Let A and B be the given points whose position vectors are \(3 \vec{a}-2 \vec{b} \text { and } 2 \vec{a}-5 \vec{b}\) respectively, with respect to the origin \(\text { i.e. } \overrightarrow{O A}=3 \vec{a}-2 \vec{b} \text { and } \overrightarrow{O B}=2 \vec{a}-5 \vec{b} \text { . }\) 
Let P and Q be the points, which divides the line joining
A and B internally and externally respectively, in the ratio 3: 2.
 
Then, by using section formula of internal division, we get
\(\overrightarrow{O P}=\frac{3 \overrightarrow{O B}+2 \overrightarrow{O A}}{3+2}=\frac{3(2 \vec{a}-5 \vec{b})+2(3 \vec{a}-2 \vec{b})}{5}\) 
\(=\frac{6 \vec{a}-15 \vec{b}+6 \vec{a}-4 \vec{b}}{5}=\frac{12 \vec{a}-19 \vec{b}}{5}\) 
\(=\frac{12}{5} \vec{a}-\frac{19}{5} \vec{b}\) 
Now,by using section formula of external division, we get
 
\(\overrightarrow{O Q}=\frac{3 \overrightarrow{O B}-2 \overrightarrow{O A}}{3-2}=\frac{3(2 \vec{a}-5 \vec{b})-2(3 \vec{a}-2 \vec{b})}{1}\) 
\(=\frac{6 \vec{a}-15 \vec{b}-6 \vec{a}+4 \vec{b}}{1}=-11 \vec{b}\)

Let given vectors are \(\vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k} \text { and } \vec{b}=a \hat{i}+6 \hat{j}-8 \hat{k}\) .
We know that vectors \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\) and \(\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) are cellinear, if
\(\frac{b_{1}}{a_{1}}=\frac{b_{2}}{a_{2}}=\frac{b_{3}}{a_{3}}\) 
\(\therefore\frac{a}{2}=\frac{6}{-3}=\frac{-8}{4}\) 
\(\Rightarrow \frac{a}{2}=-2 \Rightarrow a=-4\)

(i) If vectors \(\vec{a} \text { and } \vec{b}\) are perpendicular, then \(\vec{a} \cdot \vec{b}=0\) 
\(\Rightarrow(3 \hat{i}+2 \hat{j}+9 \hat{k}) \cdot(\hat{i}+p \hat{j}+3 \hat{k})=0\) 
\(\Rightarrow 3+2 p+27=0\) 
\(\Rightarrow p=-15\) 
(ii) We know that, the vectors \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}\) and
\(\vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\) are parallel iff
\(\Rightarrow \frac{a_{1}}{b_{1}}=\frac{a_{2}}{b_{2}}=\frac{a_{3}}{b_{3}}=l\) 
So, the given vectors \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) 
and \(\vec{b}=\hat{i}+p \hat{j}+3 \hat{k}\) will be parallel iff
\(\frac{3}{1}=\frac{2}{p}=\frac{9}{3}\) 
\(\Rightarrow 3=\frac{2}{p} \Rightarrow p=\frac{2}{3}\)

Let \(\vec{a}=3 \hat{i}-5 \hat{k}, \vec{b}=2 \hat{i}+7 \hat{j} \text { and } \vec{c}=\hat{i}+\hat{j}+\hat{k}\) be three given vectors.
Let \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) be a vector such that its dot products with \(\vec{a}, \vec{b} \text { and } \vec{c}\) are -1, 6 and 5 respectively, we have
\(\vec{r} \cdot \vec{a}=(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-5 \hat{k}) \Rightarrow-1=3 x-5 z\)  ..(i)
\(\vec{r} \cdot \vec{b}=(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+7 \hat{j}) \Rightarrow 6=2 x+7 y\)  ...(ii)
and \(\vec{r} \cdot \vec{c}=(x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \Rightarrow 5=x+y+z\)  ...(iii)
On solving Equation (i), (ii) and (iii), we get
x = 3, y = 0 and z = 2
Hence, the required vector is \(\vec{r}=3 \hat{i}+2 \hat{k}\) 

Given, \(\overrightarrow{O A}=2 \vec{a}+\vec{b} \text { and } \overrightarrow{O B}=\vec{a}-3 \vec{b}\) 
Also, it is given that C is the point which divides the line joining A and B externally in the ratio 1: 2.
Then by using section formula of external division, we get

