Maths Solve the Following Question.(5 Marks)

respectively w.r.t. the origin $\mathrm{O}$.

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

$\bar{c}=3 \hat{i}+5 \hat{j}-2 \hat{k}, \bar{d}=-2 \hat{i}+5 \hat{j}-4 \hat{k}$

$\begin{aligned} \therefore \overline{\mathrm{AB}} & =\vec{b}-\bar{a}=(-2 \hat{i}+2 \hat{j}-3 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k}) \\ & =-5 \hat{i}-2 \hat{k} \\ \overline{\mathrm{DC}} & =\bar{c}-\bar{d}=(3 \hat{i}+5 \hat{j}-2 \hat{k})-(-2 \hat{i}+5 \hat{j}-4 \hat{k}) \\ & =5 \hat{i}+2 \hat{k}=-(-5 \hat{i}-2 \hat{k})\end{aligned}$

$\therefore \overline{\mathrm{DC}}=-\overline{\mathrm{AB}}$

$\therefore \overline{\mathrm{DC}}$ is scalar multiple of $\overline{\mathrm{AB}}$

$\therefore \overline{\mathrm{DC}}$ is parallel to $\overline{\mathrm{AB}}$

Also, $|\overline{\mathrm{DC}}|=\sqrt{5^2+2^2}=\sqrt{25+4}=\sqrt{29}$

and $|\overline{\mathrm{AB}}|=\sqrt{(-5)^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}$

$\begin{aligned} & \therefore|\overline{\mathrm{DC}}|=|\overline{\mathrm{AB}}| \\ & \therefore l(\mathrm{AB})=l(\mathrm{DC})\end{aligned}$

∴ opposite sides AB and DC of ABCD are parallel and equal. ∴ ABCD is a parallelogram.

$\begin{aligned} \overline{\mathrm{AB}} & =\bar{b}-\bar{a}=(-2 \hat{i}+2 \hat{j}-3 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k}) \\ & =-5 \hat{i}-2 \hat{k} \\ \overline{\mathrm{AD}} & =\bar{d}-\bar{a}=(-2 \hat{i}+5 \hat{j}-4 \hat{k})-(3 \hat{i}+2 \hat{j}-\hat{k})\end{aligned}$

$=-5 \hat{i}+3 \hat{j}-3 \hat{k}$

$\therefore \overline{\mathrm{AB}} \times \overline{\mathrm{AD}}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -5 & 0 & -2 \\ -5 & 3 & -3\end{array}\right|$

$\begin{aligned} & =(0+6) \hat{i}-(15-10) \hat{j}+(-15+0) \hat{k} \\ & =6 \hat{i}-5 \hat{j}-15 \hat{k}\end{aligned}$

$\therefore$ area of parallelogram $=|\overline{\mathrm{AB}} \times \overline{\mathrm{AD}}|$

$\begin{aligned} & =\sqrt{6^2+(-5)^2+(-15)^2}=\sqrt{36+25+225} \\ & =\sqrt{286} \text { sq units. }\end{aligned}$

$\therefore \mathrm{Q} \equiv\left(\frac{3 \lambda+1}{\lambda+1}, \frac{44 \lambda+2}{\lambda+1}, \frac{5 \lambda+4}{\lambda+1}\right)$

.......(i)

Direction ratios of $\mathrm{PQ}$ are

$\frac{3 \lambda+1}{\lambda+1}-2, \frac{4 \lambda+2}{\lambda+1}-4, \frac{5 \lambda+4}{\lambda+1}-3$

i.e., $\frac{\lambda-1}{\lambda+1}, \frac{-2}{\lambda+1}, \frac{2 \lambda+1}{\lambda+1}$

Now, direction ratios of AB are, 3 – 1, 4 – 2, 5 – 4 i.e., 2, 2, 1.

Since $P Q$ is perpendicular to $A B$,

$2\left(\frac{\lambda-1}{\lambda+1}\right)+\frac{2(-2)}{\lambda+1}+1\left(\frac{2 \lambda+1}{\lambda+1}\right)=0$

$\therefore \frac{2 \lambda-2-4+2 \lambda+1}{\lambda+1}=0$

$\begin{aligned} & \therefore 4 \lambda-5=0 \\ & \therefore 4 \lambda=5 \\ & \therefore \lambda=\frac{5}{4} \\ & \text { Putting } \lambda=\frac{5}{4} \text { in (i). }\end{aligned}$

Coordinates of Q are,

$\begin{aligned} & \frac{3\left(\frac{5}{4}\right)+1}{\left(\frac{5}{4}\right)+1}=\frac{19}{19} \\ & \frac{\left(\frac{5}{4}\right)+2}{\left(\frac{5}{4}\right)+1}=\frac{28}{9}\end{aligned}$

$\begin{aligned} & \frac{5\left(\frac{5}{4}\right)+4}{\left(\frac{5}{4}\right)+1}=\frac{41}{9} \\ & \therefore Q \equiv\left(\frac{19}{9}, \frac{28}{9}, \frac{41}{9}\right)\end{aligned}$

$\begin{aligned} & \Rightarrow-30 n^2-45 n l-\left.15\right|^2=0 \\ & \Rightarrow 2 n^2+3 n l+\left.\right|^2=0 \\ & \Rightarrow 2 n^2+2 n l+n l+\left.\right|^2=0\end{aligned}$

