Let \(\theta \) be angle between a and b
LHS,\(\left| \overrightarrow {a } \times \overrightarrow {b } \right| ^{2 }=(\overrightarrow a\times\overrightarrow b).(\overrightarrow a \times \overrightarrow b)\)
\(=(ab\sin\theta)\overset\wedge n.(ab\sin\theta)\overset\wedge n\)
\(=(a^{ 2 }b^{2 }\sin^{2 }\theta)(\overset\wedge n.\overset\wedge n)\)
\(=a^{2 }b^{2 }\sin^{ 2 }\theta\)
\(=a^{2 }b^{ 2}(1-\cos^{ 2 }\theta)\)
\(=a^{ 2 }b^{ 2 }-(ab\cos\theta)^{ 2}\)
\(=(\overrightarrow a.\overrightarrow a)(\overrightarrow b.\overrightarrow b)-(\overrightarrow a.\overrightarrow b)^{2}\)....(i)
Also, RHS=\(\begin{vmatrix} \overrightarrow { a } .\overrightarrow { a } & \overrightarrow { a } .\overrightarrow { b } \\ \overrightarrow { a } .\overrightarrow { b } & \overrightarrow { b } .\overrightarrow { b } \end{vmatrix}\)
\(=(\overrightarrow a.\overrightarrow a).(\overrightarrow b.\overrightarrow b)-(\overrightarrow a.\overrightarrow b).(\overrightarrow a.\overrightarrow b)\)
\(=(\overrightarrow a.\overrightarrow a).(\overrightarrow b.\overrightarrow b)-(\overrightarrow a.\overrightarrow b)^{ 2}\)...(ii)
From eqn.(i) and (ii)
\(\Rightarrow\)\(RHS=LHS \) 
Hence proved.

\(\cos \theta=\frac{\overrightarrow a.\overrightarrow b}{\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| }=\frac{12}{10\times 2}=\frac{3}{5}\)
\(\cos \theta =\frac{3}{5}\)
\(\sin \theta=\sqrt{1-\cos ^{2}\theta} =\sqrt{1-\frac{9}{25}}\)
\(\sin \theta=\frac{\left| \overrightarrow { a } \times \overrightarrow { b } \right| }{\left| \overrightarrow {a } \right| \left| \overrightarrow {b } \right| }\)
\(\Rightarrow \left| \overrightarrow { a } \times \overrightarrow { b } \right| =\left| \overrightarrow { a } \right| \left| \overrightarrow { b} \right| \sin\theta\)
\(\Rightarrow \left| \overrightarrow { a } \times \overrightarrow { b } \right| =10\times2\times\frac{4}{5}=16\)

\(\overset\rightarrow {BA}\)= Position vector of A- Position vector of B
\(\overset\rightarrow{BA}=(\overset\wedge i+2\overset\wedge j-3\overset\wedge k)-(-\overset\wedge i-2\overset\wedge j+\overset\wedge k)\)
\(\overset\rightarrow {BA}=2\overset\wedge i+4\overset\wedge j-\overset\wedge k \)
\(\left| \overset\rightarrow {BA}\right| =\sqrt{2^{2}+4^{2}+(-4)^{2}}\)
\(=\sqrt{4+16+16}\)
\(=\sqrt{36}=6 \)
Direction cosines of the vector \(\overset\rightarrow{BA}\) are \((\frac{2}{6},\frac{4}{6},-\frac{4}{6})\)
\(=(\frac{1}{3},\frac{2}{3},-\frac{2}{3})\)

\(\overset\rightarrow c=\overset\rightarrow a+\overset\rightarrow b\)
\(=(2\overset\wedge i+\overset\wedge j+3\overset\wedge k)+(\overset\wedge i+2\overset\wedge j-\overset\wedge k)\)
\(=3\overset\wedge i+3\overset\wedge j+2\overset\wedge k\)
\(\overset\wedge c=\frac{\overset\rightarrow c}{\left|c \right| }\)
\(=\frac{2i+3j+2k}{\sqrt{9+9+4}}\)
\(=\frac{3}{\sqrt{22}}\overset\wedge i+\frac{3}{\sqrt{22}}\overset\wedge j+\frac{2}{\sqrt{22}}\overset\wedge k\)

(i)
(ii)

(i) Magnitude of 2\(\overrightarrow { a } \) is 6 direction is along \(\overrightarrow { a } \)
(ii) Magnitude of -5\(\overrightarrow { a } \) is 15 direction is opposite to that of \(\overrightarrow { a } \)

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

\(\theta={\pi \over3}\)

Vector of magnitude 7 along is \(7 \hat{a}=\frac{7(\hat{i}-2 \hat{j})}{\sqrt{5}}\)
\(7\hat { a } =\frac { 7 }{ \sqrt { 5 } } \hat { i } -\frac { 14 }{ \sqrt { 5 } } \hat { j } \)

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1-1-1}{\sqrt{3} \sqrt{3}}=-\frac{1}{3} \Rightarrow \theta=\cos ^{-1}\left(-\frac{1}{3}\right)\)

