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.

Given, \(\left| a\right| =\left| b \right| \)
and a.b \(=\frac{1}{2}\)
Let \(\theta \) be thae angle between \(\overrightarrow a\) and \(\overrightarrow b\)
then, \(\cos \theta=\frac{a.b}{\left| a\right| \left| b \right| }\)
\(\cos 60^{ \circ }=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right| .\left| \overrightarrow { a} \right| }\)
\(\Rightarrow \frac{1}{2}=\frac{\frac{1}{2}}{\left| \overrightarrow { a } \right|^{ 2} }\)
\(\Rightarrow \left| \overrightarrow { a } \right| ^{2 }=\frac{\frac{1}{2}}{\frac{1}{2}}=1\)
\(\Rightarrow \left| \overrightarrow { a} \right| =1 \)
\(\Rightarrow\left| \overrightarrow { a} \right| =\left| \overrightarrow { b} \right| =1\)

\((\overset\rightarrow a-\overset\rightarrow b)^{2}=(\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a+\overset\rightarrow b)\)
\(=\overset\rightarrow a.\overset\rightarrow a+\overset\rightarrow a.\overset\rightarrow b+\overset\rightarrow b.\overset\rightarrow a+\overset\rightarrow b.\overset\rightarrow b\)
(as \(\overset\rightarrow a\bot\overset\rightarrow b\), then \(\overset\rightarrow a. \overset\rightarrow b=\overset\rightarrow b.\overset\rightarrow a=0)\)
\(=\left|\overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right|\) ... (i)
\((\overset\rightarrow a-\overset\rightarrow b)^{2}=(\overset\rightarrow a-\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)\)
\(=\overset\rightarrow a.\overset\rightarrow a-\overset\rightarrow a.\overset\rightarrow b-\overset\rightarrow b.\overset\rightarrow a+\overset\rightarrow b.\overset\rightarrow b\)
\(=\left|\overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right|^{2 }\)...(ii)
By (i) and (ii) both are equal,
\((\overset\rightarrow a+\overset\rightarrow b)^{2} =(\overset\rightarrow a-\overset\rightarrow b)^{2}\)

Positive vector of \(C=\frac{2(\overset\rightarrow a-3\overset\rightarrow b)+3(2\overset\rightarrow a+\overset\rightarrow b)}{2+3}\)
\(=\frac{2\overset\rightarrow a-6\overset\rightarrow b+6\overset\rightarrow a+3\overset\rightarrow b}{5}\)
Position vector of \(C=\frac{8\overset\rightarrow a-3\overset\rightarrow b}{5}\)

(i)
(ii)

\(\left| \overrightarrow { x } \right| \)=3

Unit vector along \(-2 \hat{i}+\hat{j}-5 \hat{k}\)
\(=\frac{-2}{\sqrt{30}} \hat{i}+\frac{1}{\sqrt{30}} \hat{j}-\frac{5}{\sqrt{30}} \hat{k}\)
\(\Rightarrow \text { Direction cosines are }{-{2\over\sqrt{30}}},{1\over\sqrt{30}},{-{5\over\sqrt{30}}}\)

3,-4 and -6

\(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{\sqrt{9+4+36}}=\frac{3}{7} \vec{i}-\frac{2}{7} \vec{j}+\frac{6}{7} \vec{k}\)

Let the points be A(3, -2, -1), B(2, 3, -4), C(-1, 1, 2) and D(4, 5, λ)
\(\overrightarrow{AB}=\)(Position vector of B)-(Position vector of A)
\(\therefore \overrightarrow{AB}=-\overset\wedge i+5\overset\wedge j-3\overset\wedge k,\)
Similarly
\(\overrightarrow{AC}=-4\overset\wedge i+3\overset\wedge j+3\overset\wedge k\)
and \(\overrightarrow{AD}=\overset\wedge i+7\overset\wedge j+(\lambda+1)\overset\wedge k\)
A, B, C, D are coplanar if\([\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]=0\)
\([\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]=\left| \begin{matrix} -1 & 5 & -3 \\ -4 & 3 & 3 \\ 1 & 7 & \lambda +1 \end{matrix} \right| =0\)
\(\therefore 1(15+9)-7(-3-12)+(\lambda+1)(-3+20)=0\)
\(\Rightarrow \lambda=-\frac{146}{17}\)

