Maths Solve the Following Question.(2 Marks)

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

$\begin{aligned} & \bar{c} \cdot(\bar{a} \times \bar{c})+\bar{c} \cdot(\bar{b} \times \bar{a}) \times \bar{c} \cdot(\bar{b} \times \bar{c}) \quad \ldots[\because \bar{a} \times \bar{a}=\overline{0}] \\ = & {[\bar{a} \bar{a} \bar{c}]+[\bar{a} \bar{b} \bar{a}]+[\bar{a} \bar{b} \bar{c}]+[\bar{b} \bar{a} \bar{c}]+[\bar{b} \bar{b} \bar{a}]+} \\ & {[\bar{b} \bar{b} \bar{c}]+[\bar{c} \bar{a} \bar{c}]+[\bar{c} \bar{b} \bar{a}]+[\bar{c} \bar{b} \bar{c}] } \\ = & 0+0+[\bar{a} \bar{b} \bar{c}]-[\bar{a} \bar{b} \bar{c}]+0+0+0-[\bar{a} \bar{b} \bar{c}]+0\end{aligned}$

$=-[\vec{a} \bar{b} c \bar{c}]=$ RHS.

$\begin{array}{r}=\bar{a} \cdot\{(\vec{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{a})\}+ \\ \quad \bar{b} \cdot\{(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{a})\} \\ =\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a} \cdot(\bar{b} \times \bar{a})+\bar{a} \cdot(\bar{c} \times \bar{a})+ \\ \bar{b} \cdot(\bar{b} \times \bar{c})+\bar{b} \cdot(\bar{b} \times \bar{a})+\bar{b} \cdot(\bar{c} \times \bar{a})\end{array}$

$\begin{aligned} & =[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{a}]+[\bar{a} \bar{c} \bar{a}]+ \\ & {[\bar{b} \bar{b} \bar{c}]+[\bar{b} \bar{b} \bar{a}]+[\bar{b} \bar{c} \bar{a}]} \\ & \end{aligned}$

$\begin{aligned} & =[\bar{a} \bar{b} \bar{c}]+0+0+0+0+[\bar{a} \bar{b} \bar{c}] \\ & =2[\bar{a} \bar{b} \bar{c}] \\ & =\text { RHS. }\end{aligned}$

Then volume of the parallelopiped $=[\bar{a} \bar{b} \bar{c}]$

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

= 0(0 – 1) – 1(0 – 1) + 1(1 – 0) = 0 + 1 + 1 = 2cu units.

Also, volume of tetrahedron $=\frac{1}{6}[\bar{a} \bar{b} \bar{c}]$

$=\frac{1}{6}(2)=\frac{1}{3}$ cubic units.

$\ldots[\because \bar{a} \times \bar{b}=-\bar{b} \times \bar{a}]$

$=2 \bar{a} \times \bar{c}=\mathrm{RHS}$

i.e. $\overline{\mathrm{QP}}=\mathrm{p}$ and $\overline{\mathrm{QG}}=g$.

We know that Q, G, P are collinear and G divides segment QP internally in the ratio 1 : 2
∴ by section formula for internal division,

$\bar{g}=\frac{1 \cdot \bar{p}+2 \bar{q}}{1+2}=\frac{\bar{p}}{3} \quad \ldots[\because \bar{q}=0]$

$\begin{aligned} & \therefore \bar{p}=3 \bar{g} \\ & \therefore \overline{\mathrm{QP}}=3 \overline{\mathrm{QG}}\end{aligned}$

Let $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{G}, \mathrm{Q}$ have position vectors $\bar{a}, \bar{b}, \bar{c}, \overline{\boldsymbol{g}}, \bar{q}$ w.r.t. P. We know that $\mathrm{Q}, \mathrm{G}, \mathrm{P}$ are

collinear and G divides segment QP internally in the ratio 1 : 2.