⇒ (2n + l) (n + l) = 0 ∴ 2n + l = 0 OR n + l = 0 ∴ l = -2n OR l = -n ∴ l = -2n From (2), 3l + m + 5n = 0 ∴ -6n + m + 5n = 0 ∴ m = n i.e. (-2n, n, n) = (-2, 1, 1) ∴ l = -n ∴ -3n + m + 5n = 0 ∴ m = -2n i.e. (-n, -2n, n) = (1, 2, -1) (a1, b1, c1) = (-2, 1, 1) and (a2, b3, c3) = (1, 2, -1)

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

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

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

$=\left|-\frac{1}{6}\right|=\frac{1}{6}$

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

Let $\bar{c}=m \hat{i}+n \hat{j}$ be perpendicular to $\bar{b}$

Then $\bar{b} \cdot \bar{c}=0 \quad \therefore(4 \hat{i}+3 \hat{j}) \cdot(m \hat{i}+n \hat{j})=0$

$\therefore 4 m+3 n=0 \quad \therefore n=-\frac{4 m}{3}$

$\therefore \bar{c} m m \hat{i}-\frac{4 m}{3} \hat{j}=\frac{m}{3}(3 \hat{i}-4 \hat{j})$

$\therefore \bar{c}=p(3 \hat{i}-4 \hat{j}) \quad \ldots\left[p=\frac{m}{3}\right]$

$\therefore|\bar{c}|=p \sqrt{3^2+(-4)^2}=p \sqrt{9+16}=5 p$

Let $\bar{d}=x \hat{i}+y \hat{i}$ be the vector having projections 1 and 2

along $\bar{b}$ and $\bar{c}$.

$\therefore \frac{\bar{b} \cdot \bar{d}}{|\bar{b}|}=1$

$\therefore \frac{(4 \hat{i}+3 \hat{j})-(x \hat{i}+y \hat{j})}{5}=1$

$\therefore 4 x+3 y=5$

...(1)

Also, $\frac{\bar{c} \cdot \bar{d}}{|\bar{c}|}=2$

$\therefore \frac{(3 p \hat{i}-4 p \hat{j}) \cdot(x \hat{i}+y \hat{j})}{5 p}=2$

$\therefore 3 p x-4 p y=10 p$

$\therefore 3 x-4 y=10$

$\ldots(2)$

From (1), $3 y=5-4 x$

$\therefore y=\frac{5-4 x}{3}$

Substituting for $y$ in (2), we get

$\begin{aligned} & 3 x-4\left(\frac{5-4 x}{3}\right)=10 \\ & \therefore 9 x-20+16 x=30 \\ & \therefore 25 x=50 \quad \therefore x=2 \\ & y=\frac{5-4 x}{3}=\frac{5-4(2)}{3}=-1 \\ & \therefore \bar{d}=2 \hat{i}-\hat{j}\end{aligned}$

Hence, the required vector is $2 \hat{i}-\hat{j}$.

intersection. We find the points of intersection of $y=x^2 \ldots$ (1)

and $y=x^3 \ldots$ (2)

From (1) and (2)

$\begin{aligned} & x^3=x^2 \\ & \therefore x^3-x^2=0 \\ & \therefore x^2(x-1)=0 \\ & \therefore x=0 \text { or } x=1 \\ & \text { When } x=0, y=0 . \\ & \text { When } x=1, y=1 .\end{aligned}$

For $y=x^2, \frac{d y}{d x}=2 x$

For $y=x^3, \frac{d y}{d x}=3 x^2$

Angle at $\mathrm{O}=(0,0)$

Slope of tangent to $y=x^2$ at $O$

$=\left(\frac{d y}{d x}\right)_{\text {at } O(0,0)}=2 \times 0=0$

$\therefore$ equation of tangent to $y=x^2$ at $\mathrm{O}$ is $y=0$.

Slope of tangent to $y=x^3$ at $\mathrm{O}=\left(\frac{d y}{d x}\right)_{\text {at } \mathrm{O} 0,0)}=3 \times 0=0$

$\therefore$ equation of tangent to $y=x^3$ at $P$ is $y=0$.

∴ the tangents to both curves at (0, 0) are y = 0 ∴ angle between them is 0. Angle at P = (1, 1)

$\therefore$ equation of tangent to $y=x^3$ at $P$ is $y-1=3(x-1) y=3 x-2$

We have to find angle between y = 2x – 1 and y = 3x – 2 Lines through origin parallel to these tagents are y = 2x and y = 3x

$\therefore \frac{x}{1}=\frac{y}{2}$ and $\frac{z}{1}=\frac{y}{3}$

These lines lie in XY-plane. ∴ the direction ratios of these lines are 1, 2, 0 and 1, 3, 0. The angle θ between them is given by

$\cos \theta=\frac{(1)(1)+(2)(3)+(0)(0)}{\sqrt{1^2+2^2+0^2} \sqrt{1^2+3^2+0^2}}$

$=\frac{1+6+0}{\sqrt{5} \sqrt{10}}=\frac{7}{\sqrt{50}}=\frac{7}{5 \sqrt{2}}$

$\therefore \theta=\cos ^{-1}\left(\frac{7}{5 \sqrt{2}}\right)$

Hence, the required angles are 0 and $\cos ^{-1}\left(\frac{7}{5 \sqrt{2}}\right)$.

with each of the vectors

Then $|\bar{r}|=1$

Also, $\bar{u}=2 \hat{i}+\hat{j}-2 \hat{k}, \quad \bar{v}=\hat{i}+2 \hat{j}-2 \hat{k}, \quad \bar{w}=2 \hat{i}-2 \hat{j}+\hat{k}$

$|\bar{u}|=\sqrt{2^2+1^2+(-2)^2}=\sqrt{4+1+4}=\sqrt{9}=3$

$|\bar{v}|=\sqrt{1^2+2^2+(-2)^2}=\sqrt{1+4+4}=\sqrt{9}=3$

$|\bar{w}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3$

Angle between $\bar{r}$ and $\bar{u}$ is $\theta$.