Let O be the origin of reference.
Then, given \(\begin{aligned} \overrightarrow{O A}=2 \hat{i}-\hat{j}+\hat{k} \end{aligned}\)
\(\begin{aligned} \overrightarrow{O B}=\hat{i}+\hat{j}+2 \hat{k} \end{aligned}\)
and \(\begin{aligned} \overrightarrow{O C}=2 \hat{i}+3 \hat{k} \end{aligned}\)
Now, \(\begin{aligned} \overrightarrow{A B} & =\overrightarrow{O B}-\overrightarrow{O A} \end{aligned}\)
\(\begin{aligned} =\hat{i}+\hat{j}+2 \hat{k}-2 \hat{i}+\hat{j}-\hat{k} \end{aligned}\)
\(\begin{aligned} =-\hat{i}+2 \hat{j}+\hat{k} \end{aligned}\)
and \(\begin{aligned} \overrightarrow{A C} & =\overrightarrow{O C}-\overrightarrow{O A} \end{aligned}\)
\(\begin{aligned} =2 \hat{i}+3 \hat{k}-2 \hat{i}+\hat{j}-\hat{k}=\hat{j}+2 \hat{k} \end{aligned}\)
Now, \(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 0 & 1 & 2 \end{array}\right|\)
\(\begin{aligned} =\hat{i}(4-1)-\hat{j}(-2-0)+\hat{k}(-1-0) \end{aligned}\)
\(\begin{aligned} =3 \hat{i}+2 \hat{j}-\hat{k} \end{aligned}\)
and \(\begin{aligned} |\overrightarrow{A B} \times \overrightarrow{A C}| & =\sqrt{(3)^2+(2)^2+(-1)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{9+4+1}=\sqrt{14} \end{aligned}\)
\(\therefore\) Unit vector perpendicular to the plane ABC
\(\begin{aligned} & =\frac{\overrightarrow{A B} \times \overrightarrow{A C}}{|\overrightarrow{A B} \times \overrightarrow{A C}|} \end{aligned}\)
\(\begin{aligned} =\frac{3 \hat{i}+2 \hat{j}-\hat{k}}{\sqrt{14}} \end{aligned}\)
\(\begin{aligned} =\frac{3}{\sqrt{14}} \hat{i}+\frac{2}{\sqrt{14}} \hat{j}-\frac{1}{\sqrt{14}} \hat{k} \end{aligned}\)

The given vectors are \(\)\(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k} \text { and } \vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\)
\(\therefore \vec{a}+\vec{b}+\vec{c} =(1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7) \hat{k} \)
\(=0 \cdot \hat{i}-4 \hat{j}-1 \cdot \hat{k} \)
\(=-4 \hat{j}-\hat{k} \)

Let
\(\overset { \rightarrow }{ a } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } ,\quad \overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } -\overset { \wedge }{ k } \) 
\(\overset { \rightarrow }{ c } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) 
\(\therefore\) Volume of the paralleopiped \(=[\overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } ]\)
\(=\left| \begin{matrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix} \right| \)
\(=2(4-1)+3(2+3)+4(-1-6)\)  [Expanding by R₁ ]
\(=6+15-28=-7=7\)    [rejecting -ve sign]

Given, \(|\vec{a}+\vec{b}|=13 \text { and }|\vec{a}|=5\)
Now, \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)
\(\begin{array}{cl} \Rightarrow \quad & |\vec{a}+\vec{b}|^2=|\vec{a}|^2+0+0+|\vec{b}|^2 \end{array}\)
\(\begin{array}{cl} {\left[\because \vec{x} \cdot \vec{x}=|\vec{x}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}=0 \text { as } \vec{a} \perp \vec{b}\right]} \end{array}\)
\(\begin{array}{ll} \Rightarrow & (13)^2=(5)^2+|\vec{b}|^2 \end{array}\)
\(\begin{array}{ll} \Rightarrow & 169=25+\left.|\vec{b}|^2 \Rightarrow|69-25=| \vec{b}\right|^2 \end{array}\)
\(\begin{array}{ll} \Rightarrow & 144=|\vec{b}|^2 \Rightarrow|\vec{b}|=12 \end{array}\)
[\(\because\) length cannot be '-' ve]

The inequality holds trivially when either \(\vec{a}=\overrightarrow{0} \text { or } \vec{b}=\overrightarrow{0}\)  Actually, in such a situation we have \(|\vec{a} \cdot \vec{b}|=0=|\vec{a}||\vec{b}|\) So, let us assume that \(|\vec{a}||\neq 0 \neq| \vec{b} \mid\)
Then, we have
\(\frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}||\vec{b}|}=|\cos \theta| \leq 1 \)
\(|\vec{a} \cdot \vec{b}| \leq|\vec{a}||\vec{b}| \)

