The given vectors are: 
\(\vec{a} =\hat{i}+\hat{j}+\hat{k} ; \quad \vec{b}=2 \hat{i}-7 \hat{j}-3 \hat{k} ; \quad \vec{c}=\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k} \)
\(|\vec{a}| =\sqrt{(1)^{2}+(1)^{2}+(1)^{2}}=\sqrt{3} \)
\(|\vec{b}| =\sqrt{(2)^{2}+(-7)^{2}+(-3)^{2}} \)
\(=\sqrt{4+49+9}\)
\(=\sqrt{62} \)
\(|\vec{c}| =\sqrt{\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(\frac{1}{\sqrt{3}}\right)^{2}+\left(-\frac{1}{\sqrt{3}}\right)^{2}} \)
\(=\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=1 \)

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

The vector with the initial point P(2, 1) and terminal point Q(-5, 7) can be given by \(\)
\(\overrightarrow{\mathrm{PQ}}=(-5-2) \hat{i}+(7-1) \hat{j} \)
\(\Rightarrow \overrightarrow{\mathrm{PQ}}=-7 \hat{i}+6 \hat{j} \)
Hence, the required scalar components are -7 and 6 while the vector components are -7 \( \hat{i}+6 \hat{j} \)

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)
\(\vec{d} =\frac{160}{3} \hat{i}-\frac{5}{3} \hat{j}-\frac{70}{8} \hat{k} \)
\(=\frac{5}{3}(32 \hat{i}-\hat{j}-14 \hat{k}) \)

\(3 \vec{a}+5 \vec{b}\)

Let  \(\hat{a}=\hat{i}+\hat{j}+\hat{k} \)
\(\hat{b}=2 \hat{i}+4 \hat{j}-5 \hat{k} \) 
and \(\hat{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}\) 
now the unit vector along \(\overset { \wedge }{ b } +\overset { \wedge }{ c } \) 
\(=\frac { (\lambda +2)\overset { \wedge }{ i } +6\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { { (\lambda +2) }^{ 2 } } +36+4 } \) 
By the question
(\(\hat { i } +\hat { j } +\hat { k } \)).\(=\frac { (\lambda +2)\overset { \wedge }{ i } +6\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { { (\lambda +2) }^{ 2 } } +40 } \) =1
\(=\frac { 1 }{ \sqrt { { (\lambda +2) }^{ 2 } } +40 } (\lambda +2+6)= 6\)
\(\Rightarrow \lambda +6=\sqrt { { (\lambda +2) }^{ 2 }+40 } \)
Squaring, \({ \lambda }^{ 2 }+12\lambda +36\) 
\(={ \lambda }^{ 2 }+4\lambda +4+40\)
\(\Rightarrow 12\lambda +36=4\lambda +44\)
\(\Rightarrow 8\lambda =8\)
\(Hence\ \lambda =1\)

