If two \(\vec{a} \text { and } \vec{b}\) vectors are given in component form as \(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} \text { and }\) \(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\) then their scalar product is given as
\( \vec{a} \cdot \vec{b} =\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right) \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \\ =a_1 \hat{i} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_2 \hat{j} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)+a_3 \hat{k} \cdot\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right) \\ =a_1 b_1(\hat{i} \cdot \hat{i})+a_1 b_2(\hat{i} \cdot \hat{j})+a_1 b_3(\hat{i} \cdot \hat{k})+a_2 b_1(\hat{j} \cdot \mid \hat{i})+a_2 b_2(\hat{j} \cdot \hat{j})+a_2 b_3(\hat{j} \cdot \hat{k}) a_3 b_1(\hat{k} \cdot \hat{i})+a_3 b_2(\hat{k} \cdot \hat{j})+a_3 b_3(\hat{k} \cdot \hat{k})\)
\( = a_1 b_1+a_2 b_2+a_3 b_3 \)
Thus \( \vec{a} \cdot \vec{b}=a_1 b_1+a_2 b_2+a_3 b_3 \)

Sum of vectors \(\overrightarrow { b } =2\hat { i } +4\hat { j } -5\hat { k } \) and \(\overrightarrow { c } =\lambda \hat { i } +2\hat { j } +3\hat { k } \) 
\(=(2+\lambda )\hat { i } +6\hat { j } -2\hat { k } \)  
\(\therefore\)  Unit vector along the sum \(=\frac { (2+\lambda )\hat { i } +6\hat { j } -2\hat { k } }{ \sqrt { { (2+\lambda ) }^{ 2 }+36+4 } } \)
by the question \(\left( \hat { i } +\hat { j } +\hat { k } \right) =\left( \frac { (2+\lambda )\hat { i } +6\hat { j } -2\hat { k } }{ \sqrt { { (2+\lambda ) }^{ 2 } } +40 } \right) =1\)
\(\Rightarrow (1){ (2+\lambda ) }+(1)(6)+(1)(-2)=\sqrt { { (2+\lambda ) }^{ 2 }+40 } \)
\(\Rightarrow 2+\lambda +6-2\quad =\sqrt { { (2+\lambda ) }^{ 2 }+40 } \)
\(\Rightarrow 6+2= \sqrt { { (2+\lambda ) }^{ 2 }+40 } \)
Squaring \({ (\lambda +6) }^{ 2 }={ { (2+\lambda ) } }^{ 2 }+40\quad \)
\(\Rightarrow { \lambda }^{ 2 }+12+36=4+{ \lambda }^{ 2 }+4\lambda +40\)
\(\Rightarrow 12\lambda +36=4\lambda +44\Rightarrow 8\lambda =8\) 
Hence  \(\lambda =1\)
(ii) Unit vector along \(\overrightarrow { b } +\overrightarrow { c } =\frac { (2+1)\hat { i } +2\hat { j } -6\hat { k } }{ \sqrt { { (2+1) }^{ 2 }+36+4 } } \) 
\(=\frac { 1 }{ 7 } (3\hat { i } +6\hat { j } -2\hat { k } ).\) 

We have : \(\Rightarrow { \overset { \rightarrow }{ |a| } =1=\overset { \rightarrow }{ |b| } }\ and\ \overset { \rightarrow }{ |a| } +\overset { \rightarrow }{ |b| } =1\) 
sqaring  \(|\overset { \rightarrow }{ a } +{ \overset { \rightarrow }{ { b| }^{ 2 } } }=1\Rightarrow (\overset { \rightarrow }{ a } +{ \overset { \rightarrow }{ { b) }^{ 2 } } }\)
\(\Rightarrow \overset { \rightarrow }{ { b| }^{ 2 } } +\overset { \rightarrow }{ { b| }^{ 2 } } +2|\overset { \rightarrow }{ a| } |\overset { \rightarrow }{ b| } =1\)
\( \Rightarrow { a }^{ 2 }+{ b }^{ 2 }+2|\overset { \rightarrow }{ a| } |\overset { \rightarrow }{ b| } cos\theta =1\)
Where '\(\theta \)' is the angle between  \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \)
\(\Rightarrow 1+1+2(1)(1)cos\theta =1\)
\(\Rightarrow cos\theta =-\frac { 1 }{ 2 } \)
Hence, \(\theta =120°\)

