Given direction ratios are 1, 2, 3
Let x = 1, y = 2, z = 3
r = \(\sqrt{x^2+y^2+z^2}=\sqrt{1+4+9}=\sqrt{14}\)
The direction cosines are \({x\over r},{y\over r},{z\over r}\)
Thus, the direction cosines are \({1\over \sqrt{14}},{2\over \sqrt{14}},{3\over \sqrt{14}}\)

The direction cosines are \({2\over7},{3\over7},{6\over7}.\)
The unit vector is \({2\over7}\hat{i}+{3\over7}\hat{j}+{6\over7}\hat{k}.\)
The required vector is \({5\over7}(2\hat{i}+3\hat{j}+6\hat{k})\).

The condition is cos²\(\alpha\) + cos²\(\beta\) + cos²\(\gamma\) = 1
Here \(\alpha =30^o,\beta =45^o,\gamma=60^o\)
cos²\(\alpha\) + cos²\(\beta\) + cos²\(\gamma\) \(={3\over 4}+{1\over2}+{1\over4}\neq1.\)
There fore they are not direction angles of any vector.

The direction ratios of \(3\hat{i}+4\hat{j}-6\hat{k}\)  are 3, 4, -6.
The direction cosines are \({x\over r},{y\over r},{z\over r},\) where r = \(\sqrt{x^2+y^2+z^2}\) .
Therefore, the direction cosines are \({3\over \sqrt{61}},{4\over \sqrt{61}},{-6\over \sqrt{61}}\)

80km, 60° south of west

The vector \(\overrightarrow{OQ}\) represents a displacement of 80 km, 60° south of west.

Given \(|\overrightarrow{a}|=2,|\overrightarrow{b}|=3,|\overrightarrow{c}|=4,\)and \(\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{0}\Rightarrow \overrightarrow{a}+\overrightarrow{b}=-\overrightarrow{c}\)
\(\therefore |\overrightarrow{a}+\overrightarrow{b}|^2=|\overrightarrow{c}|^2\)
\(\Rightarrow |\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2(\overrightarrow{a}.\overrightarrow{b})=|\overrightarrow{c}|^2\)
\(\Rightarrow 4+9+2(\overrightarrow{a}.\overrightarrow{b})=16\)
\(\Rightarrow 13+2(\overrightarrow{a}.\overrightarrow{b})=16\)
\(\Rightarrow 2(\overrightarrow{a}.\overrightarrow{b})=16-13=3\)
\(\Rightarrow \overrightarrow{a}.\overrightarrow{b}={3\over2}\)
\(\Rightarrow4( \overrightarrow{a}.\overrightarrow{b})=4\times {3\over2}=6\).....(1)
Also \(\overrightarrow{b}+\overrightarrow{c}=-\overrightarrow{a}\)
\(|\overrightarrow{b}+\overrightarrow{c}|^2=|-\overrightarrow{a}|^2\)
\(|\overrightarrow{b}|^2+|\overrightarrow{c}|^2+2(\overrightarrow{b}.\overrightarrow{c})=|-\overrightarrow{a}|^2\)
\(9+16+2(\overrightarrow{b}.\overrightarrow{c})=4\)
\(25+2(\overrightarrow{b}.\overrightarrow{c})=4\)
\(2(\overrightarrow{b}.\overrightarrow{c})=4-25=-21\)
\((\overrightarrow{b}.\overrightarrow{c})={-21\over 2}\)
\(3(\overrightarrow{b}.\overrightarrow{c})=3({-21\over 2})={-63\over2}\)..(2)
Also, \(\overrightarrow{c}+\overrightarrow{a}=-\overrightarrow{b}\)
\(|\overrightarrow{c}+\overrightarrow{a}|=|-\overrightarrow{b}|\)
\(|\overrightarrow{c}+\overrightarrow{a}|^2=|-\overrightarrow{b}|^2\)
\(|\overrightarrow{c}|^2+|\overrightarrow{a}|^2+2(\overrightarrow{c}.\overrightarrow{a})=|\overrightarrow{b}|^2\)
\(16+4+2(\overrightarrow{c}.\overrightarrow{a})=9\)
\(20+2(\overrightarrow{c}.\overrightarrow{a})=9\)
\(\Rightarrow 2(\overrightarrow{c}.\overrightarrow{a})=9-20=-11\)
\((\overrightarrow{c}.\overrightarrow{a})={-11\over2}\)
\(\therefore 3(\overrightarrow{c}.\overrightarrow{a})=3({-11\over2})={-33\over2}\)....(3)
Adding (1), (2) and (3) we get,
\(4\overrightarrow{a}.\overrightarrow{b}+3\overrightarrow{b}.\overrightarrow{c}+3\overrightarrow{c}.\overrightarrow{a}=\)\(6-{63\over2}-{33\over2}={12-63-33\over 2}={12-96\over2}={-84\over2}=-42\)
\(\therefore 4\overrightarrow{a}.\overrightarrow{b}+3\overrightarrow{b}.\overrightarrow{c}+3\overrightarrow{c}.\overrightarrow{a}=-42\)