$\therefore \cos \theta=\frac{\bar{r} \cdot \bar{u}}{|\bar{r}||\bar{u}|}$

$=\frac{(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+\hat{j}-2 \hat{k})}{1 \times 3}$

$=\frac{2 x+y-2 z}{3}$

...(1)

Also, the angle between $\bar{r}$ and $\bar{v}$ and between $\bar{r}$ and $\bar{w}$ is $\theta$.

$\therefore \cos \theta=\frac{\bar{r} \cdot \bar{v}}{|\bar{r}||\bar{v}|}$

$=\frac{(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\vec{i}+2 \hat{j}-2 \hat{k})}{1 \times 3}$

$=\frac{x+y-2 z}{3}$

$\ldots$ (2)

and $\cos \theta=\frac{\bar{r} \cdot \bar{w}}{|\bar{r}||\bar{w}|}$

$=\frac{(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})}{1 \times 3}$

$=\frac{2 x-2 y+z}{3}$

... (3)

From (1) and (2), we get

$\frac{2 x+y-2 z}{3}=\frac{x+2 y-2 z}{3}$

$\begin{aligned} & \therefore 2 x+y-2 z=x+2 y-2 z \\ & \therefore x=y\end{aligned}$

From $(2)$ and $(3)$, we get

$\frac{x+2 y-2 z}{3}=\frac{2 x-2 y+z}{3}$

$\therefore x+2 y-2 z=2 x-2 y+z$.

$\therefore 3 y=3 z$

$\ldots[\because x=y]$

$\begin{aligned} & \therefore y=z \\ & \therefore x=y=z \\ & \therefore \bar{r}=x \hat{i}+y \hat{j}+z \hat{k}=x \hat{i}+x \hat{j}+x \hat{k} \\ & \therefore|\bar{r}|=\sqrt{x^2+x^2+x^2}=1 \\ & \therefore x^2+x^2+x^2=1 \quad \therefore 3 x^2=1 \\ & \therefore x^2=\frac{1}{3} \quad \therefore x= \pm \frac{1}{\sqrt{3}}\end{aligned}$

$\begin{aligned} \therefore \bar{r} & = \pm \frac{1}{\sqrt{3}} \hat{i} \pm \frac{1}{\sqrt{3}} \hat{j} \pm \frac{1}{\sqrt{3}} \hat{k} \\ & = \pm \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

Hence, the required unit vectors are $\pm \frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$.

Since, $\bar{c}$ is parallel to $\bar{b}, \bar{c}=m \bar{b}$, where $m$ is a scalar.

$\therefore \bar{c}=m(3 \hat{i}+\hat{k})$

i.e. $\bar{c}=3 m \hat{i}+m \hat{k}$

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

Since, $\bar{d}$ is perpendicular to $\bar{b}=3 \hat{i}+\hat{k}, \bar{d} \cdot \bar{b}=0$

$\begin{aligned} & \therefore(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{k})=0 \\ & \therefore 3 x+z=0 \quad \therefore z=-3 x \\ & \therefore \bar{d}=x \hat{i}+y \hat{k}-3 x \hat{k}\end{aligned}$

Now, $\bar{a}=\bar{c}+\bar{d}$ gives

$\begin{aligned} \therefore 5 \hat{i}-2 \hat{j}+5 \hat{k} & =(3 m \hat{i}+m \hat{k})+(x \hat{i}+y \hat{j}-3 x \hat{k}) \\ & =(3 m+x) \hat{i}+y \hat{j}+(m-3 x) \hat{k}\end{aligned}$

By equality of vectors 3m + x = 5 … (1) y = -2 and m – 3x = 5 From (1) and (2) 3m + x = m – 3x ∴ 2m = -4x m ∴ m = -2x Substituting m = -2x in (1), we get ∴ -6x + x = 5 ∴ -5x = 5 ∴ x = -1 ∴ m = -2x = 2

$\therefore \bar{c}=6 \hat{i}+2 \hat{k}$ is parallel to $\hat{b}$ and

$\bar{d}=-\hat{i}-2 \hat{j}+3 \hat{k}$ is perpendicular to $\bar{b}$

Hence, $\bar{a}=\bar{c}+\bar{d}$, where $\bar{c}=6 \hat{i}+2 \hat{k}$ and $\bar{d}=-\hat{i}-2 \hat{j}+3 \hat{k}$.