Here  \(\overset { \rightarrow }{ AB } =(2\overset { \wedge }{ i } -\overset { \wedge }{ j } +3\overset { \wedge }{ k } )-(\overset { \wedge }{ i } +\overset { \wedge }{ j } -\overset { \wedge }{ k } )\) 
\(=\overset { \wedge }{ i } -2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \) 
and \(\overset { \rightarrow }{ CD } =(3\overset { \wedge }{ i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } )-(2\overset { \wedge }{ i } -3\overset { \wedge }{ k } )\)         
\(=\overset { \wedge }{ i } -2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \) 
\(\therefore\) The projcetionj of \(\overset { \rightarrow }{ AB } \) along \(\overset { \rightarrow }{ CD } \)  \(=\frac { \overset { \rightarrow }{ AB } .\overset { \rightarrow }{ CD } }{ |\overset { \rightarrow }{ CD } | } \) 
\(=\frac { (\overset { \wedge }{ i } -2\overset { \wedge }{ j } +4\overset { \wedge }{ k } ).(\overset { \wedge }{ i } -2\overset { \wedge }{ j } +4\overset { \wedge }{ k } ) }{ |\overset { \wedge }{ i } -2\overset { \wedge }{ j } +4\overset { \wedge }{ k } | } \)
\(=\frac { (1)(1)+(-2)(-2)+(4)(4) }{ \sqrt { 1+4+16 } } =\frac { 1+4+16 }{ \sqrt { 21 } } =\frac { 21 }{ \sqrt { 21 } } =\sqrt { 21 } \)      

\(\vec{r} =2 \vec{a}-\vec{b}+3 \vec{c} \)
\(=2 \hat{i}+2 \hat{j}+2 \hat{k}-4 \hat{i}+2 \hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k} \)
\(\Rightarrow \vec{r}=\hat{i}-2 \hat{j}+2 \hat{k} \)
Vector of magnitude 6 units and parallel to
\((2 \hat a-\hat b+3\hat c) \text { is } 6 \hat{r}\)
\(\text { Vector }=6\left(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{1+4+4}}\right)=2 \hat{i}-4 \hat{j}+4 \hat{k}\)

Applying triangle law of addition of vectors in triangles 
APB, AQB and ARB, we get
\(\overrightarrow{A P}+\overrightarrow{P B}=\overrightarrow{A B}, \overrightarrow{A Q}+\overrightarrow{Q B}=\overrightarrow{A B}\) 
and \(\overrightarrow{A R}+\overrightarrow{R B}=\overrightarrow{A B}\) 
On adding all these, we get

\(\overrightarrow{A P}+\overrightarrow{P B}+\overrightarrow{A Q}+\overrightarrow{Q B}+\overrightarrow{A R}+\overrightarrow{R B}=3 \overrightarrow{A B}\)

\(\overrightarrow b\times \overrightarrow c=\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ -1 & 2 & 1 \\ 3 & 1 & 2 \end{matrix} \right| \)
\(=\overset\wedge i(4-1)-\overset\wedge j(-2-3)+\overset\wedge k(-1-6)\)
\(=3\overset\wedge i+5\overset\wedge j-7\overset\wedge k\)
\(\therefore \overrightarrow a.(\overrightarrow b\times\overrightarrow c)=(2\overset\wedge i+\overset\wedge j+3\overset\wedge k).(3\overset\wedge i+5\overset\wedge j-7\overset\wedge k)\)
= 6 + 5 - 21 = -10

\((\hat{k} \times \hat{i}) \cdot \hat{j}+\hat{i} \cdot \hat{k}=\hat{j} \cdot \hat{j}+\hat{i} \cdot \hat{k}\)
= 1 + 0 = 1