Let \(\overrightarrow{a}=5\hat{i}+6\hat{j}+7\hat{k}\)
\(\overrightarrow{b}=\)7 \(\hat{i}\) -8\(\hat{j}\) +9 \(\hat{k}\)
\(\overrightarrow{c}=\)3\(\hat{i}\)+20\(\hat{j}\) +5\(\hat{k}\)
Let \(\overrightarrow{a}=s\overrightarrow{b}+t \overrightarrow{c}\)
\(\Rightarrow 5\hat { i } +6\hat { j } +7\hat { k } =s(7\hat { i } -8\hat { j } +9\hat { k } )+t(3\hat { i } +20\hat { j } +5\hat { k } )\)
\(\Rightarrow 5\hat { i } +6\hat { j } +7\hat { k } =(7s+3t)\hat { i } +(-8s+20t)\hat { j } +(9s+5t)\hat { k } \)
Equating the like components, both sides we get.
                   5 = 7s + 3t  .....(1)
                  -8s + 20t = 6 ....(2)
                   9s + 5t = 7 ......(3)

                  164s  = 82    \(\Rightarrow \quad s=\frac { 82 }{ 164 } =\frac { 1 }{ 2 } \)
Substituting  \(\\ s=\frac { 1 }{ 2 } \) in (1) we get,
\(7\left( \frac { 1 }{ 2 } \right) +3t=5\quad \Rightarrow 3t=5-\frac { 7 }{ 2 } =\frac { 10-7 }{ 2 } =\frac { 3 }{ 2 } \)
\(\Rightarrow t=\frac { 3 }{ 2\times 3 } =\frac { 1 }{ 2 } \)
Substituting \(s=\frac { 1 }{ 2 } ,t=\frac { 1 }{ 2 } \) in (3) we get,
\(9\left( \frac { 1 }{ 2 } \right) +5\left( \frac { 1 }{ 2 } \right) =7\)
\(\Rightarrow \frac { 9 }{ 2 } +\frac { 5 }{ 2 } =7\)
\(\Rightarrow \frac { 14 }{ 2 } =7\)
\(\Rightarrow\) 7 = 7 which satisfies the (3) equation.
Thus, one vector is a linear combination of other two vectors.
Hence, the given vectors are co-planar.                                 

Given points are A(1, 0, 0) and B(0, 1, 0) and one set of direction ratios are a, a+b, a+b+c.
Case (i): \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(0\hat{i}+\hat{j}+0\hat{k})-(\hat{i}+0\hat{j}+0\hat{k})=-\hat{i}+\hat{j}\)
\(\therefore \) Direction ratios of the line \(\overrightarrow{AB}\) are (-1, 1, 0)
Given (-1, 1, 0) = (a, a + b, a + b + c)
Equating the like components both sides, we get
a = -1, a + b = 1, a + b + c = 0
a = -1, -1 + b = 1 \(\Rightarrow\) b = 2
-1 + 2 + c = 0 \(\Rightarrow\) c = -1
\(\therefore \) a = -1, b = 2, c = -1
Case (ii): \(\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=(\hat{i}+0\hat{j}+0\hat{k})-(0\hat{i}+\hat{j}+0\hat{k})=\hat{i}-\hat{j}\)
\(\therefore \) Direction ratios of the line\(\overrightarrow{BA}\) are (1, -1, 0)
Given (1, -1, 0) = (a, a + b, a + b + c)
Equation the like components both sides, we get
a = 1, a + b = -1, a + b + c = 0
a = 1,1 + b = -1 \(\Rightarrow\) b = -2
1 - 2 + c = 0 \(\Rightarrow\) c = 1
\(\therefore \) a = 1, b = -2, c = 1