\(\overrightarrow{O C}=\frac{2 \overrightarrow{O A}-\overrightarrow{O B}}{2-1}\) 
\(\Rightarrow \overrightarrow{O C}=\frac{2(\overrightarrow{2 a}+\vec{b})-1(\vec{a}-3 \vec{b})}{1} \quad[\text { from Eq }\) 
\(=4 \vec{a}+2 \vec{b}-\vec{a}+3 \vec{b}=3 \vec{a}+5 \vec{b}\) 
i.e. to show \(\overrightarrow{O A}=\frac{\overrightarrow{O B}+\overrightarrow{O C}}{2}\) 
Consider, \(\frac{\overrightarrow{O B}+\overrightarrow{O C}}{2}=\frac{\vec{a}-3 \vec{b}+3 \vec{a}+5 \vec{b}}{2}\) 
[from Equation (i) and (ii)]
\(=\frac{4 \vec{a}+2 \vec{b}}{2}=2 \vec{a}+\vec{b}=\overrightarrow{O A} \quad[\text { from Eq. }(i)]\) 
Thus, \(\frac{\overrightarrow{O B}+\overrightarrow{O C}}{2}=\overrightarrow{O A}\) 
Hence, A is mid-point of line segment BC.

\(\overrightarrow a+\overrightarrow b=({2\overset\wedge i+3\overset\wedge j+4\overset\wedge k}),\) and \(\overrightarrow a-\overrightarrow b=\overset\wedge j-2\overset\wedge k\)
Let \(\overrightarrow c=(\overrightarrow a+\overrightarrow b)\times (\overrightarrow a-\overrightarrow b)\)
\(=\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 2 & 3 & 4 \\ 0 & -1 & -2 \end{matrix} \right| \)
\(\Rightarrow \overrightarrow c=-2\overset\wedge i+4\overset\wedge j-2\overset\wedge k\)
\(\Rightarrow \overset\wedge c=-\frac{1}{\sqrt6}\overset\wedge i+\frac{2}{\sqrt6}\overset\wedge j-\frac{1}{\sqrt6}\overset\wedge k\)

let  \(\overset { \rightarrow }{ a } =\overset { \wedge }{ j } -2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \ \text {and}\)\(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
then  \(\overset { \rightarrow }{ |a| } =\sqrt { { 1 }^{ 2 }+{ (-2) }^{ 2 }+{ 3 }^{ 2 } } \)  
\(=\sqrt { 1+4+9 } =\sqrt { 14 } \) 
\(\overset { \rightarrow }{ |b| } =\sqrt { { 3 }^{ 2 }+{ (-2) }^{ 2 }+{ 1 }^{ 2 } } \) 
\(=\sqrt { 9+4+1 } =\sqrt { 14 } \) 
and  \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =(\overset { \wedge }{ j } -2\overset { \wedge }{ j } +3\overset { \wedge }{ k } )+(3\overset { \wedge }{ i } -2\overset { \wedge }{ j } +\overset { \wedge }{ k } )\)  
= (1)(3)+(-2)(-2)+(3)(1)
= 3+4+3 = 10
if cos \(\theta\) be the angle between \(\overset { \rightarrow }{ a }\ and\ \overset { \rightarrow }{ b } \)
then cos \(\theta\) \(=\frac { ​​\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } }{ \overset { \rightarrow }{ |a| } \overset { \rightarrow }{ |b| } } =\frac { 10 }{ \sqrt { 14 } \sqrt { 14 } } =\frac { 10 }{ 14 } =\frac { 5 }{ 7 } \) 
Hence, \(\theta\) = cos^-1 \(\frac { 5 }{ 7 } \) 
\(\therefore \overset { \rightarrow }{ a } \ and\ \overset { \rightarrow }{ b } \) may have opposite direction, their magnitudes may be different.

\(\overset { \rightarrow }{ AB } =-4\overset { \wedge }{ i } -6\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ BC } =3\overset { \wedge }{ i } +10\overset { \wedge }{ j } +52\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ CD } =-7\overset { \wedge }{ i } -5\overset { \wedge }{ j }\)
The four points are coplanar if \(\overset { \rightarrow }{ AB } ,\overset { \rightarrow }{ BC } ,\overset { \rightarrow }{ CD } \) are coplanar.
If \(\left[ \overset { \rightarrow }{ AB } ,\overset { \rightarrow }{ BC } ,\overset { \rightarrow }{ CD } \right] =0\quad i.e.if\left| \begin{matrix} -4 & -6 & -2 \\ 3 & 10 & 5 \\ -7 & -5 & 0 \end{matrix} \right| =0\)
If -4(0 + 25) + 6(0 + 35) -2(-15 + 70) = 0
If -100 + 210 - 110 = 0
If 0 = 0, which is true.