Let given position vectors are \(\overrightarrow{O A}=\vec{a}-2 \vec{b}\) and \(\overrightarrow{O B}=2 \vec{a}+\vec{b}\)
Let \(\overrightarrow{O C}\) be the position vector of a point C which divides the join of points, with position vectors \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\), externally in the ratio 2 : 1.
\(\therefore \overrightarrow{O C}=\frac{2 \overrightarrow{O B}-1 \overrightarrow{O A}}{2-1}\) [by external section formula)
\(\begin{aligned} =\frac{2(2 \vec{a}+\vec{b})-1(\vec{a}-2 \vec{b})}{1} \end{aligned}\)
\(\begin{aligned} =4 \vec{a}+2 \vec{b}-\vec{a}+2 \vec{b}=3 \vec{a}+4 \vec{b} \end{aligned}\)

\(\text { Let initial point be }(\alpha, \beta, \gamma) \text { . }\)
\(\therefore 3 \hat{i}-\hat{j}-5 \hat{k}=(0-\alpha) \hat{i}+(1-\beta) \hat{j}+(3-\gamma) \hat{k} \)
\(\Rightarrow 3=-\alpha,-1=1-\beta,-5=3-\gamma \)
\(\Rightarrow \alpha_{0}=-3, \beta=2, \gamma=8 \)
Coordinates of initial point are (-3,2,8)

Given, \(\vec{\alpha}=4 \hat{i}+5 \hat{j}-\hat{k}, \vec{\beta}=\hat{i}-4 \hat{j}+5 \hat{k}\)
and \(\vec{q}=3 \hat{i}+\hat{j}-\hat{k}\)
Also, vector \(\vec{p}\) is perpendicular to \(\alpha\) and \(\beta\).
Then, \(\vec{p}=\lambda(\vec{\alpha} \times \vec{\beta})\)       ...(i)
Now, \(\vec{\alpha} \times \vec{\beta}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 5 & -1 \\ 1 & -4 & 5 \end{array}\right|\)
\(\begin{aligned} =\hat{i}(25-4)-\hat{j}(20+1)+\hat{k}(-16-5) \end{aligned}\)
\(\begin{aligned} =\hat{i}(21)-\hat{j}(21)+\hat{k}(-21) \end{aligned}\)
\(\Rightarrow \quad \vec{\alpha} \times \vec{\beta}=21 \hat{i}-21 \hat{j}-21 \hat{k}\)
So, \(\vec{p}=21 \lambda \hat{i}-21 \lambda \hat{j}-21 \lambda \hat{k}\) [from Eq. (i)]  ...(ii)
Also, given that \(\vec{p} \cdot \vec{q}=21\)
\(\begin{aligned} \therefore \quad(21 \lambda \hat{i}-21 \lambda \hat{j}-21 \lambda \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})=21 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 63 \lambda-21 \lambda+21 \lambda=21 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad 63 \lambda=21 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \lambda=1 / 3 \end{aligned}\)
On putting \(\lambda=\frac{1}{3}\) in Eq. (ii), we get
\(\vec{p}=21 \times \frac{1}{3} \hat{i}-21 \times \frac{1}{3} \hat{j}-21 \times \frac{1}{3} \hat{k}\)
\(\therefore \quad \vec{p}=7 \hat{i}-7 \hat{j}-7 \hat{k}\)
which is the required vector.

As given \(\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } =0=\overrightarrow { a } \bot \)both \(\overrightarrow { b } \) and \(\overrightarrow { c } \)
⇒ \(\overrightarrow { a } \parallel \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Let \(\overrightarrow { a } =\lambda \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
then \(\left| \overrightarrow { a } \right| =\left| \lambda \right| \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| \)
⇒ \(\frac { \left| \overrightarrow { a } \right| }{ \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| } =\left| \lambda \right| \)
⇒ \(\left| \lambda \right| =\frac { 1 }{ sin\frac { \pi }{ 6 } } =2\)
λ = 士2
\(\overrightarrow { a } =\pm 2\left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Hence proved
(ii) Now \(\left[ \overrightarrow { a } +\overrightarrow { b } ,\overrightarrow { b } +\overrightarrow { c } ,\overrightarrow { c } +\overrightarrow { a } \right] \)
= \(\left[ \left( \overrightarrow { a } +\overrightarrow { b } \right) \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right] .\left( \overrightarrow { c } +\overrightarrow { a } \right) \)
= \(\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } +\left( \overrightarrow { b } \times \overrightarrow { c } \right) .\overrightarrow { a } \)
(As the scalar triple product = 0, if any two vectors are equal)
Hence, \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } \) = \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\overrightarrow { c } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= 2\(\overrightarrow { a } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= \(2\overrightarrow { a } \left( \pm \frac { 1 }{ 2 } \overrightarrow { a } \right) \)
= 士1
Hence proved