$\therefore \bar{p} \cdot \bar{q}=|\bar{p} \|| \bar{q} \mid \cos \frac{\pi}{3}=2 \times 2 \times \frac{1}{2}=2$

Now, $\bar{a}=3 \bar{p}-\bar{q}$

$\therefore|\bar{a}|^2=|\overline{3 p}-\bar{q}|^2$

$\begin{aligned} & =(3 \bar{p}-\bar{q}) \cdot(3 p-\bar{q}) \\ & =3 \bar{p} \cdot(3 \bar{p}-\bar{q})-\bar{q} \cdot(3 p-\bar{q}) \\ & =9 \bar{p} \cdot \bar{p}-3 \bar{p} \cdot \bar{q}-3 \bar{q} \cdot \bar{p}+\bar{q} \cdot \bar{q}\end{aligned}$

$=9|\bar{p}|^2-6 \bar{p} \cdot \bar{q}+|\bar{q}|^2 \quad \ldots .[\because \bar{q} \cdot \bar{p}=\bar{p} \cdot \bar{q}]$

$=9 \times 4-6 \times 2+4 \quad \ldots[\because \bar{p} \cdot \bar{q}=2]$

$=28$

$\therefore|\bar{a}|=\sqrt{28}$

Also $\bar{b}=\bar{p}+3 \bar{q}$

$\therefore|\bar{b}|^2=|\bar{p}+3 \bar{q}|^2$

$\begin{aligned} & =(\bar{p}+3 \bar{q}) \cdot(\bar{p}+3 \bar{q}) \\ & =\bar{p} \cdot(\bar{p}+3 \bar{q})+3 \bar{q} \cdot(\bar{p}+3 \bar{q})\end{aligned}$

$=\bar{p} \cdot \bar{p}+3 \bar{p} \cdot \bar{q}+3 \bar{q} \cdot \bar{p}+9 q \cdot \bar{q} \quad \ldots[\because \bar{p} \cdot \bar{q}=\bar{q} \cdot \bar{p}]$

$=|\bar{p}|^2+3 \bar{p} \cdot \bar{q}+3 \bar{p} \cdot \bar{q}+9|\bar{q}|^2$

$=4+12+36 \quad \ldots[\because \bar{p} \cdot \bar{q}=2]$

$=52$

Ratio of lengths of the sides

$=\frac{|\bar{a}|}{|\bar{b}|}=\frac{\sqrt{28}}{\sqrt{52}}=\frac{2 \sqrt{7}}{2 \sqrt{13}}=\frac{\sqrt{7}}{\sqrt{13}}$

Hence, the ratio of the lengths of the sides is $\sqrt{7}: \sqrt{13}$.

Let $A, B, C, D, E, F, P, Q$ have

position vectors $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{e}, \bar{f}, \bar{p}$,

$\bar{q}$ respectively

$\because A B C D$ is a parallelogram

$\therefore \overline{A B}=\overline{D C}$

$\therefore \bar{b}-\bar{a}=\bar{c}-\bar{d}$

$\therefore \bar{c}=\bar{b}+\bar{d}-\bar{a}$

$\ldots$ (1)

$E$ is the midpoint of $B C$

$\therefore \bar{e}=\frac{\bar{b}+\bar{c}}{2} \quad \therefore 2 \bar{e}=\bar{b}+\bar{c}$

$\ldots(2)$

$\mathrm{F}$ is the mid-point of $\mathrm{CD}$

$\therefore \bar{f}=\frac{\bar{c}+\bar{d}}{2} \quad \therefore 2 \bar{f}=\bar{c}+\bar{d}$

$\ldots(3)$

$2 \bar{e}=\bar{b}+\bar{c} \quad \ldots[$ By (2)]

$=\bar{b}+(\bar{b}+\bar{d}-\bar{a}) \quad \ldots$ [By (1)]

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

LHS is the position vector of the point on AE and RHS is the position vector of the point on DB. But AE and DB meet at Q.

$\therefore \bar{q}=\frac{2 \bar{b}+\bar{d}}{2+1}$

$\therefore \mathrm{Q}$ divides $\mathrm{DB}$ in the ratio $2: 1$

$\ldots(4)$

$2 \bar{f}=\bar{c}+\bar{d} \quad \ldots[$ By (3)]

$=(\bar{b}+\bar{d}-\bar{a})+\bar{d}$

$\ldots[$ Вy $(1)]$

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

LHS is the position vector of the point on AF and RHS is the position vector of the point on DB.

But AF and DB meet at P.

$\therefore \bar{p}=\frac{\bar{b}+2 \bar{d}}{1+2}$

∴ P divides DB in the ratio 1 : 2 … (5) From (4) and (5), if follows that P and Q trisect DB.

Suppose $\bar{p}=x \bar{a}+y \bar{b}+z \bar{c}$.

Then, $\hat{\mathrm{i}}+4 \hat{\mathrm{j}}-4 \hat{\mathrm{k}}=x(2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}})+y(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+4 \hat{k})+z(-\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})$

$\therefore \hat{\mathrm{i}}+4 \hat{j}-4 \hat{k}=(2 x+2 y-z) \hat{i}+(-x-2 y+3 z) \hat{j}+(3 x+4 y-5 z) \hat{k}$

By equality of vectors, $ \begin{aligned} & 2 x+2 y-z=1 \\ & -x-2 y+3 z=4 \\ & 3 x+4 y-5 z=-4 \end{aligned} $

We have to solve these equations by using Cramer’s Rule.

