(2)

We have, A(1, –1, 2), B(5, 7, –6), C(3, 4, –10) and D(–1, –4, –2)

$\overrightarrow{AC}$ and $\overrightarrow{BD}$ are diagonals of ABCD.

Now, $\overrightarrow{AC}=2\hat{i}+5\hat{j}–12\hat{k}$

and $\overrightarrow{BD}=6\hat{i}+11\hat{j}–4\hat{k}$

Area, of ABCD = $\frac{1}{2}$$\left|\right|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 5& –12\\ 6& 11& –4\end{array}\left|\right|$

$=\frac{1}{2}|112\hat{i}–64\hat{j}–8\hat{k}|=\frac{1}{2}\times \sqrt{16704}$

$=\frac{1}{2}\times 24\sqrt{29}=12\sqrt{29}$.

(2)

Consider $\overrightarrow{a}+\overrightarrow{b}+\hat{i}$

$=2\hat{i}+5\hat{j}–\hat{k}+2\hat{i}–2\hat{j}+2\hat{k}+\hat{i}$

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

Now, $(\overrightarrow{c}+\hat{i})\times (\overrightarrow{a}+\overrightarrow{b}+\hat{i})=\overrightarrow{a}\times (\overrightarrow{c}+\hat{i})$

$\Rightarrow (\overrightarrow{c}+\hat{i})\times (5\hat{i}+3\hat{j}+\hat{k})=(2\hat{i}+5\hat{j}–\hat{k})\times (\overrightarrow{c}+\hat{i})$

$\Rightarrow (\overrightarrow{c}+\hat{i})\times (5\hat{i}+3\hat{j}+\hat{k}+2\hat{i}+5\hat{j}–\hat{k})=0$

$\Rightarrow (\overrightarrow{c}+\hat{i})\times (7\hat{i}+8\hat{j})=0$

$\Rightarrow \overrightarrow{c}+\hat{i}=\lambda (7\hat{i}+8\hat{j})$

$\overrightarrow{a}·\overrightarrow{c}+\overrightarrow{a}·\hat{i}=\lambda \overrightarrow{a}·(7\hat{i}+8\hat{j})$

$\Rightarrow –29+2=\lambda (14+40) \Rightarrow \lambda =\frac{–1}{2}$

$\therefore \overrightarrow{c}+\hat{i}=\frac{–7}{2}\hat{i}–4\hat{j}$

$\Rightarrow \overrightarrow{c}=\frac{–9}{2}\hat{i}–4\hat{j}$

$\therefore \overrightarrow{c}·(–2\hat{i}+\hat{j}+\hat{k})=9–4=5$.

(2)

Consider $|\overrightarrow{c}–\overrightarrow{a}{|}^2$

    $=|\overrightarrow{c}{|}^2+|\overrightarrow{a}{|}^2–2\overrightarrow{a}·\overrightarrow{c}=|\overrightarrow{c}{|}^2+4–2(\overrightarrow{b}\times \overrightarrow{c})·\overrightarrow{c}=|\overrightarrow{c}{|}^2+4–0$

$\therefore |\overrightarrow{c}–\overrightarrow{a}{|}^2=|\overrightarrow{c}{|}^2+4$          ... (i)

Now, $\left|\overrightarrow{a}\right|=|\overrightarrow{b}\times \overrightarrow{c}| \Rightarrow 2=\left|\overrightarrow{b}\right|\left|\overrightarrow{c}\right|\sin \alpha , \alpha \in [0,\frac{\pi }{3}]$

    $=3\left|\overrightarrow{c}\right| \sin \alpha$

$\Rightarrow \left|\overrightarrow{c}\right|=\frac{2}{3}\mathrm{cosec} \alpha$

$\Rightarrow |\overrightarrow{c}{|}_{\min }=\frac{2}{3}\times \frac{2}{\sqrt{3}}$              $(\because \alpha \in [0,\frac{\pi }{3}\left]\right)$

$\Rightarrow 27|\overrightarrow{c}–\overrightarrow{a}{|}_{\min }^2=27(\frac{16}{27}+4)$

    =16 + 108 = 124.