Let O be the origin and let \(\overrightarrow{OA}\), \(​​\overrightarrow{OB}\), and\(​​\overrightarrow{OC}\) be the vectors 2\(\hat{i}\) + 3\(\hat{j}\) − 5\(\hat{k}\), 3\(\hat{i}\) + \(\hat{j}\) − 2\(\hat{k}\) and, 6\(\hat{i}\) − 5\(\hat{j}\) + 7\(\hat{k}\) respectively. Then
\(\overrightarrow{AB}=\hat{i}-2\hat{j}+3\hat{k} \ and \ \overrightarrow{AC}=4\hat{i}-8\hat{j}+12\hat{k}\).
Thus \(\overrightarrow{AC}=4\overrightarrow{AB}\) and hence \(​​\overrightarrow{AB}\) and\(​​\overrightarrow{AC}\) are parallel. They have a common point namely A. Thus, the three points are collinear.
Alternative method
Let O be the point of reference.
Let \(\overrightarrow {OA} = 2\hat i+3\hat j-5\hat k, \) \(\overrightarrow {OB} = 3 \hat j+\hat j-2\hat k\ and\ \overrightarrow {OC} = 6\hat i-5\hat j+7\hat k \)
\(\overrightarrow {AB} = \hat i- 2\hat j+3\hat k; \overrightarrow {BC} = 3\hat i-6\hat j+9\hat k; \overrightarrow {CA} = -4\hat i+8\hat j-12 \hat k\\ |\overrightarrow {AB}| = \sqrt 14; |\overrightarrow {BC}|= \sqrt 126 = 3 \sqrt 14; |\overrightarrow {CA}|= \sqrt 224 = 4 \sqrt 4\)
Thus, AC = AB + BC.
Hence A, B, C are lying on the same line. That is, they are collinear.

Given  \(\overrightarrow{PO}\) +\(\overrightarrow{OQ}\) = \(\overrightarrow{QO}\) +\(\overrightarrow{OR}\)
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) [By triangle law of addition]
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) and Q is a common point.
Hence, the points P, Q, R are collinear.

Let \(\overrightarrow{a}\) and \(\overrightarrow{b}\) be the position vectors of the points A and B.
\(\Rightarrow \overrightarrow{OA}=\overrightarrow{a}\) and \( \overrightarrow{OB}=\overrightarrow{b}\).
Let P divides the line segment AB in the ratio 1:2 and Q divides the line segment AB in the ratio 2 : 1
\(\therefore \overrightarrow{OP}={1.(\overrightarrow{OB})+2(\overrightarrow{OA})\over 1+2}={1(\overrightarrow{b})+2(\overrightarrow{a})\over 3}={\overrightarrow{b}+2\overrightarrow{a}\over 3}\)
and \( \overrightarrow{OQ}={2(\overrightarrow{OB})+1(\overrightarrow{OA})\over 2+1}={2\overrightarrow{b}+\overrightarrow{a}\over 3}={\overrightarrow{a}+2\overrightarrow{b}\over 3}\)
Hence, the required position vectors are \({\overrightarrow{b}+2\overrightarrow{a}\over 3}\)and \({\overrightarrow{a}+2\overrightarrow{b}\over 3}\).