$D=\left|\begin{array}{ccc}2 & 2 & -1 \\ -1 & -2 & 3 \\ 3 & 4 & -5\end{array}\right|$

= 2(10 – 12) – 2(5 – 9) – 1(-4 + 6) = -4 + 8 – 2 = 2 ≠ 0

$D_x=\left|\begin{array}{ccc}1 & 2 & -1 \\ 4 & -2 & 3 \\ -4 & 4 & -5\end{array}\right|$

= 1(10 – 12) – 2(-20 + 12) – 1 (16 – 8) = -2 + 16 – 8 = 6

$D_y=\left|\begin{array}{ccc}2 & 1 & -1 \\ -1 & 4 & 3 \\ 3 & -4 & -5\end{array}\right|$

= 2(-20 + 12) – 1(5 – 9) – 1(4 – 12) = -16 – 4 – 8 = -28

$D_z=\left|\begin{array}{ccc}2 & 2 & 1 \\ -1 & -2 & 4 \\ 3 & 4 & -4\end{array}\right|$

= 2(8 – 16) – 2(4 – 12) + 1(-4 + 6) = -16 – 16 + 2 = -30

$\begin{aligned} & \therefore x=\frac{D_x}{D}=\frac{6}{2}=3 \\ & \therefore y=\frac{D_y}{D}=\frac{-28}{2}=-14 \\ & \therefore z=\frac{D_z}{D}=\frac{-30}{2}=-15 \\ & \therefore \bar{p}=3 \bar{a}-14 \bar{b}-3 \bar{c}\end{aligned}$

Let $A B C D$ be a parallelogram with

$\overline{\mathrm{AB}}=3 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overline{\mathrm{BC}}=-2 \hat{j}+7 \hat{k}$

Then $\overline{\mathrm{AC}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=(3 \hat{i}+4 \hat{j}-5 \hat{k})+(-2 \hat{j}+7 \hat{k})$

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

$\therefore|\overline{\mathrm{AC}}|=\sqrt{3^2+2^2+2^2}=\sqrt{9+4+4}=\sqrt{17}$

$\therefore$ unit vector along $\overline{\mathrm{AC}}=\frac{\overline{\mathrm{AC}}}{|\overline{\mathrm{AC}}|}$

$=\frac{1}{\sqrt{17}}(3 \hat{i}+2 \hat{j}+2 \hat{k})$

Also, $\overline{\mathrm{BD}}=\overline{\mathrm{BA}}+\overline{\mathrm{AD}}=-\overline{\mathrm{AB}}+\overline{\mathrm{BC}}=\overline{\mathrm{BC}}-\overline{\mathrm{AB}}$

$\begin{aligned} & =(-2 \hat{j}+7 \hat{k})-(3 \hat{i}+4 \hat{j}-5 \hat{k}) \\ & =-3 \hat{i}-6 \hat{j}+12 \hat{k} \\ & =3(-\hat{i}-2 \hat{j}+4 \hat{k})\end{aligned}$

$\therefore|\overline{\mathrm{BD}}|=3 \sqrt{(-1)^2+(-2)^2+4^2}=3 \sqrt{1+4+16}=3 \sqrt{21}$

$\therefore$ unit vector along $\overline{\mathrm{BD}}=\frac{\overline{\mathrm{BD}}}{|\overline{\mathrm{BD}}|}$

$\begin{aligned} & =\frac{3(-\hat{i}-2 \hat{j}+4 \hat{k})}{3 \sqrt{21}} \\ & =\frac{1}{\sqrt{21}}(-\hat{i}-2 \hat{j}+4 \hat{k})\end{aligned}$

Hence, the unit vectors parallel to the diagonals are

$\frac{1}{\sqrt{17}}(3 \hat{i}+2 \hat{j}+2 \hat{k})$ and $\frac{1}{\sqrt{21}}(-\hat{i}-2 \hat{j}+4 \hat{k})$ respectively.

$\mathrm{DC}$ is parallel to $\mathrm{AB}$ and $\mathrm{DC}=3 \mathrm{AB}$.

$\because \overline{\mathrm{AB}}=\bar{p} \quad \therefore \overline{\mathrm{DC}}=3 \bar{p}$

$\mathrm{M}$ is the midpoint of $\mathrm{DC}$

$\therefore \overline{\mathrm{DM}}=\overline{\mathrm{MC}}=\frac{1}{2} \overline{\mathrm{DC}}=\frac{3-}{2} \bar{p}_{\mathrm{M}}$

$\overline{\mathrm{AM}}=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}+\overline{\mathrm{CM}}$

$=\overline{\mathrm{AB}}+\overline{\mathrm{BC}}-\overline{\mathrm{MC}}$

$=\bar{p}+\bar{q}-\frac{3 p}{2}$

$=-\frac{1-}{2^p}$

2. $\overline{\mathrm{BD}}=\overline{\mathrm{BC}}+\overline{\mathrm{CD}}=\overline{\mathrm{BC}}-\overline{\mathrm{DC}}=\bar{q}-3 \bar{p}$

3.$\overline{\mathrm{MB}}=\overline{\mathrm{MC}}+\overline{\mathrm{CB}}=\overline{\mathrm{MC}}-\overline{\mathrm{BC}}=\frac{3}{2} \bar{p}-\bar{q}$

4. $\begin{aligned} \overline{\mathrm{DA}} & =\overline{\mathrm{DC}}+\overline{\mathrm{CB}}+\overline{\mathrm{BA}}=\overline{\mathrm{DC}}-\overline{\mathrm{BC}}-\overline{\mathrm{AB}} \\ & =3 \bar{p}-\bar{q}-\bar{p}=2 \bar{p}-\bar{q}\end{aligned}$