-15

\(\lambda=-4\)

\(\left| \overrightarrow { a } \right| \ and\ \left| \overrightarrow { b } \right| \) =4,2

\((\overrightarrow { a } -\overrightarrow { d } )\) x \((\overrightarrow { b } -\overrightarrow { c } )\) 
\(=\overrightarrow { a } *\overrightarrow { b } -\overrightarrow { a } *\overrightarrow { c } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } *\overrightarrow { c } \)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } -\overrightarrow { c } \)
\([\because \overrightarrow { a } *\overrightarrow { b } =\overrightarrow { c } -\overrightarrow { d } and\overrightarrow { a } *\overrightarrow { c } =\overrightarrow { b } *\overrightarrow { d } ]\)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } +\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { c } *\overrightarrow { d } =\overrightarrow { 0 } \) 
Hence, \((\overrightarrow { a } -\overrightarrow { d } )\) is parallel to \((\overrightarrow { b } -\overrightarrow { c } )\)

\(\text {Let vector be } \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
According to given condition \( x-y+z=4 \text { ; }\)
\(2 x+y-3 z=0 ; x+y+z=2 \)
\(x-y+z=4 \)
\(2 x+y-3 z=0 \)
\(x+y+z=2 \)
From (ii), y = 3z − 2x
Substituting in (i) and (iii), we get
\(x-3 z+2 x+z=4 \Rightarrow 3 x-2 z=4 \)
\(x+3 z-2 x+z=2 \Rightarrow-x+4 z=2 \)
Solving (v) and (vi), we get
\(x=2, z=1\)
Substituting in (iv), we get
\(y=-1\)
\(\therefore \text { Vector is } 2 \hat{i}-\hat{j}+\hat{k} \)

Let \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)
Now, projection vector of \(\vec{a} \text { on } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}\)
Here, \(\begin{aligned} \vec{a} \cdot \vec{b} & =(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}-2 \hat{j}+\hat{k}) \end{aligned}\)
\(\begin{aligned} =2 \times 1+(-1) \times(-2)+1 \times 1 \end{aligned}\)
= 2 + 2 + 1 = 5
and \(|\vec{b}|=\sqrt{(1)^2+(-2)^2+(1)^2}=\sqrt{1+4+1}=\sqrt{6}\)
\(\therefore\) Projection vector of \(\vec{a} \text { on } \vec{b}=\frac{5}{6}(\hat{i}-2 \hat{j}+\hat{k})\)

We have, \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) 
and \(\vec{c}=3 \hat{i}+\hat{j}\) 
Now, \(\vec{a}+\lambda \vec{b}=(2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\) 
\( =2 \hat{i}+2 \hat{j}+3 \hat{k}-\lambda \hat{i}+2 \lambda \hat{j}+\lambda \hat{k} \) 
\(=(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}\) 
Since, \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\).
\(\therefore \ (\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0\) 
\(\Rightarrow [(2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}] \cdot(3 \hat{i}+\hat{j})=0\)
\(\Rightarrow(2-\lambda)(3)+(2+2 \lambda)(1)+(3+\lambda)(0)=0\)
\(\Rightarrow 6-3 \lambda+2+2 \lambda=0\)
\(\Rightarrow\)\(-\lambda+8=0\)
\(\therefore\) \(\lambda=8\)

We have,\(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k} \text { and } \vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\) 
Now,\(\vec{a}+\vec{b}=5 \hat{i}-\hat{j}-3 \hat{k}+\hat{i}+3 \hat{j}-5 \hat{k}=6 \hat{i}+2 \hat{j}-8 \hat{k}\) 
and \(\vec{a}-\vec{b}=5 \hat{i}-\hat{j}-3 \hat{k}-\hat{i}-3 \hat{j}+5 \hat{k}=4 \hat{i}-4 \hat{j}+2 \hat{k}\) 
Now,\((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=(6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot(4 \hat{i}-4 \hat{j}+2 \hat{k})\) 
= 24 - 8 -16 = 0
Hence,\((\vec{a}+\vec{b}) \text { and }(\vec{a}-\vec{b})\) are perpendicular vectors.