Given \(\vec{a}\)= \(\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2\hat{j}+3\hat{k}\)
\(\vec { a } +\vec { b } =2\hat { i } +3\hat { j } +4\hat { k } \)
\(\vec { a } -\vec { b } =-\hat{j}-2\hat{k}\)
A unit vector which is perpendicular to \((\vec { a } +\vec { b } )\) and \((\vec { a } +\vec { b } )\) is
\(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 4 \\ 0 & -1 & -1 \end{matrix} \right| =\hat { i } (-6+4)-\hat { j } (-4+0)+\hat { k } (-2+0)\)
= -2 \(\hat{i}\)+4\(\hat{j}\)-2\(\hat{k}\)
Its magnitude is \(\sqrt { { (-2) }^{ 2 }+{ 4 }^{ 2 }+{ (-2) }^{ 2 } } =\sqrt { 4+16+4 } =\sqrt { 24 } =\sqrt { 4\times 6 } =2\sqrt { 6 } \)
\(\therefore\) The unit vector which is perpendicular to\((\vec { a } +\vec { b } )\) and \((\vec { a } +\vec { b } )\) is
\(\pm \frac { (-2\hat { i } +4\hat { j } -2\hat { k } ) }{ 2\sqrt { 6 } } =\pm \frac { (-\hat { i } +2\hat { j } -\hat { k } ) }{ \sqrt { 6 } } \)

Let  \(\vec{a}\)= \(\hat{i}+2\hat{j}+\hat{k}\)
 \(\vec{b}\)= \(\hat{i}+3\hat{j}+4\hat{k}\)
A unit vector which is perpendicular to the vector \(\vec{a}\) and \(\vec{b}\) is \(\frac { \vec { a } \times \vec { b } }{ \left| \vec { a } \times \vec { b } \right| } \)
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & 1 \\ 1 & 3 & 4 \end{matrix} \right| \)= \(\hat{i}\)(8-3)-\(\hat{j}\)(4-1)+ \(\hat{k}\)(3-2) = 5\(\hat{i}\)-3\(\hat{j}\)+\(\hat{k}\)
\(\left| \vec { a } \times \vec { b } \right| =\sqrt { { 5 }^{ 2 }+{ (-3) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 25+9+1 } =\sqrt { 35 } \)
A unit vector which is perpendicular to the vector \(\vec{a}\) and \(\vec{b}\) is \(\frac { 5\hat { i } -3\hat { j } +\hat { k } }{ \sqrt { 35 } } \)
Hence, a vector of magnitude 10\(\sqrt{3}\) , which is perpendicular to the vectors \(\vec{a}\) and \(\vec{b}\) is \(\pm \frac { 10\sqrt { 3 } }{ \sqrt { 35 } } \left( 5\hat { i } -3\hat { j } +\hat { k } \right) \)

\(tan {\theta \over 2}={sin{\theta \over2}\over cos {\theta \over 2}}={{1\over 2}|\overrightarrow{a}-\overrightarrow{b}|\over {1\over2}|\overrightarrow{a}+\overrightarrow{b}| }\)       
\(tan {\theta \over 2}={|\overrightarrow{a}-\overrightarrow{b}|\over|\overrightarrow{a}+\overrightarrow{b}|}\)         
Hence, proved.

Let  \(\overrightarrow{a}\)and \(\overrightarrow{b}\) be the unit vectors and \(\theta\) is the angle between \(\overrightarrow{a}\) and \(\overrightarrow{b}\).
Consider \(|\overrightarrow{a}-\overrightarrow{b}|^2=|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2(\overrightarrow{a}.\overrightarrow{b})\)      \([\because |\overrightarrow{a}|=1;|\overrightarrow{b}|=1]\)
\(=1+1-2|\overrightarrow{a}||\overrightarrow{b}|cos \theta =2+2cos \theta\)
\(=2(1+cos \theta)=2.2cos^2{\theta \over2}=4cos^2{\theta \over2}\)
\(\therefore |\overrightarrow{a}+\overrightarrow{b}|=2cos{\theta \over2}\)
\(\Rightarrow cos{\theta \over2}={1\over 2}|\overrightarrow{a}+\overrightarrow{b}|\)

