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

Given, \(\left| a\right| =\left| b \right| \)
and a.b \(=\frac{1}{2}\)
Let \(\theta \) be thae angle between \(\overrightarrow a\) and \(\overrightarrow b\)
then, \(\cos \theta=\frac{a.b}{\left| a\right| \left| b \right| }\)
\(\cos 60^{ \circ }=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right| .\left| \overrightarrow { a} \right| }\)
\(\Rightarrow \frac{1}{2}=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right|^{ 2} }\)
\(\Rightarrow \left| \overrightarrow { a } \right| ^{2 }=\frac{\frac{1}{2}}{\frac{1}{2}}=1\)
\(\Rightarrow \left| \overrightarrow { a} \right| =1 \)
\(\Rightarrow\left| \overrightarrow { a} \right| =\left| \overrightarrow { b} \right| =1\)

Required projection :
\(\frac{(\overset\wedge i+3\overset\wedge j+7\overset\wedge k).(2\overset\wedge i-3\overset\wedge j+6\overset\wedge k)}{\left| 2\overset\wedge i-6\overset\wedge j+6\overset\wedge k \right| }\)
\(=\frac{35}{\sqrt49}=\frac{35}{7}=5\)

\(\overset\rightarrow c=\overset\rightarrow a-\overset\rightarrow b\)
\(=(3\overset\wedge i-2\overset\wedge j+3\overset\wedge k)-(\overset\wedge i-4\overset\wedge j-\overset\wedge k)\)
\(\Rightarrow \overset\rightarrow c=2 \overset\wedge i+2\overset\wedge j+4\overset\wedge k\)
\(\overset\wedge c=\frac{\overset\rightarrow c}{\left| c \right| }=\frac{2\overset\wedge i+2\overset\wedge j+4\overset\wedge k}{\sqrt{4+4+16}}\)
\(=\frac{2\overset\wedge i+2\overset\wedge j+4\overset\wedge k}{\sqrt{24}}\)
\(=\frac{2(\overset\wedge i+\overset\wedge j+2\overset\wedge k)}{2\sqrt6}\)
\(\overset\wedge c=\frac{\overset\wedge i+\overset\wedge j+2\overset\wedge k}{\sqrt{6}}\)
\(\Rightarrow\)Required vector =\(9\overset\wedge c=\frac{9\overset\wedge i+9\overset\wedge j+18\overset\wedge k}{\sqrt6}\)

Scalar quantity: Work done
Vector quantity: Force, velocity, displacement.

Given, \(\vec{a}=\vec{b} \Rightarrow x \hat{i}+2 \hat{j}-z \hat{k}=3 \hat{i}-y \hat{j}+\hat{k}\)
On comparing the coefficient of components, we get
x = 3, y = -2, z = -1
Now, x + y + z = 3 - 2 - 1 = 0

\(\frac{3}{1}=\frac{2}{p}=\frac{9}{3} \Rightarrow p=\frac{2}{3}\)

\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{\sqrt{9+4+36}}=\frac{3}{7} \vec{i}-\frac{2}{7} \vec{j}+\frac{6}{7} \vec{k}\)

Given, \(\vec{a}=3 \hat{i}-\hat{j} \text { and } \vec{b}=2 \hat{i}+\hat{j}-3 \hat{k}\)
Let \(\overrightarrow{b_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{b}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) are two vectors such that \(\overrightarrow{b_1}+\overrightarrow{b_2}=\vec{b}, \overrightarrow{b_1} \| \vec{a}\) and \(\overrightarrow{b_2} \perp \vec{a}\)
Consider, \(\overrightarrow{b_1}+\overrightarrow{b_2}=\vec{b}\)
\(\Rightarrow \quad\left(x_1+x_2\right) \hat{i}+\left(y_1+y_2\right) \hat{j}+\left(z_1+z_2\right) \hat{k}=2 \hat{i}+\hat{j}-3 \hat{k}\)
On comparing the coefficients of \(\hat{i}, \hat{j} \text { and } \hat{k}\) both sides, we get
\(\Rightarrow\)   x₁ + x₂ = 2                                                 ...(i)
        y₂ + y₂ = 1                                                 ...(ii)
and  z₁ + z₂ = -3                                                ...(iii)
Now, consider, \(\vec{b}_1 \| \vec{a}\)
\(\Rightarrow \quad \frac{x_1}{3}=\frac{y_1}{-1}=\frac{z_1}{0}=\lambda\) (say)
\(\Rightarrow \quad x_1=3 \lambda, y_1=-\lambda \text { and } z_1=0\)     ...(iv)
On substituting the values of x, y and z, from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get
\(x_2=2-3 \lambda, y_2=1+\lambda \text { and } z_2=-3\)  ...(v)
Since, \(\overrightarrow{b_2} \perp \vec{a}\), therefore \(\vec{b}_2 \cdot \vec{a}=0\)
\(\begin{aligned} \Rightarrow & 3 x_2-y_2 =0 \end{aligned} \)
\(\Rightarrow 3(2-3 \lambda)-(1+\lambda) =0\)                          [from Eq. (v)]
\(\Rightarrow 6-9 \lambda-1-\lambda =0 \Rightarrow 5-10 \lambda=0\)
\(\Rightarrow \quad \lambda=\frac{1}{2}\)
On substituting \(\lambda=\frac{1}{2}\) in Eqs. (iv) and (v), we get
\(\begin{aligned} x_1=\frac{3}{2}, y_1=\frac{-1}{2}, z_1=0 \end{aligned}\)
and \(\begin{aligned} x_2=\frac{1}{2}, y_2=\frac{3}{2} \text { and } z_2=-3 \end{aligned}\)
Hence, \(\begin{aligned} \vec{b}_1+\vec{b}_2 & =\left(\frac{3}{2} \hat{i}-\frac{1}{2} \hat{j}\right)+\left(\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}-3 \hat{k}\right) \end{aligned}\)
\(\begin{aligned} =2 \hat{i}+\hat{j}-3 \hat{k}=\vec{b} \end{aligned}\),
where \(\vec{b}_1 \| \vec{a} \text { and } \overrightarrow{b_2} \perp \vec{a}\)