Given \(\overrightarrow{O A}=(\hat{i}+\hat{j}+\hat{k}), \overrightarrow{O B}=(2 \hat{i}+5 \hat{j})\)
\(\overrightarrow{O C}=(3 \hat{i}+2 \hat{j}-3 \hat{k}) \text { and } \overrightarrow{O D}=(\hat{i}-6 \hat{j}-\hat{k})\)
Angle between \(\overrightarrow{A B} \text { and } \overrightarrow{C D}\)  is given by
\(\cos \theta=\frac{\overrightarrow{A B} \cdot \overrightarrow{C D}}{|\overrightarrow{A B}| \cdot|\overrightarrow{C D}|}\)      ...(i)
Here, \(\begin{aligned} \overrightarrow{A B} & =(2-1) \hat{i}+(5-1) \hat{j}+(0-1) \hat{k} \end{aligned}\)
\(\begin{aligned} =\hat{i}+4 \hat{j}-\hat{k} \end{aligned}\),
\(\begin{aligned} \overrightarrow{C D} & =(1-3) \hat{i}+(-6-2) \hat{j}+\{-1-(-3)\} \hat{k} \end{aligned}\)
\(\begin{aligned} =-2 \hat{i}-8 \hat{j}+2 \hat{k} \end{aligned}\)
\(\begin{aligned} |\overrightarrow{A B}|=\sqrt{1^2+4^2+(-1)^2}=\sqrt{18}=\sqrt{9 \times 2}=3 \sqrt{2} \end{aligned}\)
and \(\begin{aligned} |\overrightarrow{C D}|=\sqrt{(-2)^2+(-8)^2+2^2} \end{aligned}\)
\(=\sqrt{72}=\sqrt{36 \times 2}=6 \sqrt{2}\)
Now, \(\cos \theta=\frac{(\hat{i}+4 \hat{j}-\hat{k}) \cdot(-2 \hat{i}-8 \hat{j}+2 \hat{k})}{3 \sqrt{2} \times 6 \sqrt{2}}\)
                                                [from Eq. (i)]
\(=\frac{1(-2)+4(-8)+(-1)(2)}{3 \times 6 \times 2}=-1\)
\(\cos \theta=-1 \Rightarrow \theta=180^{\circ}=\pi\)
So, angle between \(\overrightarrow{A B} \text { and } \overrightarrow{C D}\) is \(\pi\).
Also, since angle between \(\overrightarrow{A B} \text { and } \overrightarrow{C D}\) is 180°, they are in opposite directions.

Since, \(\overrightarrow{A B} \text { and } \overrightarrow{C D}\) are parallel to the same line m, they are collinear.

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:
\(\overrightarrow{\mathrm{BA}} \cdot \overrightarrow{\mathrm{BC}}=|\overrightarrow{\mathrm{BA}} \| \overrightarrow{\mathrm{BC}}| \cos (\angle \mathrm{ABC}) \)
\(\therefore 10=\sqrt{17} \times \sqrt{6} \cos (\angle \mathrm{ABC}) \)
\(\Rightarrow \cos (\angle \mathrm{ABC})=\frac{10}{\sqrt{17} \times \sqrt{6}} \)
\(\Rightarrow \angle \mathrm{ABC}=\cos ^{-1}\left(\frac{10}{\sqrt{102}}\right) \)

(i) (c): Position vector of \(\overrightarrow{A B} \) 
\(=(2-1) \hat{i} \dot{+}(1-1) \hat{j}+(3-1) \hat{k}=\hat{i}+2 \hat{k}\)
(ii) (b): Position vector of  \(\overrightarrow{A C} \)
\(=(3-1) \hat{i}+(2-1) \hat{j}+(2-1) \hat{k}=2 \hat{i}+\hat{j}+\hat{k}\)
(iii) (d): Position vector of \(\overrightarrow{AD} \)
\(=(3-1) \hat{i}+(3-1) \hat{j}+(4-1) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
(iv) (b): Area of \(\Delta A B C\) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{array}\right|=\hat{i}(0-2)-\hat{j}(1-4)+\hat{k}(1-0)\)
\(=-2 \hat{i}+3 \hat{j}+\hat{k}\)
\( \Rightarrow |\overrightarrow{A B} \times \overrightarrow{A C}| =\sqrt{(-2)^{2}+3^{2}+1^{2}} \)
\(=\sqrt{4+9+1}=\sqrt{14}\)
Area of \(\Delta A B C\) \(=\frac{1}{2} \sqrt{14} \text { sq. units }\)
(v) (a): Unit vector along \(\overrightarrow{A D}=\frac{\overrightarrow{A D}}{|\overrightarrow{A D}|}\)
\(=\frac{2 \hat{i}+2 \hat{j}+3 k}{\sqrt{2^{2}+2^{2}+3^{2}}}=\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{4+4+9}}=\frac{1}{\sqrt{17}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\)