Let \(\overrightarrow{a}=\hat{i}-\hat{j}\) and \(\overrightarrow{b}=\hat{j}-\hat{k}\)
\(\overrightarrow{a}.\overrightarrow{b}=(\hat{i}-\hat{j}).(\hat{j}-\hat{k})=\)1(0)-1(1)+0(-1)=-1
\(|\overrightarrow{a}|=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\)
\(|\overrightarrow{b}|=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\)
Let \(\theta\) be the angle between the vectors \(\overrightarrow{a}\) and \(\overrightarrow{b}\)
\(\therefore cos \theta ={\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow{a}|.|\overrightarrow{b}|}\)
\(\Rightarrow cos \theta ={-1\over \sqrt{2}\sqrt{2}}=-{1\over 2}\)
\(\Rightarrow cos \theta =-cos ({\pi\over 3})\)
\(\Rightarrow cos \theta =-cos (\pi-{\pi\over 3})=cos({2\pi\over 3})\)
\(\Rightarrow \theta ={2\pi\over 3}\)

\((\overrightarrow{a}+\lambda \overrightarrow{b}).\overrightarrow{c}=0\Rightarrow \overrightarrow{a}.\overrightarrow{c}+\lambda \overrightarrow{b}.\overrightarrow{c}=0\)
\(\Rightarrow (6+2)+\lambda (-3+2)=0\)
\(\Rightarrow \lambda=8\).

\(\vec{b} \times \vec{c}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & x & 1 \\ 1 & -1 & 4 \end{array}\right|\)
\(=\hat{i}(4 x+1)-\hat{j}(8-1)+\hat{k}(-2-x) \)
\(=\hat{i}(4 x+1)-7 \hat{j}+\hat{k}(-2-x) \)
\(\vec{a} \cdot(\vec{b} \times \vec{c}) =(4 x+1)-7-2-x=70 \)
\(=(\hat{i}+\hat{j}+\hat{k}) \propto(4 x+1) \hat{i}-7 \hat{j}+\hat{k}(-2-x) \)
\(4 x+1-7-2-x =70 \)
\(3 x-8 =70 \)
\(3 x =78 \)
\(x =\frac{78}{3}=26 \)

\(\text { Projection }=\frac{5-3-3 \lambda}{\sqrt{1+9+\lambda^{2}}}=\frac{5-3-3 \lambda}{\sqrt{25+1+9}}\)
\(\sqrt{10+\lambda^{2}} =\sqrt{35} \)
\(\lambda^{2}+10 =35 \)
\(\lambda^{2} =25 \)
\(\lambda =\pm 5 \)

\(A B=B C=2 \text { (ie) } \vec{a}=\vec{b}=2\)
\(\therefore A B C \text { is isosceles triangle }\)
\(\therefore \angle A+\angle B =180^{\circ}-120^{\circ} \)
\(2 \angle A =60^{\circ} \)
\(\angle A =30^{\circ} \)

\(\cos 60^{\circ} =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \)
\(\frac{1}{2} =\frac{\frac{1}{2}}{|\vec{a} \| \vec{a}|} \)
\(|\vec{a}|^{2} =1 \Rightarrow|a|=1 \)

\(\left(\frac{3+2+x_{1}}{3},\right. \left.\frac{4+3+x_{2}}{3}, \frac{-4+4+x_{3}}{3}\right)=(1,2,3) \)
\(3+2+x_{1} =3 ; 4+3+x_{2}=6 ;-4+4+x_{3}=9 \)
\(5+x_{1}=3 7+x_{2}=6 \)
\(x_{1}=-2 \quad x_{2}=-1 \)
\(\therefore \text { Third vertex is }(-2,-1,9)\)

\(\vec{m}=\frac{|\vec{b}+2 \vec{a}|}{1+2}=\frac{2 \vec{a}+\vec{b}}{3}\)

\(\text { In } \Delta \mathrm{BCD}, \overrightarrow{\mathrm{BD}}=\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CD}}\)
\(=\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}\)

\(\vec{a}+\vec{b} =2 \hat{i}-\hat{j} \)
\(|\vec{a}+\vec{b}| =\sqrt{4+1}=\sqrt{5} \)
\(\text { Unit vector }=\frac{2 \hat{i}-\hat{j}}{\sqrt{5}}\)

\(\underbrace{\overrightarrow{A B}}+ \overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A C}+\overrightarrow{C D}}+\overrightarrow{D A} \)
\(=\underbrace{\overrightarrow{A D}+\overrightarrow{D A}} \)
\(=\overrightarrow{A A}=\overrightarrow{0} . \)