$\therefore \bar{g}=\frac{1 \cdot \bar{p}+2 \bar{q}}{1+2}=\frac{2 \bar{q}}{3}$

$[\because \bar{p}=\overline{0}]$

$\therefore 3 \bar{g}=2 \bar{q}$

$\therefore \frac{3(\bar{a}+\bar{b}+\bar{c})}{3}=2 \bar{q}$

$\therefore \bar{a}+\bar{b}+\bar{c}=2 \bar{q}$

$\therefore \overline{\mathrm{PA}}+\overline{\mathrm{PB}}+\overline{\mathrm{PC}}=2 \overline{\mathrm{PQ}}$

Then $\bar{a}=-\hat{i}+6 \hat{j}+5 \hat{k}, \bar{p}=\frac{-14 \hat{i}+39 \hat{j}+28 \hat{k}}{5}$

Now, $\mathrm{P}$ divides $\mathrm{AB}$ internally in the ratio $3: 2$

$\therefore \bar{p}=\frac{3 \bar{b}+2 \bar{a}}{5}$

$\therefore 5 \bar{p}=3 \bar{b}+2 \bar{a} \quad \therefore 3 \bar{b}=5 \bar{p}-2 \bar{a}$

$\therefore 3 \vec{b}=5\left(\frac{-14 \hat{i}+39 \hat{j}+28 \hat{k}}{5}\right)-2(-\hat{i}+6 \hat{j}+5 \hat{k})$

$\begin{aligned} & =-14 \hat{i}+39 \hat{j}+28 \hat{k}+2 \hat{i}-12 \hat{j}-10 \hat{k} \\ & =-12 \hat{i}+27 \hat{i}+18 \hat{k}\end{aligned}$

$\therefore \bar{b}=-4 \hat{i}+9 \hat{j}+6 \hat{k}$

∴ coordinates of B are (-4, 9, 6).

Taking dot product of both sides with itself, we get

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

$=|\bar{a}|^2+0+0+|\bar{b}|^2 \quad \ldots[\bar{a} \cdot \bar{b}=\bar{b} \cdot \bar{a}=0]$

$=1+1=2 \quad \ldots[|\bar{a}|=|\bar{b}|=1]$

$\therefore|\bar{c}|=\sqrt{2}$.

$\begin{aligned} \mathrm{LHS} & =\overline{\mathrm{AB}}+\overline{\mathrm{AE}}+\overline{\mathrm{BC}}+\overline{\mathrm{DC}}+\overline{\mathrm{ED}} \\ & =(\overline{\mathrm{AB}}+\overline{\mathrm{BC}})+(\overline{\mathrm{AE}}+\overline{\mathrm{ED}}+\overline{\mathrm{DC}})\end{aligned}$

$\begin{aligned} & =\overline{\mathrm{AC}}+\overline{\mathrm{AC}} \\ & =2 \overline{\mathrm{AC}}=\mathrm{RHS}\end{aligned}$

$
|\bar{m}|=\sqrt{(5)^2+(3)^2+(-4)^2}=\sqrt{50}
$
$
|\bar{n}|=\sqrt{(7)^2+(0)^2+(1)^2}=\sqrt{50}
$
If $\theta$ is the angle between $\bar{m}$ and $\bar{n}$ then
$
\begin{aligned}
& \cos \theta=\frac{\bar{m} \cdot \bar{n}}{|\bar{m}||\bar{n}|}=\frac{31}{\sqrt{50} \times \sqrt{50}}=\frac{31}{50} \\
& \therefore \theta=\cos ^{-1}\left(\frac{31}{50}\right)
\end{aligned}
$

$=\overline{0}=$ RHS.

$\begin{aligned} & =(0-0) \hat{i}-(0-0) \hat{j}+(2-(-2)) \hat{k} \\ & =4 \hat{k}\end{aligned}$

$\therefore(\bar{a} \times \bar{b}) \times \bar{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 0 & 4 \\ 2 & 1 & -2\end{array}\right|$

$=(0-4) \hat{i}-(0-8) \hat{j}+(0-0) \hat{k}$

$=-4 \hat{i}+8 \hat{j}$

$\bar{a} \times(\bar{b} \times \bar{c}) \neq(\bar{a} \times \bar{b}) \times \bar{c}$

$\begin{aligned} & =(-4-0) \hat{i}-(-2-0) \hat{j}+(1-4) \hat{k} \\ & =-4 \hat{i}+2 \hat{j}-3 \hat{k}\end{aligned}$