\(=\frac{\hat{j}+6 \hat{k}}{\sqrt{37}}\)

\(\overset\rightarrow a \times \overset\rightarrow b\)= \(\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & -1 & 2 \\ 2 & -1 & -1 \end{matrix} \right| \)
\(=i(1+2)-j(-1-4)+k(1-2)\)
= 3i + 5j + k
Area of \(\left| \overrightarrow { a } \times \overrightarrow { b } \right| =\sqrt{9+25+1}\)
\(=\sqrt{35} sq.units\)

\(\overrightarrow a.\overrightarrow b =0\)
\(\left| \overrightarrow { a } \right| =0 or \left| \overrightarrow {b } \right| =0\)
or \(\overrightarrow a\bot \overrightarrow b\) ...(i)
\(\overrightarrow a\times\overrightarrow b=0\)
\(\left| \overrightarrow {a } \right| =0 or \left| \overrightarrow {b } \right| =0\)
or \(a\parallel b\) ...(ii)
By eqn. (i) and (ii),
It is conclude that \(\left| \overrightarrow { a} \right| =0 or \left| \overrightarrow { b } \right| =0\)
\(\left| \overrightarrow { a} \right| \bot \left| \overrightarrow { b } \right| \) and \(a \parallel b\) is not possible.

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=\overset\wedge i+2\overset\wedge j+\overset\wedge k+3\overset\wedge i+\overset\wedge j-\overset\wedge k\)
\(\Rightarrow \overset\rightarrow c=4\overset\wedge i+3\overset\wedge j\)
\(\Rightarrow \overset\rightarrow d=\overset\rightarrow a-\overset\rightarrow b\)
=(i+2j+k)-(3i+j-k)
\(\Rightarrow\overset\rightarrow d =-2\overset\wedge i+\overset\wedge j+2\overset\wedge k\)
Projection \(\overset\rightarrow c \) on \(\overset\rightarrow d\)\(=\frac{\overset\rightarrow c.\overset\rightarrow d}{\left| \overset\rightarrow d \right| }\)
\(=\frac{(4\overset\wedge i+3\overset\wedge j).(-2\overset\wedge i+\overset\wedge j+2\overset\wedge k)}{\left|-2\overset\wedge i+\overset\wedge j+2\overset\wedge k \right| }\)
\(=\frac{-8+3}{\sqrt{4+1+4}}=-\frac{5}{3}\)

\(\left|\overset\rightarrow a+\overset\rightarrow b \right|^{ 2 } +\left|\overset\rightarrow a-\overset\rightarrow b \right|^{ 2 }=2{(\left| \overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right| ^{ 2 })}\)
\(\Rightarrow\left| \overset\rightarrow b \right| ^{ 2 }=2,116\)
\(\Rightarrow\left| \overset\rightarrow b \right| =46\)

As area of parallelogram
= \(\left| \begin{matrix} \widehat { i } & \widehat { j } & \widehat { k } \\ 2 & 1 & -2 \\ 3 & 1 & -1 \end{matrix} \right| \)
= \(\left| \widehat { i } -4\widehat { j } -\widehat { k } \right| \)
= \(\sqrt { 1+16+1 } \)
= \(3\sqrt { 2 } \) sq units

As sin \(\theta \) = \(\frac { |\vec { a } \times \vec { b } | }{ |\vec { b } ||\vec { b } | } \)
\(=\frac { 1 }{ \sqrt { 2 } .\sqrt { 2 } } =\frac { 1 }{ 2 } \)
\(\Rightarrow \theta =\frac { \pi }{ 6 } \)

As \({ \left| \sqrt { 3 } \vec { a } -\vec { b } \right| }^{ 2 }=({ \sqrt { 3 } \vec { a } -\vec { b } ) }^{ 2 }\)
\(=3\vec { { a }^{ 2 } } +\vec { { b }^{ 2 } } -2\sqrt { 3 } \vec { a } .\vec { b } \)
\(1=3+1-2\sqrt { 3 } \vec { a } .\vec { b } \)
\(\Rightarrow \vec { a } .\vec { b } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore cos\theta =\frac { \vec { a } .\vec { b } }{ |\vec { b } ||\vec { b } | } =\frac { \sqrt { 3 } }{ 2 } \)\(\Rightarrow \theta ={ 30 }^{ 0 }\)