{j}+3 y \hat{k}\end{aligned}$

By equality of vectors,
y = -2 ….(1)
2x – 2y = 2 … (2)
3y = 0 … (3)
From (1), y = -2
From (3), y = 0 This is not possible.
Hence, the points A, B, C, D are not coplanar.

P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.

Then $\bar{a}=4 \hat{i}+5 \hat{j}+2 \hat{k}, \bar{b}=3 \hat{i}+2 \hat{j}+4 \hat{k}, \bar{c}=5 \hat{i}+8 \hat{j}+0 \hat{k}$

$\overline{\mathrm{AB}}=\overline{\mathrm{b}}-\overline{\mathrm{a}}$

$\begin{aligned} & =(3 \hat{i}+2 \hat{j}+4 \hat{k})-(4 \hat{i}+5 \hat{j}+2 \hat{k}) \\ & =-\hat{i}-3 \hat{j}+2 \hat{k} i . e \\ & =-(\hat{i}+3 \hat{j}-2 \hat{k}) \\ & \text { and } \overline{B C}=\bar{c}-\bar{b} \\ & =(5 \hat{i}+8 \hat{j}+0 \hat{k})-(3 \hat{i}+2 \hat{j}+4 \hat{k}) \\ & =2 \hat{i}+6 \hat{j}-4 \hat{k} \\ & =2(\hat{i}+3 \hat{j}-2 \hat{k})\end{aligned}$

$=2 \cdot \overline{\mathrm{AB}} \ldots[\mathrm{By}(1)]$

$\therefore \overline{\mathrm{BC}}$ is a non-zero scalar multiple of $\overline{\mathrm{AB}}$

∴ they are parallel to each other. But they have the point B in common.

$\therefore \overline{\mathrm{BC}}$ and $\overline{\mathrm{AB}}$ are collinear vectors.

Hence, the points A, B and C are collinear.

A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively.

$\begin{aligned} & \text { Then, } \bar{a}=3 \hat{i}+2 \hat{j}-4 \hat{k}, \bar{b}=9 \hat{i}+8 \hat{j}-10 \hat{k} \\ & \bar{c}=-2 \hat{i}-3 \hat{j}+\hat{k}\end{aligned}$

$\begin{aligned} \overline{\mathrm{AB}} & =\bar{b}-\bar{a} \\ & =(9 \hat{i}+8 \hat{j}-10 \hat{k})-(3 \hat{i}+2 \hat{j}-4 \hat{k}) \\ & =6 \hat{i}+6 \hat{j}-6 \hat{k}\end{aligned}$

$\ldots(1)$

and $\overline{\mathrm{BC}}=\bar{c}-\bar{b}$

$\begin{aligned} & =(-2 \hat{i}-3 \hat{j}+\hat{k})-(9 \hat{i}+8 \hat{j}-10 \hat{k}) \\ & =-11 \hat{i}-11 \hat{j}+11 \hat{k} \\ & =-11(\hat{i}+\hat{j}-\hat{k})=-\frac{11}{6}(6 \hat{i}+6 \hat{j}-6 \hat{k})\end{aligned}$

$=-\frac{11}{6} \overline{\mathrm{AB}}$

... [By (1)]

$\therefore \overline{\mathrm{BC}}$ is a non-zero scalar multiple of $\overline{\mathrm{AB}}$

∴ they are parallel to each other

But they have the point B in common.

$\therefore \overline{\mathrm{BC}}$ and $\overline{\mathrm{AB}}$ are collinear vectors.

Hence, the points A, B and C are collinear.

Given: $\overline{\mathrm{OA}}=\bar{a}_r \overline{\mathrm{OB}}=\bar{a}$ Let $\mathrm{AD}, \mathrm{BE}, \mathrm{OC}$ meet at $\mathrm{M}$.

Then M bisects AD, BE, OC.

$\overline{\mathrm{AB}}=\overline{\mathrm{AO}}+\overline{\mathrm{OB}}=-\overline{\mathrm{OA}}+\overline{\mathrm{OB}}=-\bar{a}+\bar{b}=\bar{b}-\bar{a}$

∵ OABM is a parallelogram

$\begin{aligned} & \therefore \overline{\mathrm{OM}}=\overline{\mathrm{AB}}=\bar{b}-\bar{a} \\ & \overline{\mathrm{OC}}=2 \overline{\mathrm{OM}}=2(\bar{b}-\bar{a})=2 \bar{b}-2 \bar{a}\end{aligned}$

$\begin{aligned} \overline{\mathrm{OD}} & =\overline{\mathrm{OC}}+\overline{\mathrm{CD}} \\ & =\overline{\mathrm{OC}}-\overline{\mathrm{DC}}\end{aligned}$

$\begin{aligned} & =\overline{\mathrm{OC}}-\overline{\mathrm{OA}} \quad \ldots[\because \mathrm{OA}=\mathrm{DC} \text { and } \mathrm{OA} \| \mathrm{DC}] \\ & =2 \bar{b}-2 \bar{a}-\bar{a} \\ & =2 \bar{b}-3 \bar{a}\end{aligned}$

$\begin{aligned} \overline{\mathrm{OE}} & =\overline{\mathrm{OM}}+\overline{\mathrm{ME}} \\ & =(\bar{b}-\bar{a})-\overline{\mathrm{EM}}\end{aligned}$

$\begin{aligned} & =\bar{b}-\bar{a}-\bar{a} \quad \ldots[\because \mathrm{EM}=\mathrm{OA} \text { and } \mathrm{EM} \| \mathrm{OA}] \\ & =\bar{b}-2 \bar{a}\end{aligned}$

$\begin{aligned} & =\bar{b}-\bar{a}-\bar{a} \quad \ldots[\because \mathrm{EM}=\mathrm{OA} \text { and } \mathrm{EM} \| \mathrm{OA}] \\ & =\bar{b}-2 \bar{a}\end{aligned}$

Hence, the position vectors of $\mathrm{C}, \mathrm{D}$ and $\mathrm{E}$ are $2 \bar{b}-2 \bar{a}_t 2 \bar{b}-3 \bar{a}$ and $\bar{b}-2 \bar{a}$ respectively.