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.

\(\overrightarrow a \times \overrightarrow b=\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 1 & 2 & -1 \\ 3 & 1 & -1 \end{matrix} \right| \)
\(=\overset\wedge i(-2+1)-\overset\wedge j(-1+3)+\overset\wedge k(1-6)\)
\(=-i-2j-5k\)
\(\left| \overrightarrow { a} \times \overrightarrow { b } \right| =\sqrt{1^{ 2}+2^{ 2}+5^{ 2}}\)
\(=\sqrt{1+4+25}=\sqrt{30}\)

\((\overset\rightarrow a-\overset\rightarrow b)^{2}=(\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a+\overset\rightarrow b)\)
\(=\overset\rightarrow a.\overset\rightarrow a+\overset\rightarrow a.\overset\rightarrow b+\overset\rightarrow b.\overset\rightarrow a+\overset\rightarrow b.\overset\rightarrow b\)
(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,
\((\overset\rightarrow a+\overset\rightarrow b)^{2} =(\overset\rightarrow a-\overset\rightarrow b)^{2}\)

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

Scalar quantity: Work done
Vector quantity: Force, velocity, displacement.

Let \(\overset\rightarrow{OP}=3\overset\rightarrow a-2\overset\rightarrow b\)
\(\overset\rightarrow{OQ}=2\overset\rightarrow a+3\overset\rightarrow b\)
The position vector of the point R dividing the join of P and Q internally in the ratio 2:1 is
\(\overset\rightarrow{OR}=\frac{2(2\overset\rightarrow a+3\overset\rightarrow b)+(3\overset\rightarrow a-2\overset\rightarrow b)}{2+1}\)
\(=\frac{4\overset\rightarrow a+6\overset\rightarrow b+3\overset\rightarrow a-2\overset\rightarrow b}{3}\)
\(=\frac{7\overset\rightarrow a}{3}+\frac{4 \overset\rightarrow b}{3}\)

\(10\over\sqrt6\)

\(\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 14 \\ 1 & -\lambda & 7 \end{array}\right|=\overrightarrow{0}\)
\( \Rightarrow \hat{i}(42+14 \lambda)-\hat{j}(14-14) +\hat{k}(-2 \lambda-6)=\overrightarrow{0} \Rightarrow \lambda=-3 \)

\(\vec{r}=\vec{a}-\vec{b}=-2 \hat{i}+\hat{j}+4 \hat{k},|\vec{r}|=\sqrt{4+1+16}=\sqrt{21} \)
\(\text { Now, unit vector } \hat{r}=\frac{\vec{r}}{|\vec{r}|}=\frac{-2}{\sqrt{21}} \hat{i}+\frac{1}{\sqrt{21}} \hat{j}+\frac{4}{\sqrt{21}} \hat{k} \text { ; } \)

\(\text { If } a \text { and } \vec{b} \text { are perpendicular, then } \vec{a} \cdot \vec{b}=0\)
\(\Rightarrow 2-2 \lambda+3=0 \Rightarrow \lambda=\frac{5}{2}\)

A,B,C,D are coplanar, if 
\(\overrightarrow { AB } .\left( \overrightarrow { AC } \times \overrightarrow { AD } \right) =0\)
\(\overrightarrow { AB } =-4\hat { i } -6\hat { j } -2\hat { k } \)
\(\overrightarrow { AC } =-\hat { i } +4\hat { j } +3\hat { k } \)
\(\overrightarrow { AD } =-8\hat { i } -\hat { j } +3\hat { k } \)
They are coplanar
= -4(15) + 6(21) - 2(33) = 0