(2)

Gicen, A(3, 1, –1), $B(\frac{5}{3},\frac{7}{3},\frac{1}{3})$, C(2, 2, 1) and $D(\frac{10}{3},\frac{2}{3},\frac{–1}{3})$ are vertices of quadrilateral ABCD

$\overrightarrow{AC}=(2–3)\hat{i}+(2–1)\hat{j}+(1+1)\hat{k}=–\hat{i}+\hat{j}+2\hat{k}$

$\overrightarrow{BD}=(\frac{10}{3}–\frac{5}{3})\hat{i}+(\frac{2}{3}–\frac{7}{3})\hat{j}+(\frac{–1}{3}–\frac{1}{3})\hat{k}=\frac{5}{3}\hat{i}–\frac{5}{3}\hat{j}–\frac{2}{3}\hat{k}$

Area of quadrilateral ABCD = $\frac{1}{2}$$|\overrightarrow{AC}\times \overrightarrow{BD}|$

$=\frac{1}{2}$$\left|\right|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ –1& 1& 2\\ 5/3& –5/3& –2/3\end{array}\left|\right|$$=\frac{1}{2}$$\left|\hat{i}\right(\frac{8}{3})–\hat{j}(\frac{–8}{3})+\hat{k}(0\left)\right|$

$=\frac{1}{2}\sqrt{\frac{64}{9}+\frac{64}{9}}=\frac{4\sqrt{2}}{3}\mathrm{sq}. \mathrm{units}$.

(4)

We have, $\overrightarrow{a}=2\hat{i}+\hat{j}–\hat{k}$

and $\overrightarrow{b}=\left(\right(\overrightarrow{a}\times (\hat{i}+\hat{j}))\times \hat{i})\times \hat{i}$

  $=$$\left(\right|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& –1\\ 1& 1& 0\end{array}|\times \hat{i})$$\times \hat{i}=\left(\right(\hat{i}–\hat{j}+\hat{k})\times \hat{i})$

   $=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& –1& 1\\ 1& 0& 0\end{array}\right|$$\times \hat{i}=(\hat{j}+\hat{k})\times \hat{i}=–\hat{k}+\hat{j}$

$\Rightarrow \overrightarrow{b}=\hat{j}–\hat{k}$

So, projection of $\overrightarrow{a}$ on $\overrightarrow{b}$ = $\frac{\overrightarrow{a}·\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\therefore$  Square of projection = 2.

(3)

We have, $\left|\right(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|$$=|\overrightarrow{a}\times \overrightarrow{b}|\left|\overrightarrow{c}\right| \sin 60°$

Now, $\overrightarrow{a}\times \overrightarrow{b}=$$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 6& 1& –1\\ 1& 1& 0\end{array}\right|$$=\hat{i}–\hat{j}+5\hat{k}$

So, $|\overrightarrow{a}\times \overrightarrow{b}|$$=\sqrt{1+1+25}=\sqrt{27}=3\sqrt{3}$

Also, $|\overrightarrow{c}-\overrightarrow{a}|$$=2\sqrt{2}$        [Given]

$\Rightarrow |\overrightarrow{c}{|}^2+|\overrightarrow{a}{|}^2–2|\overrightarrow{c}·\overrightarrow{a}|=8$

$\Rightarrow |\overrightarrow{c}{|}^2+38–12|\overrightarrow{c}|=8$          $[\because \overrightarrow{c}·\overrightarrow{a}=6|\overrightarrow{c}\left|\right]$

$\Rightarrow |\overrightarrow{c}{|}^2–12|\overrightarrow{c}|+30=0$

$\Rightarrow \left|\overrightarrow{c}\right|=\frac{12+\sqrt{24}}{2}$          $[\because |\overrightarrow{c}|\ge 6]$

$\Rightarrow \left|\overrightarrow{c}\right|=6+\sqrt{6}$

$\therefore$$\left|\right(\overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c}|$$=\sqrt{27}\times (6+\sqrt{6})\times \frac{\sqrt{3}}{2}=\frac{9}{2}(6+\sqrt{6})$.

(3)

Given, $\overrightarrow{a}=\alpha t\hat{i}+6\hat{j}–3\hat{k}$ and $\overrightarrow{b}=t\hat{i}–2\hat{j}–2\alpha t\hat{k}$

So, $\overrightarrow{a}·\overrightarrow{b}<0 \forall t\in R$      [$\therefore$  $\overrightarrow{a}$ and $\overrightarrow{b}$ are inclined at an obtuse angle]

$\Rightarrow \alpha t^2–12+6\alpha t<0 \forall t\in R \Rightarrow \alpha <0 \mathrm{and} D<0$

$\Rightarrow 36{\alpha }^2+48\alpha <0 \Rightarrow 12\alpha (3\alpha +4)<0$

$\Rightarrow \frac{–4}{3}<\alpha <0$

Also, for $\alpha =0, \overrightarrow{a}·\overrightarrow{b}<0 \Rightarrow \alpha \in (–\frac{4}{3},0]$.