The sum of the given vectors is
\(\vec{a}+\vec{b}(=\vec{c}, \text { say })=4 \hat{i}+3 \hat{j}-2 \hat{k}\)
\(\text { and } \quad|\vec{c}|=\sqrt{4^{2}+3^{2}+(-2)^{2}}=\sqrt{29}\)
Thus, the required unit vector is
\(\hat{c}=\frac{1}{|\vec{c}|} \vec{c}=\frac{1}{\sqrt{29}}(4 \hat{i}+3 \hat{j}-2 \hat{k})=\frac{4}{\sqrt{29}} \hat{i}+\frac{3}{\sqrt{29}} \hat{j}-\frac{2}{\sqrt{29}} \hat{k}\)

The given vectors are \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k} \text { and } \vec{b}=-\hat{i}+\hat{j}-\hat{k}\)
\(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k} \)
\(\vec{b}=-\hat{i}+\hat{j}-\hat{k} \)
\(\therefore \vec{a}+\vec{b}=(2-1) \hat{i}+(-1+1) \hat{j}+(2-1) \hat{k}=1 \hat{i}+0 \hat{j}+1 \hat{k}=\hat{i}+\hat{k} \)
\(|\vec{a}+\vec{b}|=\sqrt{1^{2}+1^{2}}=\sqrt{2} \)
Hence, the unit vector in the direction of \((\vec{a}+\vec{b}){\text {is }}\)
\(\frac{(\vec{a}+\vec{b})}{|\vec{a}+\vec{b}|}=\frac{\hat{i}+\hat{k}}{\sqrt{2}}=\frac{1}{2} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}\)

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.

\( \vec{a}=\hat{i}+\hat{j}+\hat{k} \text { and } \vec{b}=\hat{j}-\hat{k} ; \vec{a} \times \vec{c}=\vec{b} \text { and }\vec{a} \cdot \vec{c}=3\)
\(\text {Let } \vec{c}=x \hat{i}+y \hat{j}+z \hat{k}\)
\( \vec{a} \cdot \vec{c}=3 \Rightarrow(\hat{i}+\hat{j}+\hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=3\)
\(\Rightarrow x+y+z=3\)
\(\text {Also, } \vec{a} \times \vec{c}=\vec{b} \Rightarrow\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ x & y & z \end{array}\right|=\hat{j}-\hat{k}\)
\(\Rightarrow(z-y) \hat{i}-(z-x) \hat{j}+(y-x) \hat{k}=\hat{j}-\hat{k}\)
\(\Rightarrow z-y=0,-(z-x)=1 \text { and } y-x=-1\)
\(\Rightarrow z=y \)
\(x-z=1\)
\(\text {and } \ y-x=-1\)
Solving for x, y, z we get \(x=\frac{5}{3}, y=\frac{2}{3} \text { and } z=\frac{2}{3}\)
Substituting for x, y, z \( \operatorname{in}(i) \text { , we get } \vec{c}=\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k} \text { . }\)

As given \(\overrightarrow { a } .\overrightarrow { b } =\overrightarrow { a } .\overrightarrow { c } =0=\overrightarrow { a } \bot \)both \(\overrightarrow { b } \) and \(\overrightarrow { c } \)
⇒ \(\overrightarrow { a } \parallel \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Let \(\overrightarrow { a } =\lambda \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
then \(\left| \overrightarrow { a } \right| =\left| \lambda \right| \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| \)
⇒ \(\frac { \left| \overrightarrow { a } \right| }{ \left| \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right| } =\left| \lambda \right| \)
⇒ \(\left| \lambda \right| =\frac { 1 }{ sin\frac { \pi }{ 6 } } =2\)
λ = 士2
\(\overrightarrow { a } =\pm 2\left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
Hence proved
(ii) Now \(\left[ \overrightarrow { a } +\overrightarrow { b } ,\overrightarrow { b } +\overrightarrow { c } ,\overrightarrow { c } +\overrightarrow { a } \right] \)
= \(\left[ \left( \overrightarrow { a } +\overrightarrow { b } \right) \times \left( \overrightarrow { b } \times \overrightarrow { c } \right) \right] .\left( \overrightarrow { c } +\overrightarrow { a } \right) \)
= \(\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } +\left( \overrightarrow { b } \times \overrightarrow { c } \right) .\overrightarrow { a } \)
(As the scalar triple product = 0, if any two vectors are equal)
Hence, \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\left( \overrightarrow { a } \times \overrightarrow { b } \right) .\overrightarrow { c } \) = \(\overrightarrow { a } .\left( \overrightarrow { b } \times \overrightarrow { c } \right) +\overrightarrow { c } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= 2\(\overrightarrow { a } \left( \overrightarrow { b } \times \overrightarrow { c } \right) \)
= \(2\overrightarrow { a } \left( \pm \frac { 1 }{ 2 } \overrightarrow { a } \right) \)
= 士1
Hence proved