Now from (1) we have
$
\begin{aligned}
& \overline{A P} \perp \overline{B C} \quad \text { and } \overline{B P} \perp \overline{A C} \\
& \overline{A P} \perp \overline{B C}=0 \quad \text { and } \overline{B P} \perp \overline{A C}=0 \\
\therefore & (\bar{p}-\bar{a}) \cdot(\bar{c}-\bar{b})=0 \text { and }(\bar{p}-\bar{b}) \cdot(\bar{c}-\bar{a})=0 \\
\therefore & \bar{p} \cdot \bar{c}-\bar{p} \cdot \bar{b}-\bar{a} \cdot \bar{c}+\bar{a} \cdot \bar{b}=0 \\
& \bar{p} \cdot \bar{c}-\bar{p} \cdot \bar{a}-\bar{b} \cdot \bar{c}+\bar{b} \cdot \bar{a}=0
\end{aligned}
$
Therefore, subtracting equation (2) from equation (3), we get
$
\begin{array}{ll}
& -\bar{p} \cdot \bar{a}+\bar{p} \cdot \bar{b}-\bar{b} \cdot \bar{c}+\bar{a} \cdot \bar{c}=0 \quad(\text { Since } \bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}) \\
\therefore \quad & \bar{p}(\bar{b}-\bar{a})-\bar{c}(\bar{b}-\bar{a})=0 \\
\therefore & (\bar{p}-\bar{c}) \cdot(\bar{b}-\bar{a})=0 \\
\therefore & \overline{C P} \cdot \overline{A B}=0 \\
\therefore & \overline{C P} \perp \overline{A B}
\end{array}
$
Hence the proof.

trapezium ABCD, with side AD || side BC.

Then the vectors $\overline{\mathrm{AD}}$ and $\overline{\mathrm{BC}}$ are parallel.

$\therefore$ there exists a scalar $k$

such that $\overline{\mathrm{AD}}=k \cdot \overline{\mathrm{BC}}$

$\therefore \overline{\mathrm{AD}}+\overline{\mathrm{BC}}=k \cdot \overline{\mathrm{BC}}+\overline{\mathrm{BC}}$

= (k + 1)BC …(1)

Let $\bar{m}$ and $\bar{n}$ be the position vectors of the midpoints $\mathrm{M}$ and $\mathrm{N}$ of the non-parallel sides

AB and DC respectively. Then seg MN is the median of the trapezium. By the midpoint formula,

$\bar{m}=\frac{\bar{a}+\bar{b}}{2}$ and $\bar{n}=\frac{\bar{d}+\bar{c}}{2}$

$\therefore \overline{\mathrm{MN}}=\bar{n}-\bar{m}$

$=\left(\frac{\bar{d}+\bar{c}}{2}\right)-\left(\frac{\bar{a}+\bar{b}}{2}\right)$

$=\frac{1}{2}(\bar{d}+\bar{c}-\bar{a}-\bar{b})$

$=\frac{1}{2}[(\bar{d}-\bar{a})+(\bar{c}-\bar{b})]$

$=\frac{\overline{\mathrm{AD}}+\overline{\mathrm{BC}}}{2}$

$\ldots(2)$

$=\frac{(k+1) \overline{B C}}{2}$

$\ldots[B y(1)]$

Thus $\overline{\mathrm{MN}}$ is a scalar multiple of $\overline{\mathrm{BC}}$

$\therefore \overline{\mathrm{MN}}$ and $\overline{\mathrm{BC}}$ are parallel vectors

$\therefore \overline{\mathrm{MN}} \| \overline{\mathrm{BC}}$ where $\overline{\mathrm{BC}} \| \overline{\mathrm{AD}}$

∴ the median MN is parallel to the parallel sides AD and BC of the trapezium.

Now $\overline{\mathrm{AD}}$ and $\overline{\mathrm{BC}}$ are collinear

$\therefore|\overline{\mathrm{AD}}+\overline{\mathrm{BC}}|=|\overline{\mathrm{AD}}|+|\overline{\mathrm{BC}}|=\mathrm{AD}+\mathrm{BC}$

$\therefore$ from (2), we have

$\therefore \overline{\mathrm{MN}}=\frac{\overline{\mathrm{AD}}+\overline{\mathrm{BC}}}{2}$

$\therefore \mathrm{MN}=\frac{1}{2}(\mathrm{AD}+B C)$

parallelogram ABCD. Then AB = DC and side AB || side DC.

$\begin{aligned} & \therefore \overline{\mathrm{AB}}=\overline{\mathrm{DC}} \\ & \therefore \bar{b}-\bar{a}=\bar{c}-\bar{d} \\ & \therefore \bar{a}+\bar{c}=\bar{b}+\bar{d} \\ & \therefore \frac{\bar{a}+\bar{c}}{2}=\frac{\bar{b}+\bar{d}}{2}\end{aligned}$

$\ldots$... (1)

The position vectors of the midpoints of the diagonals $\mathrm{AC}$ and $\mathrm{BD}$ are $(\bar{a}+\bar{c}) / 2$ and $(\bar{b}+$$\bar{d}) / 2$. By (1), they are equal.

∴ the midpoints of the diagonals AC and BD are the same.
This shows that the diagonals AC and BD bisect each other.

Conversely, suppose that the diagonals AC and BD
of □ ABCD bisect each other,
i. e. they have the same midpoint.
∴ the position vectors of these midpoints are equal.