Let \(\overset { \rightarrow }{ a } ={ a }_{ 1 }\overset { \wedge }{ i } +{ a }_{ 2 }\overset { \wedge }{ j } +{ a }_{ 3 }\overset { \wedge }{ k } \)
\(\therefore\)   \(\overset { \rightarrow }{ a } .\overset { \wedge }{ i } =({ a }_{ 1 }\overset { \wedge }{ i } +{ a }_{ 2 }\overset { \wedge }{ j } +{ a }_{ 3 }\overset { \wedge }{ k } ).\overset { \wedge }{ i } \)
\(={ a }_{ 1 }(1)+{ a }_{ 2 }(0)+{ a }_{ 3 }(0)={ a }_{ 1 }\)
\(\therefore\) \((\overset { \rightarrow }{ a } .\overset { \wedge }{ i } )\overset { \wedge }{ i } ={ a }_{ 1 }\overset { \wedge }{ i } \)
Similarly, \((\overset { \rightarrow }{ a } .\overset { \wedge }{ j } )\overset { \wedge }{ j } ={ a }_{ 2 }\overset { \wedge }{ j } and\quad (\overset { \rightarrow }{ a } .\overset { \wedge }{ k } ).\overset { \wedge }{ k } ={ a }_{ 3 }\overset { \wedge }{ k } \)
Adding \((\overset { \rightarrow }{ a } .\overset { \wedge }{ i } )\overset { \wedge }{ i } +(\overset { \rightarrow }{ a } .\overset { \wedge }{ j } )\overset { \wedge }{ j } +(\overset { \rightarrow }{ a } .\overset { \wedge }{ k } ).\overset { \wedge }{ k } \) 
\(={ a }_{ 1 }\overset { \wedge }{ i } +{ a }_{ 2 }\overset { \wedge }{ j } +{ a }_{ 3 }\overset { \wedge }{ k } \) 
\(=\overset { \rightarrow }{ a } \), which is true

Part (C) is the coorrect answer.
Reson:\(\hat{i} \cdot(\hat{i} * \hat{k})+\hat{j} \cdot(\hat{i} * \hat{k})+\hat{k} \cdot(\hat{i} * \hat{j})\) 
\(=\hat{i} . \hat{i}+\hat{j}-(\hat{j})+\hat{k} \cdot \hat{k}=\hat{i} \cdot \hat{i}-\hat{j} \cdot \hat{j}+\hat{k} \cdot \hat{k}\) 
= 1 - 1 + 1 = 1

Here \(\overset { \rightarrow }{ |a| } =\overset { \rightarrow }{ |b| } =\overset { \rightarrow }{ |c| } \) 
and \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } =0\)..(1)
Let \(\alpha,\beta,\gamma\) be the angle, which \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{b }+\overset { \rightarrow }{ c } \) makes with \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{b },\overset { \rightarrow }{ c } \) respectively
\(\therefore\) cos \(\alpha\) = \(\frac { \overset { \rightarrow }{ a } .(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ) }{ \overset { \rightarrow }{ |a } ||\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } | } \) 
\(=\frac { \overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } ) }{ \overset { \rightarrow }{ |a } ||\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } | } \) 
\(=\frac { \overset { \rightarrow }{ |a } |+0+0 }{ \overset { \rightarrow }{ |a } ||\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } | } \) 
\(=\frac { \overset { \rightarrow }{ |a } | }{ |\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } | } \) ..(2)
Similarly cos \(\beta\) = \(\frac { \overset { \rightarrow }{ |b } | }{ |\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } | } \) .(3)
and cos \(\gamma\) = \(\frac { \overset { \rightarrow }{ |c } | }{ |\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } | } \) ...(4)
From (2), (3) and (4), cos \(\alpha\) = cos \(\beta\) =  cos \(\gamma\) 
\(\rightarrow\)  \(\alpha\) = \(\beta\) =  \(\gamma\) 
Hence, the vector  \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \) is equally inclined to \(\vec{a} \vec{b} \text { and } \vec{c}\) 