(i) (b) : \(\overrightarrow{A B}=(-2 \hat{i}+4 \hat{j}+\hat{k})-(\hat{i}+\hat{j}+\hat{k})=-3 \hat{i}+3 \hat{j}\)
\(\therefore \overrightarrow{A B}=\sqrt{(-3)^{2}+3^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\)
Distance between House (A) and ATM (B) is 3\(\sqrt 2\) units.
(ii) (c): \(\overrightarrow{B C}=(-\hat{i}+5 \hat{j}+5 \hat{k})-(-2 \hat{i}+4 \hat{j}+\hat{k})=\hat{i}+\hat{j}+4 \hat{k}\)
\( \therefore |\overrightarrow{B C}| =\sqrt{1^{2}+1^{2}+4^{2}}=\sqrt{1+1+16} \)
\(=\sqrt{18}=3 \sqrt{2}\)
Distance between ATM (B) and School (C) is 3\(\sqrt 2\) units.
(iii) (a): \(\overrightarrow{C D}=(2 \hat{i}+2 \hat{j}+5 \hat{k})-(-\hat{i}+5 \hat{j}+5 \hat{k})=3 \hat{i}-3 \hat{j}\)
\(\therefore |\overrightarrow{C D}|=\sqrt{3^{2}+(-3)^{2}}=\sqrt{9+9}=3 \sqrt{2}\)
Distance between School (C) and Shopping mall (D) is 3\(\sqrt 2\) units.
(iv) (d): Total distance travelled by Ritika
\( =|\overrightarrow{A B}|+|\overrightarrow{B C}|+|\overrightarrow{C D}|=(3 \sqrt{2}+3 \sqrt{2}+3 \sqrt{2}) \text { units } \)
\(=9 \sqrt{2} \text { units }\)
(v) (c): Distance between house and shopping mall is \(|\overrightarrow{A D}|\)
Now, \(\overrightarrow{A D}=\hat{i}+\hat{j}+4 \hat{k}\)
\(\therefore|\overrightarrow{A D}|=\sqrt{1^{2}+1^{2}+4^{2}}=\sqrt{1+1+16}=\sqrt{18}=3 \sqrt{2}\)
Thus, extra distance travelled by Ritika in reaching shopping mall = \((9 \sqrt{2}-3 \sqrt{2})\) units = \(6 \sqrt{2} \) units.

To find projection of \((\vec{b}+\vec{c}) \text { on } \vec{a}\)
Given, \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Consider, \(\begin{aligned} (\vec{b}+\vec{c}) & =(\hat{i}+2 \hat{j}-2 \hat{k})+(2 \hat{i}-\hat{j}+4 \hat{k}) \end{aligned}\)
\(\begin{aligned} =3 \hat{i}+\hat{j}+2 \hat{k} \end{aligned}\)
Now, the projection of \(\vec{b}+\vec{c} \text { on } \vec{a}\) is given by
\(\begin{aligned} \frac{(\vec{b}+\vec{c}) \vec{a}}{|\vec{a}|}=\frac{(3 \hat{i}+\hat{j}+2 \hat{k})(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{2^2+(-2)^2+1^2}} \end{aligned}\)
\(\begin{aligned} \frac{6-2+2}{\sqrt{4+4+1}}=\frac{6}{\sqrt{9}}=\frac{6}{3}=2 \end{aligned}\)

(i) Collinear vectors : \(\vec{a}, \vec{c} \text { and } \vec{d}\)
(ii) Equal vectors : \(\vec{a} \text { and } \vec{c} \text { . }\)
(iii) Coinitial vectors : \(\vec{b}, \vec{c} \text { and } \vec{d}\)

(i) Time-scalar
(ii) Volume-scalar
(iii) Force-vector
(iv) Speed-scalar
(v) Density-scalar
(vi) Velocity-vector

The vector \(\overrightarrow{\mathrm{OP}}\) represents the required displacement