$\begin{aligned} & \therefore \frac{\bar{a}+\bar{c}}{2}=\frac{\bar{b}+\bar{d}}{2} \therefore \bar{a}+\bar{c}=\bar{b}+\bar{d} \\ & \therefore \bar{b}-\bar{a}=\bar{c}-\bar{d} \therefore \overline{\mathrm{AB}}=\overline{\mathrm{DC}} \\ & \therefore \overline{\mathrm{AB}} \| \overline{\mathrm{DC}} \text { and }|\overline{\mathrm{AB}}|=|\overline{\mathrm{DC}}|\end{aligned}$

∴ side AB || side DC and AB = DC.
∴ □ ABCD is a parallelogram.

Let AD and BE intersect at P.

Let $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{P}$ have position vectos $\bar{a}, \bar{b}, \bar{c}_t, \bar{d}, \bar{e}_{,} \bar{p}$ respectively.

D and E divide segments BC and CA internally in the ratio 2 : 3. By the section formula for internal division,

$\begin{aligned} & \bar{d}=\frac{2 \bar{c}+3 \bar{b}}{2+3} \\ & \therefore 5 \bar{d}=2 \bar{c}+3 \bar{b}\end{aligned}$

$\ldots(1)$

and $\bar{e}=\frac{2 \bar{a}+3 \bar{c}}{2+3}$

$\therefore 5 \vec{e}=2 a+3 \vec{c}$

$\cdots(2)$

From (1), $5 \bar{d}-3 \bar{b}=2 \bar{c} \quad \therefore 15 \bar{d}-9 \bar{b}=6 \bar{c}$

From (2), $5 \bar{e}-2 \bar{a}=3 \bar{c} \quad \therefore 10 \bar{e}-4 \bar{a}=6 \bar{c}$

Equating both values of $6 \bar{c}$, we get

$\begin{aligned} & 15 \bar{d}-9 \bar{b}=10 \bar{e}-4 \bar{a} \\ & \therefore 15 \bar{d}+4 \bar{a}=10 \bar{e}+9 \bar{b} \\ & \therefore \frac{15 \bar{d}+4 \bar{a}}{15+4}=\frac{10 \bar{e}+9 \bar{b}}{10+9}\end{aligned}$

LHS is the position vector of the point which divides segment AD internally in the ratio 15 : 4.
RHS is the position vector of the point which divides segment BE internally in the ratio 10 : 9.
But P is the point of intersection of AD and BE.
∴ P divides AD internally in the ratio 15 : 4 and P divides BE internally in the ratio 10 : 9.
Hence, the position vector of the point of interaction of

$\mathrm{AD}$ and $\mathrm{BE}$ is $\bar{p}=\frac{15 \bar{d}+4 \bar{a}}{19}=\frac{10 \bar{e}+9 \bar{b}}{19}$ and it divides $A D$ internally in the ratio $15: 4$ and $B E$

internally in the ratio 10 : 9.

Let $\bar{a}, \bar{b}, \bar{c}, \bar{d}, \bar{p}, \bar{q}, \bar{r}$ and $s$ be the position vectors of the points $A, B, C, D, P, Q, R$ and $S$ respectively. Since P, Q, R and S are the midpoints of the sides AB, BC, CD and DA respectively,

$\bar{p}=\frac{\bar{a}+\bar{b}}{2}, \bar{q}=\frac{\bar{b}+\bar{c}}{2}, \bar{r}=\frac{\bar{c}+\bar{d}}{2}$ and $\bar{s}=\frac{\bar{d}+\bar{a}}{2}$

$\therefore \overline{\mathrm{PQ}}=\bar{q}-\bar{p}$

$=\left(\frac{\bar{b}+\bar{c}}{2}\right)-\left(\frac{\bar{a}+\bar{b}}{2}\right)$

$=\frac{1}{2}(\bar{b}+\bar{c}-\bar{a}-\bar{b})=\frac{1}{2}(\bar{c}-\bar{a})$

$\overline{\mathrm{SR}}=\bar{r}-\bar{s}$

$\begin{aligned} & =\left(\frac{\bar{c}+\bar{d}}{2}\right)-\left(\frac{\bar{d}+\bar{a}}{2}\right) \\ & =\frac{1}{2}(\bar{c}+\bar{d}-\bar{d}-\bar{a})=\frac{1}{2}(\bar{c}-\bar{a})\end{aligned}$

Similarly, $\overline{\mathrm{QR}} \mid \overline{\mathrm{PS}}$

$\therefore \overline{\mathrm{PQ}}=\overline{\mathrm{SR}} \quad \therefore \overline{\mathrm{PQ}} \| \overline{\mathrm{SR}}$

∴ □PQRS is a parallelogram.

Then $\bar{a}=3 \hat{i}+0 \cdot \hat{j}+p \hat{k}, \bar{b}=-\hat{i}+q \hat{j}+3 \hat{k}$ and $\bar{c}=-3 \hat{i}+3 \hat{j}+0 \hat{k}$.

As the points A, B, C are collinear, suppose the point C divides line segment AB in the ratio λ : 1.