We have : \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } = 0\) and  \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\) 
\(\therefore\) \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } \Rightarrow \overset { \rightarrow }{ a } .\overset { \rightarrow }{ a } -\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\) 
\(\Rightarrow \) \(\overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) =0\) 
\(\Rightarrow \) \(\overset { \rightarrow }{ a } =0\ or\ \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } =0\ or\ \overset { \rightarrow }{ a } \bot \left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) \)
\(\Rightarrow \) Either \(\overset { \rightarrow }{ a } =0\ or\ \overset { \rightarrow }{ a } =\overset { \rightarrow }{ b } \ or\ \overset { \rightarrow }{ a } \bot \left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) \)
hence we conclude that \(\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } \)

\(\text { Let } \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \)
\(\therefore|\vec{a}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \)
\(\text { Hence, the direction cosines of }\vec{a} \text { are }\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\) 

The unit vector \(\hat{a} \ {\text {in}} \) the direction of vector \( \vec{a}=\hat{i}+\hat{j}+2 \hat{k} \text { is given by } \hat{a}=\frac{\vec{a}}{|a|} \text { . }\)
\(|\vec{a}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6} \)
\(\therefore \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k} \)

(i) scalar    
(ii) Scalar
(iii) vector    
(iv) Vector
(v) Scalar

Let the 1^st paycheck be x (integer).
Mrs. Rodger got a weekly raise of 145.
So after completing the 1st week she will get (x+145).
Similarly after completing the 2^nd week she will get (x +145) + 145.
= (x + 145 + 145)
= (x + 290)
So in this way end of every week her salary will increase by 145.

Let \(\overset { \rightarrow }{ d } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \) ..(1)
Now  \(\overset { \rightarrow }{ d } \) is perp.to \(\overset { \rightarrow }{ a } \) 
\(\therefore\) \(\overset { \rightarrow }{ d } \).\(\overset { \rightarrow }{ a } \) = 0
\(\Rightarrow (x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } ).(\overset { \wedge }{ i } +4\overset { \wedge }{ j } +2\overset { \wedge }{ k } )=0\) ...(2)
\(\Rightarrow x+4y+2z=0\)
and \(\overset { \rightarrow }{ d } \) is prep.to \(\overset { \rightarrow }{ b } \)
\(\therefore\) \(\overset { \rightarrow }{ d } \).\(\overset { \rightarrow }{ b } \) = 0
\(\therefore(\hat{i} \hat{i}+\hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})=0\)
\(\Rightarrow\) 3x - 2y + 7z = 0..(3)
Also  \(\overset { \rightarrow }{ c } \).\(\overset { \rightarrow }{ d } \) = 15
\(\Rightarrow\) \((2\overset { \wedge }{ i } -\overset { \wedge }{ j } +4\overset { \wedge }{ k } ).(x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } )=15\) 
2x - y + 4z = 15 ......(4)
(3)-3(2) gives: -14y + z = 0        ..(5)
(4)-2(2) gives :  -9y = 15        ....(6)
from (6). y = -\(\frac { 5 }{ 3 } \) 
putting in (5), -14(\(\frac {- 5 }{ 3 } \)) + z = 0
\(\Rightarrow\) z = -\(\frac {- 70 }{ 3 } \)
Putting in (2), \(x -\frac {- 20 }{ 3 } -\frac {140}{3}=0 \rightarrow x =\frac {160}{3}\) 
putting in (1)
\(\vec{d} =\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{8} \hat{k} \)
\(=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \)

The vertices of △ABC are given as A(1, 2, 3), B(−1, 0, 0), and C(0, 1, 2) .
Also, it is aiven that \( \square \mathrm{ABC} \text { is the anale between the vectors } \overrightarrow{\mathrm{BA}} \text { and } \overrightarrow{\mathrm{BC}}\)
\(\overrightarrow{\mathrm{BA}}=\{1-(-1)\} \hat{i}+(2-0) \hat{j}+(3-0) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k} \)
\(\overrightarrow{\mathrm{BC}}=\{0-(-1)\} \hat{i}+(1-0) \hat{j}+(2-0) \hat{k}=\hat{i}+\hat{j}+2 \hat{k} \)
\(\therefore \overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+\hat{j}+2 \hat{k})=2 \times 1+2 \times 1+3 \times 2=2+2+6=10 \)
\(|\overrightarrow{\mathrm{BA}}|=\sqrt{2^{2}+2^{2}+3^{2}}=\sqrt{4+4+9}=\sqrt{17} \)
\(|\overrightarrow{\mathrm{BC}}|=\sqrt{1+1+2^{2}}=\sqrt{6} \)
Now, it is known that:
\(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=|\overrightarrow{\mathrm{BA}} \| \overrightarrow{\mathrm{BC}}| \cos (\angle \mathrm{ABC}) \)
\(\therefore 10=\sqrt{17} \times \sqrt{6} \cos (\angle \mathrm{ABC}) \)
\(\Rightarrow \cos (\angle \mathrm{ABC})=\frac{10}{\sqrt{17} \times \sqrt{6}} \)
\(\Rightarrow \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right) \)

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