Let x = 3, y = -1, z = 3
\(\therefore r= \sqrt{x^2+y^2+z^2}=\sqrt{9+1+9}=\sqrt{19}\)
Hence, the direction consines are \({3\over \sqrt{19}},{-1\over \sqrt{19}},{3\over \sqrt{19}}\)

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 \({3\over \sqrt{61}},{4\over \sqrt{61}},{-6\over \sqrt{61}}\)

(i)

(ii) 

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

The given vector is 5\(\hat{i}\) - 3\(\hat{j}\) - 48\(\hat{k}\)
The direction ratios are 5, -3, -48.
r = \(\sqrt{x^2+y^2+z^2}=\sqrt{5^2+(-3)^2+(-48)^2}\)
\(=\sqrt{25+9+2304}=\sqrt{2338}\)
Hence, the direction cosines are \({5\over \sqrt{2338}},{-3\over \sqrt{2338}},{-48\over \sqrt{2338}}\)

Let \(\overrightarrow{a},\overrightarrow{b} , \overrightarrow{c}\in V\), where V is the set of all vectors.
Let R be the relation defined by\(\overrightarrow{a}=\overrightarrow{b} \)
(i) Reflexive : \(\overrightarrow{a}=\overrightarrow{a} \Rightarrow aRa \Rightarrow R\) is Reflexive.
(ii) Symmetric: \(\overrightarrow{a}=\overrightarrow{b} \Rightarrow \overrightarrow{b} =\overrightarrow{a}\)
\(\therefore aRb \overrightarrow{a} \Rightarrow bRa \Rightarrow R\) is Symmetric.
(iii) Transitive : \(\overrightarrow{a}=\overrightarrow{b},\overrightarrow{b}=\overrightarrow{c} \Rightarrow\overrightarrow{a}=\overrightarrow{c}\)
\(\therefore aRb, bRc\Rightarrow aRc \)
\(\therefore \) R is transitive.
\(\therefore \) The relation R is an equivalence relation on V.
Hence Proved.

Let O be the origin. It is given that
\(\overrightarrow{OA}=2\overrightarrow{a}+4\overrightarrow{b}\ and\ \overrightarrow{OB}=2\overrightarrow{a}- 8\overrightarrow{b}\)
Let C and D be the points which divide the segment AB in the ratio 1 : 3 internally and externally respectively. Then
\(\overrightarrow{OC}={3\overrightarrow{OA}+\overrightarrow{OB}\over3+1}={3(2\overrightarrow{a}+4\overrightarrow{b})+(2\overrightarrow{a}+8\overrightarrow{b})\over 4}=2\overrightarrow{a}+\overrightarrow{b}.\)
\(\overrightarrow{OC}={3\overrightarrow{OA}-\overrightarrow{OB}\over3-1}={3(2\overrightarrow{a}+4\overrightarrow{b})-(2\overrightarrow{a}+8\overrightarrow{b})\over 2}=2\overrightarrow{a}+10\overrightarrow{b}.\)

Given that the position vector of the \(\triangle\)ABC is \(\vec{a}\), \(\vec{b}\)and \(\vec{c}\)
\(\therefore \vec { OA } =\vec { a } ,\vec { OB } =\vec { b } \ and \ \vec { OC } =\vec { c } \)
\(\vec { AB } =\vec { OB } -\vec { OA } =\vec { b } -\vec { a } \)
\(\vec { AC } =\vec { OC } -\vec { OA } =\vec { c } -\vec { a } \)
\(\therefore \vec { AB } \times \vec { AC } =(\vec { b } -\vec { a } )\times (\vec { c } -\vec { a } )=\vec { b } \times \vec { c } -\vec { b } \times \vec { a } -\vec { a } \times \vec { c } +\vec { a } \times \vec { a } \)
= \(\vec { b } \times \vec { c } +\vec { a } \times \vec { b } +\vec { c } \times \vec { a } +\vec { 0 } \)
\(\left| \vec { AB } \times \vec { AC } \right| =\left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| \)
\(\therefore\) Area of \(\triangle\)ABC = \(\frac { 1 }{ 2 } \left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| \)
Condition for the points A, B, C to be collinear is area of \(\triangle\)ABC = 0
\(\Rightarrow \frac { 1 }{ 2 } \left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| =0\)
\(\Rightarrow \left| \vec { a } \times \vec { b } +\vec { b } \times \vec { c } +\vec { c } \times \vec { a } \right| =0\) which is the required condition.