$\therefore \bar{a} \times(\bar{b} \times \bar{c})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 0 \\ -4 & 2 & -3\end{array}\right|$

$\begin{aligned} & =(6-0) \hat{i}-(-3-0) \hat{j}+(2-8) \hat{k} \\ & =6 \hat{i}+3 \hat{j}-6 \hat{k}\end{aligned}$

$=\bar{a} \cdot(\bar{c} \times \bar{a})+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \bar{b} \cdot(\bar{b} \times \bar{c})-$

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

$=0+\bar{a} \cdot(\bar{b} \times \bar{c})+2 \bar{b} \cdot(\bar{c} \times \bar{a})+2 \times 0-0-0$

$\begin{aligned} & =\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]+2\left[\begin{array}{lll}\bar{b} & \bar{c} & \bar{a}\end{array}\right] \\ & =\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]+2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]=3\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]\end{aligned}$

$\begin{aligned} & =\bar{a} \cdot(\bar{b} \times \bar{a})+\bar{a} \cdot(\bar{b} \times \bar{b})+\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a} \cdot(\bar{c} \times \bar{a})+ \\ & \bar{a} \cdot(\bar{c} \times \bar{b})+\bar{a} \cdot(\bar{c} \times \bar{c}) \\ & \end{aligned}$

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

$=0$.

Then, $\bar{a}, \bar{b}, \bar{c}$ are coplanar

$\begin{aligned} & \therefore[\bar{a} \bar{b} \bar{c}]=0 \\ & \therefore\left|\begin{array}{rrr}-3 & 4 & -2 \\ 1 & 0 & 2 \\ 1 & -p & 0\end{array}\right|=0\end{aligned}$

∴ -3(0 + 2p) – 4(0 – 2) – 2(-p – 0) = 0 ∴ -6p + 8 + 2p = 0 ∴ -4p = -8 P = 2.

$\therefore[\bar{a} \bar{b} \bar{c}]=\left|\begin{array}{rrr}3 & 0 & 5 \\ 4 & 2 & -3 \\ 3 & 1 & 4\end{array}\right|$

= 3(8 + 3) – 0(16 + 9) + 5(4 – 6) = 33 – 0 – 10 = 23

$\therefore$ volume of the parallelopiped $=[\bar{a} \bar{b} \bar{c}]$

= 23 cubic units.

= 3(9 + 2) + 1 (6 – 5) + 4(4 + 15) = 33 + 1 + 76 = 110.

Then $|\bar{a} \cdot \bar{b}|=|\bar{a} \times \bar{b}|$ gives

$|a b \cos \theta|=|a b \sin \theta|$

$\therefore-a b \cos \theta=a b \sin \theta \quad \ldots[\because \bar{a} \cdot \bar{b}<0]$

$\therefore-1=\tan \theta$

$\therefore \tan \theta=-\tan \frac{\pi}{4}=\tan \left(\pi-\frac{\pi}{4}\right)$

$\therefore \tan \theta=\tan \frac{3 \pi}{4} \quad \therefore \theta=\frac{3 \pi}{4}$

Hence, the angle between $\bar{a}$ and $\bar{b}$ is $\frac{3 \pi}{4}$.

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

$=(3+6) \hat{i}-(-9-2) \hat{j}+(9-1) \hat{k}$

$=9 \hat{i}+11 \hat{j}+8 \hat{k}$

and $|\bar{a} \times \bar{b}|=\sqrt{9^2+11^2+8^2}=\sqrt{81+121+64}=\sqrt{266}$

Area of the parallelogram having diagonals $\bar{a}$ and $\bar{b}$

$=\frac{1}{2}|\bar{a} \times \bar{b}|=\frac{1}{2} \sqrt{266}$ sq units.

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

$\begin{aligned} & =(6+3) \hat{i}-(-6-1) \hat{j}+(-6+2) \hat{k} \\ & =9 \hat{i}+7 \hat{j}-4 \hat{k}\end{aligned}$

$|\bar{a} \times \bar{b}|=\sqrt{9^2+7^2+(-4)^2}=\sqrt{81+49+16}=\sqrt{146}$

Area of the parallelogram whose adjacent sides are $\bar{a}$ and $\bar{b}$ is $|\bar{a} \times \bar{b}|=\sqrt{146}$ sq units.