Let \(x\overset\wedge i+y\overset\wedge j+z\overset\wedge k \) be the unit vector along \(\overrightarrow c.\)
Since \(-\overset\wedge i+\overset\wedge j-\overset\wedge k\)bisects the angle between \(\overrightarrow c\) and \(3\overset\wedge i+4\overset\wedge j.\) Therefore,
\(\lambda(-\overset\wedge i+\overset\wedge j-\overset\wedge k)=(x\overset\wedge i+y\overset\wedge j+z \overset\wedge k)+\frac{3\overset\wedge i+4\overset\wedge j}{5}\)
\(x+\frac{3}{5}=-\lambda\)
\(y+\frac{4}{5}=-\lambda\)
and \(z=-\lambda\)
Now, \(x^{ 2}+y^{2 }+z^{2 }=1\)\([\because x\overset\wedge i+y\overset\wedge j+z\overset\wedge k\) is a unit vector\(]\)
\(\Rightarrow ({-\lambda-\frac{3}{5}})^{ 2}+({\lambda-\frac{4}{5}})^{ 2}+\lambda^{ 2}\)
= 1
\(\Rightarrow 3\lambda^{ 2}-\frac{2}{5}\lambda=0\)
\(\Rightarrow \lambda=0 or \lambda =\frac{2}{15}\)
But \(\lambda\ne0,\) because \(\lambda=0\) implies that the givem vectors are parallel.
\(\Rightarrow \lambda=\frac{-2}{15}\)
\(y=\frac{-10}{15}\)
and \(z=\frac{-2}{15}\)
Hence, \(x\overset\wedge i+y\overset\wedge j+z\overset\wedge k=\frac{1}{15}(11\overset\wedge i+10\overset\wedge j+2\overset\wedge k)\)

Here \(\overset { \rightarrow }{ PQ } \) = P.V of Q - P.V of P
\(=(-\overset { \wedge }{ i } +0\overset { \wedge }{ j } +\overset { \wedge }{ 8k } )\quad -(\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } )\) 
\(=-2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +6\overset { \wedge }{ k } \)
\(\therefore \overset { \rightarrow }{ QP } =-\overset { \rightarrow }{ PQ } =-2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +6\overset { \wedge }{ k } \)
\(\therefore \overset { \rightarrow }{ |QP| } =\sqrt { 4+9+36 } =\sqrt { 49 } =7\)
\(\therefore\) Unit vector in the direction of \(\overset { \rightarrow }{ QP } \)  is given by:
\(\overset { \rightarrow }{ QP } =\frac { \overset { \rightarrow }{ QP } }{ \overset { \rightarrow }{ |QP } | } =\frac { 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } -6\overset { \wedge }{ k } }{ 7 } \) 
Hence, the reqd.vector of magnitude 11 in the direction of \(\overset { \rightarrow }{ QP } \) is:
\(11\overset { \rightarrow }{ QP } =11\frac { 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } -6\overset { \wedge }{ k } }{ 7 } =\frac { 22 }{ 7 } \overset { \wedge }{ i } +\frac { 33 }{ 7 } \overset { \wedge }{ j } -\frac { 66 }{ 7 } \overset { \wedge }{ k } \) .

We have:
\(\overset { \rightarrow }{ |a| } =1\) , \(\overset { \rightarrow }{ |b| } =2\)  and   \(\overset { \rightarrow }{ a } \).\(\overset { \rightarrow }{ b } \)=1.
\(\theta=\cos ^{-1}\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)=\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}\)

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

If \(\theta \) be the angle between \(\overset { \rightarrow }{ a } \) and \(\overset { \rightarrow }{ b } \), then
cos \(\theta \) = \(\frac { \overset { \rightarrow }{ a } ,\overset { ​​​​\rightarrow }{ b } }{ |\overset { \rightarrow }{ a } ||\overset { \rightarrow }{ b } | } =\frac { \sqrt { 6 } }{ \sqrt { 3(2) } } =\frac { \left( \sqrt { 3 } \right) \left( \sqrt { 2 } \right) }{ \left( \sqrt { 3 } \right) \left( 2 \right) } \)
= \(\frac { 1 }{ \sqrt { 2 } } =cos\frac { \pi }{ 4 } \)
Hence, '\(\theta \)' = \(\frac { \pi }{ 4 } \) 

(i) true (ii) False
(iii) False (iv) True

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.

We have : \(\overset { \rightarrow }{ |a| } =\overset { \rightarrow }{ |b| } =\overset { \rightarrow }{ |c| } =1\) ..(1)
Now   \(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { 0 } \)..(2) 
Squaring,   \({ \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \right) }^{ 2 }=0\)
\(\Rightarrow \) \(\overset { \rightarrow }{ { |a| }^{ 2 } } +\overset { \rightarrow }{ { |b| }^{ 2 } } +\overset { \rightarrow }{ { |c| }^{ 2 } } +2\left( \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } \right) =0\)
\(\Rightarrow \) \({ (1) }^{ 2 }+{ (2) }^{ 2 }+{ (3) }^{ 2 }+2\left( \overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } \right) =0\) 
hence, \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } +\overset { \rightarrow }{ b } .\overset { \rightarrow }{ c } +\overset { \rightarrow }{ c } .\overset { \rightarrow }{ a } =-\frac { 3 }{ 2 } \) .

\(\overrightarrow { a } .\overrightarrow { b } +\overrightarrow { b } .\overrightarrow { c } +\overrightarrow { c } .\overrightarrow { a } = -25\)

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