Note that the direction ratio’s a, b, c of a vector \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\) are just the respective components x, y and z of the vector. So, for the given vector, we have a = 1, b = 1 and c = –2. Further, if l, m and n are the direction cosines of the given vector, then
\(l=\frac{a}{|\vec{r}|}=\frac{1}{\sqrt{6}}, \ m=\frac{b}{|\vec{r}|}=\frac{1}{\sqrt{6}}, \ n=\frac{c}{|\vec{r}|}=\frac{-2}{\sqrt{6}} \text { as }|\vec{r}|=\sqrt{6}\)
Thus, the direction cosines are \(\left(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}},-\frac{2}{\sqrt{6}}\right) \text { . }\)

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

Let \(\overset { \rightarrow }{ a } \) = \(\overset { \wedge }{ i } -\overset { \wedge }{ 2j } +\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } -\overset { \wedge }{ 3j } +\overset { \wedge }{ 5k } \) 
here \(\overset { \rightarrow }{ |a| } =\sqrt { { 1 }^{ 2 }+{ (-2) }^{ 2 }+{ 1 }^{ 2 } } \) 
\(=\sqrt { 1+4+1 } =\sqrt { 6 } \)
and \(\overset { \rightarrow }{ |b| } =\sqrt { { 1 }^{ 2 }+{ (3) }^{ 2 }+{ 5 }^{ 2 } } \) 
\(=\sqrt { 1+9+25 } =\sqrt { 35 } \) 
Clearly  \(\overset { \rightarrow }{ a } \) \(\neq \) 0,  \(\overset { \rightarrow }{ b } \)  \(\neq \) 0 But 
\(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\left( \overset { \wedge }{ i } -\overset { \wedge }{ 2j } +\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ i } +\overset { \wedge }{ 3j } +\overset { \wedge }{ 5k } \right) \\ \) 
= (1) (1) + (-2) (3) + (1) (5)
= 1 - 6 + 5 = 0
Hence, \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\text{ through}\ \overset { \rightarrow }{ a } \neq \overset { \rightarrow }{ 0 } ,\ \overset { \rightarrow }{ b } \neq \overset { \rightarrow }{ 0 } \).

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, '\(\theta \)' = \(\frac { \pi }{ 4 } \) 

The unit vector \(\hat{a} \ {\text {in}} \) the direction of vector \( \vec{a}=\hat{i}+\hat{j}+2 \hat{k} \text { is given by } \hat{a}=\frac{\vec{a}}{|a|} \text { . }\)
\(|\vec{a}|=\sqrt{1^{2}+1^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6} \)
\(\therefore \hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{\hat{i}+\hat{j}+2 \hat{k}}{\sqrt{6}}=\frac{1}{\sqrt{6}} \hat{i}+\frac{1}{\sqrt{6}} \hat{j}+\frac{2}{\sqrt{6}} \hat{k} \)

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.

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

Let  \(\hat{a}=\hat{i}+\hat{j}+\hat{k} \)
\(\hat{b}=2 \hat{i}+4 \hat{j}-5 \hat{k} \) 
and \(\hat{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}\) 
now the unit vector along \(\overset { \wedge }{ b } +\overset { \wedge }{ c } \) 
\(=\frac { (\lambda +2)\overset { \wedge }{ i } +6\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { { (\lambda +2) }^{ 2 } } +36+4 } \) 
By the question
(\(\hat { i } +\hat { j } +\hat { k } \)).\(=\frac { (\lambda +2)\overset { \wedge }{ i } +6\overset { \wedge }{ j } -2\overset { \wedge }{ k } }{ \sqrt { { (\lambda +2) }^{ 2 } } +40 } \) =1
\(=\frac { 1 }{ \sqrt { { (\lambda +2) }^{ 2 } } +40 } (\lambda +2+6)= 6\)
\(\Rightarrow \lambda +6=\sqrt { { (\lambda +2) }^{ 2 }+40 } \)
Squaring, \({ \lambda }^{ 2 }+12\lambda +36\) 
\(={ \lambda }^{ 2 }+4\lambda +4+40\)
\(\Rightarrow 12\lambda +36=4\lambda +44\)
\(\Rightarrow 8\lambda =8\)
\(Hence\ \lambda =1\)

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