$\ldots[\because \bar{a} \times \bar{a}=\bar{b} \times \bar{b}=\overline{0}$ and $-(\bar{b} \times \bar{a})=\bar{a} \times \bar{b}]$

$=\mathrm{RHS}$

$\therefore 2(\bar{a}-\bar{b}) \times 2(\bar{a}+\bar{b})=8(\bar{a} \times \bar{b})$.

Then $\bar{u} \cdot \bar{v}=12$ gives

$|\bar{u} \| \bar{v}| \cos \theta=12$

∴ 10 × 2 × cos θ = 12

$\therefore \cos \theta=\frac{3}{5}$, where $0 \leqslant \theta \leqslant \frac{\pi}{2}$

$\begin{aligned} \sin \theta & =\sqrt{1-\cos ^2 \theta} \\ & =\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}}\end{aligned}$

$=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Now, $|\bar{u} \times \bar{v}|=|\bar{u}||\bar{v}| \sin \theta$

$\therefore \bar{u} \times \bar{v}=10 \times 2 \times \frac{4}{5}=16$

Then $|\bar{u} \times \bar{v}|=8$ gives

$|\bar{u}||\bar{v}| \sin \theta=8$

∴ 2 × 5 × sin θ = 8

$\begin{aligned} & \therefore \sin \theta=\frac{4}{5} \\ & \cos \theta= \pm \sqrt{1-\sin ^2 \theta} \quad \ldots[\because 0 \leqslant \theta \leqslant \pi]\end{aligned}$

$\begin{aligned} & = \pm \sqrt{1-\left(\frac{4}{5}\right)^2}= \pm \sqrt{1-\frac{16}{25}} \\ & = \pm \sqrt{\frac{9}{25}}= \pm \frac{3}{5}\end{aligned}$

Now, $\bar{u} \cdot \bar{v}=|\bar{u}||\bar{v}| \cos \theta$

$\therefore \overline{u \cdot v}=2 \times 5 \times\left( \pm \frac{3}{5}\right)= \pm 6$

$\begin{aligned} \therefore \bar{a} \times \bar{b} & =\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & -2 & 1\end{array}\right| \\ & =(1-6) \hat{i}-(2+3) \hat{j}+(-4-1) \hat{k} \\ & =-5 \hat{i}-5 \hat{j}-5 \hat{k}\end{aligned}$

$\begin{aligned} \therefore|\bar{a} \times \bar{b}| & =\sqrt{(-5)^2+(-5)^2+(-5)^2} \\ & =\sqrt{25+25+25}=\sqrt{75}=5 \sqrt{3}\end{aligned}$

$\therefore$ unit vectors perpendicular to both the vectors $\bar{a}$ and $\bar{b}$.

$=\frac{ \pm(\bar{a} \times \bar{b})}{|\vec{a} \times \bar{b}|}=\frac{ \pm(-5 \hat{i}-5 \hat{j}-5 \hat{k}}{5 \sqrt{3}}$

$=\frac{ \pm(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}$

∴ required vectors of magnitude 5 units

$= \pm \frac{5}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned} & \because \bar{a} \times \bar{b}=2 \hat{i}+\hat{j}+2 \hat{k} \\ & \therefore|\bar{a} \times \bar{b}|=\sqrt{2^2+1^2+2^2}=\sqrt{4+1+4}=3 \\ & \therefore|\bar{a}||\bar{b}| \sin \theta=3\end{aligned}$

... (1)

$\because \bar{a} \cdot \bar{b}=\sqrt{3}$

$\therefore|\bar{a}||\bar{b}| \cos \theta=\sqrt{3}$

... (2)

Dividing (1) by (2), we get

$\frac{|\bar{a}||\bar{b}| \sin \theta}{|\bar{a}||\bar{b}| \cos \theta}=\frac{3}{\sqrt{3}} \quad \therefore \tan \theta=\sqrt{3}=\tan 60^{\circ}$

∴ θ = 60°.