Let \(\overrightarrow{a}\) = \(\hat{i}+3\hat{j}+7\hat{k}\) and \(\overrightarrow{b}=\)\(2\hat{i}+6\hat{j}+3\hat{k}\)
\(\overrightarrow{a}.\)\(\overrightarrow{b}=\)\((\hat{i}+3\hat{j}+7\hat{k})\).\((2\hat{i}+6\hat{j}+3\hat{k})\) = 1(2) + 3(6) + 7(3) = 2 + 18 + 21 = 41
\(|\overrightarrow{b}|=\)\(\sqrt{2^2+6^2+3^2}=\sqrt{4+36+9}=\sqrt{49}=7\)
Now, projection of \(\overrightarrow{a}\) on \(\overrightarrow{b}\) = \({\overrightarrow{a}.\overrightarrow{b}\over |\overrightarrow{b}| }={41\over7}\)

Trivially they form a triangle. It is enough to prove one angle is \({\pi\over2}\). So find the sides of the triangle.
Let O be the point of reference and A, B, C be (4, - 3, 1), (2, - 4, 5) and (1, - 1, 0) respectively.
\(\overrightarrow{OA}=4\hat{i}-3\hat{j}+\hat{k}\),\(\overrightarrow{OB}=2\hat{i}-4\hat{j}+5\hat{k}\), \(\overrightarrow{OC}=\hat{i}-\hat{j}\)
Now, \(\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=-2\hat{i}-\hat{j}+4\hat{k}\)
Similarly, \(\overrightarrow{BC}=-\hat{i}+3\hat{j}-5\hat{k};\overrightarrow{CA}=3\hat{i}-2\hat{j}+\hat{k}\)
Clearly, \(\overrightarrow{AB}.\overrightarrow{CA}=0\)
Thus one angle is \({\pi\over2}\). Hence they form a right angled triangle.

Given \(\overrightarrow { OP } =\hat { i } +\hat { j } +\hat { k } \)
\(\overrightarrow { OQ } =2\hat { i } +5\hat { j } \)
\(\overrightarrow { OR } =3\hat { i } +2\hat { j } -3\hat { k }\)
and \(\overrightarrow { OS } =\hat { i } -6\hat { j } -\hat { k } \)
\(\overrightarrow { PQ } =\overrightarrow { OQ } -\overrightarrow { OP } =(2\hat { i } +5\hat { j } )-(\hat { i } +\hat { j } +\hat { k } )=\hat { i } +4\hat { j } -\hat { k } \)
\(\overrightarrow { RS } =\overrightarrow { OS } -\overrightarrow { OR } =(\hat { i } -6\hat { j } -\hat { k } )-(3\hat { i } +2\hat { j } -3\hat { k } )\)
\(=-2\hat { i } -8\hat { j } +2\hat { k } =-2(\hat { i } +4-\hat { k } )=-2\overrightarrow { PQ } \)
\(\therefore \overrightarrow { RQ } =\lambda \overrightarrow { PQ } \) where \(\lambda =-2\)
\(\therefore \overrightarrow { RQ } ||\overrightarrow { PQ } \)