∴ by the section formula,

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

$\begin{aligned} \therefore & -3 \hat{i}+3 \hat{j}+0 \cdot \hat{k} \\ = & \frac{\lambda(-\hat{i}+q \hat{j}+3 \hat{k})+(3 \hat{i}+0 \cdot \hat{j}+p \hat{k})}{\lambda+1}\end{aligned}$

$\begin{aligned} \therefore & (\lambda+1)(-3 \hat{i}+3 \hat{j}+0 \cdot \hat{k}) \\ & =(-\lambda \hat{i}+\lambda \hat{j}+3 \lambda \hat{k})+(3 \hat{i}+0 \hat{j}+p \hat{k}) \\ \therefore \quad & -3(\lambda+1) \hat{i}+3(\hat{\lambda}+1) \hat{j}+0 \hat{k} \\ & =(-\lambda+3) \hat{i}+\lambda 2 \hat{j}+(3 \lambda+p) \hat{k}\end{aligned}$

By equality of vectors, we have,
-3(λ + 1) = -λ + 3 … (1)
3(λ + 1 ) = λ q … (2)
0 = 3λ + p … (3)
From equation (1), -3λ – 3 = -λ + 3
∴ -2λ = 6 ∴ λ = -3
∴ C divides segment AB externally in the ratio 3 : 1.

2.Putting λ = -3 in equation (2), we get
3(-3 + 1) = -3q
∴ -6 = -3q ∴ q = 2
Also, putting λ = -3 in equation (3), we get
0 = -9 + p ∴ p = 9
Hence p = 9 and q = 2.

$-5 \hat{i}+2 \hat{j}-5 \hat{k}$ respectively.

If $\mathrm{R}(\bar{r})$ divides the line segment $\mathrm{PQ}$ internally in the ratio $3: 2$, by section formula for

internal division,

$\bar{r}=\frac{3 \bar{q}+2 \bar{p}}{3+2}=\frac{3(-5 \hat{i}+2 \hat{j}-5 \hat{k})+2(2 \hat{i}-\hat{j}+3 \hat{k})}{5}$

$=\frac{-11 \hat{i}+4 \hat{j}-9 \hat{k}}{5}=\frac{1}{5}(-11 \hat{i}+4 \hat{j}-9 \hat{k})$

$\therefore$ coordinates of $\mathrm{R}=\left(-\frac{11}{5}, \frac{4}{5},-\frac{9}{5}\right)$

Hence, the position vector of $R$ is $\frac{1}{5}(-11 \hat{i}+4 \hat{j}-9 \hat{k})$

and the coordinates of $\mathrm{R}$ are $\left(-\frac{11}{5}, \frac{4}{5},-\frac{9}{5}\right)$.

2.If $R(\bar{r})$ divides the line segment joining $P$ and $Q$ externally in the ratio $3: 2$, by section

formula for external division,

$\bar{r}=\frac{3 \bar{q}-2 \bar{p}}{3-2}=\frac{3(-5 \hat{i}+2 \hat{j}-5 \hat{k})-2(2 \hat{i}-\hat{j}+3 \hat{k})}{3-2}$

$=-19 \hat{i}+8 \hat{j}-21 \hat{k}$

∴ coordinates of R = (-19, 8, -21).

Hence, the position vector of $\mathrm{R}$ is $-19 \hat{i}+8 \hat{j}-21 \hat{k}$ and coordinates of $\mathrm{R}$ are $(-19,8,-21)$.

$\bar{c}=3 \vec{i}+\hat{j}-2 \hat{k}$ and $\bar{p}=-\hat{i}-3 \hat{j}+4 \hat{k}$.

Suppose $\bar{p}=x \bar{a}+y \bar{b}+\bar{z}$.

Then, $-\hat{i}-3 \hat{j}+4 \hat{k}=x(2 \hat{i}+\hat{j}-4 \hat{k})+y(2 \hat{i}-\hat{j}+3 \hat{k})+$

$z(3 \hat{i}+\hat{j}-2 \hat{k})$

$\therefore-\hat{i}-3 \hat{j}+4 \hat{k}=(2 x+2 y+3 z) \hat{i}+(x-y+z) \hat{j}+$

$(-4 x+3 y-2 z) \hat{k}$

By equality of vectors,
2x + 2y + 3 = -1
x – y + z = -3
-4x + 3y – 2z = 4
We have to solve these equations by using Cramer’s Rule

$\begin{aligned} & D=\left|\begin{array}{rrr}2 & 2 & 3 \\ 1 & -1 & 1 \\ -4 & 3 & -2\end{array}\right| \\ & =2(2-3)-2(-2+4)+3(3-4) \\ & =-2-4-3=-9 \neq 0\end{aligned}$

$\begin{aligned} D_x & =\left|\begin{array}{rrr}-1 & 2 & 3 \\ -3 & -1 & 1 \\ 4 & 3 & -2\end{array}\right| \\ & =-1(2-3)-2(6-4)+3(-9+4) \\ & =1-4-15=-18\end{aligned}$

$\begin{aligned} D_y & =\left|\begin{array}{rrr}2 & -1 & 3 \\ 1 & -3 & 1 \\ -4 & 4 & -2\end{array}\right| \\ & =2(6-4)+1(-2+4)+3(4-12) \\ & =4+2-24=-18\end{aligned}$

$\begin{aligned} & D_z=\left|\begin{array}{rrr}2 & 2 & -1 \\ 1 & -1 & -3 \\ -4 & 3 & 4\end{array}\right| \\ & =2(-4+9)-2(4-12)-1(3-4) \\ & =10+16+1=27\end{aligned}$

$\begin{aligned} & \therefore x=\frac{D_y}{\mathrm{D}}=\frac{-18}{-9}=2 \\ & \therefore y=\frac{D_y}{D}=\frac{-18}{-9}=2 \\ & \therefore z=\frac{D_2}{\mathrm{D}}=\frac{27}{-9}=-3 \\ & \therefore \bar{p}=2 \bar{a}+2 \bar{b}-3 \bar{c} .\end{aligned}$