Then $\bar{a} \times \bar{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 1 & 0\end{array}\right|$

$\begin{aligned} & =(0-2) \hat{i}-(0-2) \hat{j}+(0-1) \hat{k} \\ & =-2 \hat{i}+2 \hat{j}-\hat{k}\end{aligned}$

$\therefore|\vec{a} \times \bar{b}|=\sqrt{(-2)^2+2^2+(-1)^2}=\sqrt{4+4+1}=\sqrt{9}=3$

Unit vector perpendicular to both $\bar{a}$ and $\bar{b}$

$= \pm \frac{\bar{a} \times \bar{b}}{|\bar{a} \times \bar{b}|}= \pm\left(\frac{-2 \hat{i}+2 \hat{j}-\hat{k}}{3}\right)$

$= \pm\left(-\frac{2}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\right)$

$\begin{aligned} & \therefore \vec{a}+\bar{b}=(2 \hat{i}+3 \hat{j}-\hat{k})+(\hat{i}-4 \hat{j}+2 \hat{k})=3 \hat{i}-\hat{j}+\hat{k} \\ & \text { and } \bar{a}-\bar{b}=(2 \hat{i}+3 \hat{j}-\hat{k})-(\hat{i}-4 \vec{j}+2 \hat{k})=\hat{i}+7 \hat{j}-3 \hat{k}\end{aligned}$

$\begin{aligned} \therefore(\bar{a}+\bar{b}) \times(\bar{a}-\bar{b}) & =\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ 1 & 7 & -3\end{array}\right| \\ & =(3-7) \hat{i}-(-9-1) \hat{j}+(21+1) \hat{k} \\ & =-4 \hat{i}+10 \hat{j}+22 \hat{k} .\end{aligned}$

Then l = cos α, m = cos β, n = cos γ Here, α = 90°, β = 135° and γ = 45° ∴ l = cos 90° = 0

$m=\cos 135^{\circ}=\cos \left(180^{\circ}-45^{\circ}\right)=-\cos 45^{\circ}=-\frac{1}{\sqrt{2}}$ and $n=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$

$\therefore$ the direction cosines of the line are $0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}$.

Projection of $\bar{a}$ on $\mathrm{X}$-axis

$=\frac{\bar{a} \cdot \hat{i}}{|\hat{i}|}=\frac{(p \hat{i}+q \hat{i}+r \hat{k}) \cdot \hat{i}}{1}=p=2$

Similarly, projections of $\bar{a}$ on $Y$ - and $Z$-axes are 3 and 4 respectively.

∴ sum of these projections = 2 + 3 + 4 = 9.

$\therefore$ the coefficients of $\hat{i}, \hat{j}, \hat{k}$ are proportional

$\begin{aligned} & \therefore \frac{2}{4}=\frac{-q}{-5}=\frac{3}{6} \\ & \therefore \frac{q}{5}=\frac{1}{2} \\ & \therefore q=\frac{5}{2} .\end{aligned}$

Given: $\mathrm{A}=(1,5,0) \therefore \bar{a}=\hat{i}+5 \hat{j}$

Now, $\overline{\mathrm{AB}}=2 \hat{i}-4 \hat{j}+7 \hat{k}$

$\begin{aligned} & \therefore \bar{b}-\bar{a}=2 \hat{i}-4 \hat{j}+7 \hat{k} \\ & \therefore \bar{b}=(2 \hat{i}-4 \hat{j}+7 \hat{k})+\bar{a} \\ & =(2 \hat{i}-4 \hat{j}+7 \hat{k})+(\hat{i}+5 \hat{j}) \\ & =3 \hat{i}+\hat{j}+7 \hat{k}\end{aligned}$

Hence, the terminal point B = (3, 1, 7).

with $\overline{A B}=2 \hat{i}+2 \hat{j}+3 \hat{k}_i \overline{B C}=3 \hat{i}+3 \hat{j}+2 \hat{k}, \overline{A C}=3 \hat{i}+4 \hat{k}$

Now,$\overline{A B}+\overline{B C}$

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

$=-\hat{i}+5 \hat{j}+5 \hat{k} \neq 3 \hat{i}+4 \hat{k}=\overline{\mathrm{AC}}$

Hence, the three vectors do not form a triangle.