Let  \(\overrightarrow {a}=\hat{i}\) − 2\(\hat{j}\) + 3\(\hat{k}\), \(\overrightarrow{b}=\) -2\(\hat{i}\) + 3\(\hat{j}\) - 4\(\hat{k}\), \(\overrightarrow{c}=\) -\(\hat{j}\) + 2\(\hat{k}\) .
Let \(\overrightarrow {a}=s\overrightarrow{b}+t\overrightarrow{c}\)
\(\Rightarrow \hat{i}-2\hat{j}+3\hat{k}=s(-2\hat{i}+3\hat{j}-4\hat{k})+t(-\hat{j}+2\hat{k})\)
\(\Rightarrow \hat{i}-2\hat{j}+3\hat{k}=(-2s)\hat{i}+(3s-t)\hat{j}+(-4s +2t)\hat{k}\)
Equating the like components both sides, we get
-2s = 1 ....(1)
3s - t = -2 .....(2)
-4s + 2t = 3 ......(3)
From(1), s = \(-{1\over2}\)
Substituting s = \(-{1\over2}\) in (2) we get,
3\(({-1\over2})-t=-2 \Rightarrow -{3\over2}-t=-2\)
\(-t=-2+{3\over2}\)
\(-t={-4+3\over2}={-1\over2}\)
\(t={1\over2}\)
Substituting s = \(-{1\over2}\),\(t={1\over2}\) in (3) we get,
\(-4({-1\over2})+2({1\over2})=+3\)
\(\Rightarrow 2+1=3\)
\(\Rightarrow{3=3}\)
which satisfies the (3) equation.
Thus, one vector is a linear combination of other two vectors.
Hence, the given vectors are co-planar.

Let \(5\hat{i}+6\hat{j}+7\hat{k}=s(7\hat{i}-8\hat{j}+9\hat{k})+t(3\hat{i}+20\hat{j}+5\hat{k})\)
Equating the components, we have
7s + 3t = 5
-8s + 20t = 6
9s + 5t = 7
Solving first two equations, we get, s = t = \({1\over2},\) which satisfies the third equation.
Thus one vector is a linear combination of other two vectors.
Hence the given vectors are coplanar.

Let the position vector of the vertices of the quadrilateral ABCD be \(\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\)and\(\overrightarrow{d}\) respectively.
\(\therefore \overrightarrow{OA}=\overrightarrow{a}, \overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}\) and \(\overrightarrow{OD}=\overrightarrow{d}.\)
Since E and F are the mid-points of AC and BD respectively, we have
\(\overrightarrow{OE}={\overrightarrow{a}+\overrightarrow{c}\over 2}\) and \(\overrightarrow{OF}={\overrightarrow{b}+\overrightarrow{d}\over 2}\)
To prove that \(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}=4\overrightarrow{EF}\)
\(LHS=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{CB}+\overrightarrow{CD}\)
\(=\overrightarrow{OB}-\overrightarrow{OA}+\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OB}-\overrightarrow{OC}+\overrightarrow{OD}-\overrightarrow{OC}\)
\(=\overrightarrow{b}-\overrightarrow{a}+\overrightarrow{d}-\overrightarrow{a}+\overrightarrow{b}-\overrightarrow{c}+\overrightarrow{d}-\overrightarrow{c}\)

\(=-2\overrightarrow{a}+2\overrightarrow{b}-2\overrightarrow{c}+2\overrightarrow{d}\)
\(=2[(\overrightarrow{b}+\overrightarrow{d})-(\overrightarrow{a}+\overrightarrow{c})]\)
\(=2[2\overrightarrow{OF}+2\overrightarrow{OE})\) [From (1)]
\(=4[\overrightarrow{OF}-\overrightarrow{OE}]=4.\overrightarrow{EF}=RHS\)
Hence proved.

\(\text { W.K.T }|\vec{a} \cdot \vec{b}|^{2}+|\vec{a} \times \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2}\)
\(|\vec{a} \times \vec{b}|^{2} =13^{2} \cdot 5^{2}-|\vec{a} \cdot \vec{b}|^{2} \)
\(=(169)(25)-3600=4225-3600 =625 \)
\(|\vec{a} \times \vec{b}| =25 \)

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

\(\vec{r}=\frac{n \vec{a}+m \vec{b}}{m+n} \text { internally }\)

\(\text { Given } \vec{r}=\frac{9 \vec{a}+7 \vec{b}}{16}\)
Comparing (1) & (2),
we get n = 9; m = 7
7 : 9 internally

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