(i) (b): Here, (4, 10) are the coordinates of A.
\(\therefore\) P.V. of A = \(4 \hat{i}+10 \hat{j}\)
(ii) (c): Here, (9, 7) are the coordinates of B.
\(\therefore\) P.V. of B \(=9 \hat{i}+7 \hat{j}\)
(iii) (b): Here, P.V. of A  = \(4 \hat{i}+10 \hat{j}\) and P.V. of \(C=4 \hat{i}+2 \hat{j}\)
\(\therefore \ \overrightarrow{A C}=(4-4) \hat{i}+(2-10) \hat{j}=-8 \hat{j}\)
(iv) (a): Here \(\vec{A}=\hat{i}+2 \hat{j}+3 \hat{k}\)
\( \therefore \ |\vec{A}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \)
\(\therefore \ \hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{\hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{14}}=\frac{1}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)
(v) (d): We have, \(\vec{A}=4 \hat{i}+3 \hat{j}\) and \( \vec{B}=3 \hat{i}+4 \hat{j}\)
\(\therefore|\vec{A}|=\sqrt{4^{2}+3^{2}}=\sqrt{16+9}=\sqrt{25}=5\)
and \(|\vec{B}|=\sqrt{3^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5\)
Thus, \(|\vec{A}|+|\vec{B}|=5+5=10\)

(i) (a) : \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}, \vec{b}=3 \hat{i}-3 \hat{j}-2 \hat{k}\)  and \(\vec{c}=-2 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\therefore \ \vec{a}+\vec{b}+\vec{c}=2 \hat{i}+3 \hat{j}+6 \hat{k}\)
(ii) (c): Using triangle law of addition in \(\Delta\)ABC, we get
\(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=\overrightarrow{0}\) which can be rewritten as
\(\overrightarrow{A B}+\overrightarrow{B C}-\overrightarrow{A C}=\overrightarrow{0} \text { or } \overrightarrow{A B}-\overrightarrow{C B}+\overrightarrow{C A}=\overrightarrow{0}\)
(iii) (c) : We have, A(1, 4, 2), B(3, -3, -2) and C(-2, 2, 6)
Now, \(\overrightarrow{A B}=\vec{b}-\vec{a}=2 \hat{i}-7 \hat{j}-4 \hat{k}\)
and \(\overrightarrow{A C}=\vec{c}-\vec{a}=-3 \hat{i}-2 \hat{j}+4 \hat{k}\)
\(\therefore \overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & -7 & -4 \\ -3 & -2 & 4 \end{array}\right|\)
\(=\hat{i}(-28-8)-\hat{j}(8-12)+\hat{k}(-4-21)=-36 \hat{i}+4 \hat{j}-25 \hat{k}\)
Now, \(|\overrightarrow{A B} \times \overrightarrow{A C}|=\sqrt{(-36)^{2}+4^{2}+(-25)^{2}}\)
\(=\sqrt{1296+16+625}=\sqrt{1937}\)
\(\therefore \text { Area of } \Delta A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|=\frac{1}{2} \sqrt{1937} \text { sq. units }\)
(iv) (d): If the given points lie on the straight line, then the points will be collinear and so area of \( \Delta A B C=0 \)
\(\Rightarrow |\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|=0\)
If a, b, c are the position vectors of the three vertices A, Band C of \( \Delta A B C \), then area of triangle
\(\left.=\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|\right]\)
(v) (b): Here, \( |\vec{a}| =\sqrt{2^{2}+3^{2}+6^{2}}=\sqrt{4+9+36} \)
\(=\sqrt{49}=7\)
\(\therefore\) Unit vector in the direction of vector \(\vec{a}\) is
\(\hat{a}=\frac{2 \hat{i}+3 \hat{j}+6 \hat{k}}{7}=\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\)