(2)

We have, $\overrightarrow{a}=4\hat{i}–\hat{j}+\hat{k}$ and $\overrightarrow{b}=11\hat{i}–\hat{j}+\hat{k}$

$\overrightarrow{a}·\overrightarrow{b}=44+1+1=46, |\overrightarrow{a}{|}^2=18, |\overrightarrow{b}{|}^2=123$

$(\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=\overrightarrow{c}\times (–2\overrightarrow{a}+3\overrightarrow{b})$          ... (i)          [Given]

and $(2\overrightarrow{a}+3\overrightarrow{b})·\overrightarrow{c}=1670$          ... (ii)

$\therefore (\overrightarrow{a}+\overrightarrow{b})\times \overrightarrow{c}=(2\overrightarrow{a}–3\overrightarrow{b})\times \overrightarrow{c}$          [From (i)]

$\Rightarrow (–\overrightarrow{a}+4\overrightarrow{b})\times \overrightarrow{c}=0 \Rightarrow \overrightarrow{c}=\lambda (4\overrightarrow{b}–\overrightarrow{a})$

$\Rightarrow (2\overrightarrow{a}+3\overrightarrow{b})·\lambda (4\overrightarrow{b}–\overrightarrow{a})=1670$          [From (ii)]

$\Rightarrow \lambda (5\overrightarrow{a}·\overrightarrow{b}–2|\overrightarrow{a}{|}^2+12\left|\overrightarrow{b}{|}^2\right)=1670$

$\Rightarrow \lambda =\frac{1670}{5\times 46–2\times 18+12\times 123}=1$

Now, $\overrightarrow{c}=4\overrightarrow{b}–\overrightarrow{a}=4(11\hat{i}–\hat{j}+\hat{k})–(4\hat{i}–\hat{j}+\hat{k})$

        $=40\hat{i}–3\hat{j}+3\hat{k}$

$\therefore |\overrightarrow{c}{|}^2=1600+9+9=1618$.

(1)

We have, $\overrightarrow{c}=\overrightarrow{a}–\overrightarrow{b}$

$\Rightarrow \overrightarrow{c}=(\alpha –5)\hat{i}+\hat{j}–2\hat{k}$

Let ABC be the given triangle.

Area of $△ABC=\frac{1}{2}|\overrightarrow{a}\times \overrightarrow{b}|=\frac{1}{2}$$\left|\right|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ \alpha & 4& 2\\ 5& 3& 4\end{array}\left|\right|$

$\Rightarrow \frac{1}{2}$$|10\hat{i}–(4\alpha –10)\hat{j}+(3\alpha –20\left)\hat{k}\right|$$=5\sqrt{6}$          [Given]

$=100+{(4\alpha –10)}^2+{(3\alpha –20)}^2=600$

$\Rightarrow 25{\alpha }^2–200\alpha =0 \Rightarrow \alpha =8$      $[\because \alpha .]$

$\Rightarrow |\overrightarrow{c}{|}^2=9+1+4=14$.

(1)

We have, $|\overrightarrow{OA}\times \overrightarrow{OC}|$$=15$

$\Rightarrow$$|2\overrightarrow{a}\times 3\overrightarrow{b}|$$=15 \Rightarrow$$|\overrightarrow{a}\times \overrightarrow{b}|$$=\frac{5}{2}$          ... (i)

Area of quadrilateral $OABC=\frac{1}{2}$$|\overrightarrow{OB}\times \overrightarrow{AC}|$

$=\frac{1}{2}$$\left|\right(6\overrightarrow{a}+5\overrightarrow{b})\times (3\overrightarrow{b}–2\overrightarrow{a}\left)\right|$$=\frac{1}{2}$$\left|18\right(\overrightarrow{a}\times \overrightarrow{b})–10(\overrightarrow{b}\times \overrightarrow{a}\left)\right|$

$=\frac{1}{2}$$\left|18\right(\overrightarrow{a}\times \overrightarrow{b})+10(\overrightarrow{a}\times \overrightarrow{b}\left)\right|$$=14$$|\overrightarrow{a}\times \overrightarrow{b}|$

$=14\times \frac{5}{2} [\mathrm{usin}g (i\left)\right]$

= 35 sq. units.