Let \(\vec{c}=\hat{a}+\hat{b}\). Then, according to given condition \(\vec{c}\) is a unit vector i.e., \(|\vec{c}|=1\)
To show \(|\hat{a}-\hat{b}|=\sqrt{3}\)
Consider, \(\begin{aligned} \vec{c} & =\hat{a}+\hat{b} \Rightarrow|\vec{c}|=|\hat{a}+\hat{b}| \end{aligned}\)
\(\begin{array}{ll} \Rightarrow & 1=|\hat{a}+\hat{b}| \Rightarrow|\hat{a}+\hat{b}|^2=1 \end{array}\)
\(\begin{array}{ll} \Rightarrow & (\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})=1 \end{array}\)
\(\begin{array}{lrl} \Rightarrow & |\hat{a}|^2+2 \hat{a} \cdot \hat{b}+|\hat{b}|^2=1 \end{array}\)
\(\begin{array}{lrl} \Rightarrow & 1+2 \hat{a} \cdot \hat{b}+1=1 \Rightarrow 2 \hat{a} \cdot \hat{b}=-1 \end{array}\)      ...(i)
Now consider, \(\begin{aligned} |\hat{a}-\hat{b}|^2 & =(\hat{a}-\hat{b}) \cdot(\hat{a}-\hat{b}) \end{aligned}\)
\(\begin{aligned} =|\hat{a}|^2-2 \hat{a} \cdot \hat{b}+|\hat{b}|^2 \end{aligned}\)
= 1 - (-1) + 1      [using Eq. (i)]
\(\begin{array}{ll} \Rightarrow & |\hat{a}-\hat{b}|^2=3 \end{array}\)
\(\begin{array}{ll} \Rightarrow & |\hat{a}-\hat{b}|=\sqrt{3} \end{array}\)
[taking positive square root, as magnitude cannot be negative]
Hence proved.

Let angle between \(\overrightarrow a\) and \(\overrightarrow b\) be \(\theta\)
Given,\((\overrightarrow a+3\overrightarrow b)\bot (7\overrightarrow a-5\overrightarrow b)\)
\(\Rightarrow (\overrightarrow a+3\overrightarrow b).(7\overrightarrow a-5\overrightarrow b)=0\)
\(\Rightarrow 7\left| \overrightarrow {b } \right| ^{ 2 }+16(\overrightarrow a.\overrightarrow b)-15\left| \overrightarrow {b } \right| ^{2 }=0\)
\(\Rightarrow 7+16\cos\theta-15=0[\because\left| \overrightarrow { a } \right| ^{ 2}=\left| \overrightarrow {b } \right| ^{ 2 }=1]\)
\(\Rightarrow \cos\theta=\frac{8}{12}=\frac{1}{2}\)
\(\Rightarrow \theta=\frac{\pi}{3}\)
Also given that \((\overrightarrow a-4\overrightarrow b)\bot(7\overrightarrow a-2\overrightarrow b)\)
\(\Rightarrow (\overrightarrow a-4\overrightarrow b).(7\overrightarrow a-2\overrightarrow b)=0 \)
\(\Rightarrow7\left| \overrightarrow { a }  \right| ^{  2}+8\left| \overrightarrow {  b}  \right| ^{ 2 }-30(\overrightarrow a.\overrightarrow b)=0\)
\(\Rightarrow15-30\cos\theta=0\)
\(\Rightarrow \cos \theta=\frac{1}{2}\)
\(\Rightarrow \theta=\frac{\pi}{3}\)

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, \(x\overset\wedge i+y\overset\wedge j+z\overset\wedge k=\frac{1}{15}(11\overset\wedge i+10\overset\wedge j+2\overset\wedge k)\)

Taking \(LHS=\overrightarrow a.\{(\overrightarrow b+\overrightarrow c)\times\overrightarrow d\}\)
\(=\overrightarrow a.\{(\overrightarrow b \times\overrightarrow d)+(\overrightarrow c\times\overrightarrow d)\}\)
\(=\overrightarrow a.(\overrightarrow b\times \overrightarrow d)+\overrightarrow a.(\overrightarrow c\times \overrightarrow d)\)
\(=\left[ \begin{matrix} \overrightarrow { a } , & \overrightarrow { b } , & \overrightarrow { d } \end{matrix} \right] +\left[ \begin{matrix} \overrightarrow { a } , & \overrightarrow { c } , & \overrightarrow { d } \end{matrix} \right] \)

Since the vector \(\overrightarrow p,\overrightarrow q \ and\ \overrightarrow r \) are coplanar,
\(\therefore [\overrightarrow p,\overrightarrow q,\overrightarrow r]=0\)
i.e., \(\left| \begin{matrix} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{matrix} \right| =0\)
 \(R_2\rightarrow R_2-R_1\)
and \(R_3\rightarrow R_3-R_1\)
or\(\left| \begin{matrix} a & 1 & 1 \\ 1-a & b-1 & 0 \\ 1-a & 0 & c-1 \end{matrix} \right| =0\)
\(\Rightarrow a(b-1)(c-1)-1(1-a)(c-1)-1(1-a)(b-1)=0\)
i.e., \(a(1-b)(1-c)+(1-a)(1-c)+(1-a)(1-b)=0\)
Dividing both the sides by (1-a)(1-b)(1-c), we get
\(\frac{a}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0\)
i.e., \(-1+\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}=0\)

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.

\(|\vec{a} \times \vec{b}|=8 \Rightarrow|\vec{a} \| \vec{b}| \sin \theta=8 \)
\(\Rightarrow \sin \theta=\frac{4}{5} \Rightarrow \cos \theta=\frac{3}{5} \)
\(\Rightarrow \vec{a} \cdot \vec{b}=|\vec{a} \| \vec{b}| \cos \theta=2.5 \cdot \frac{3}{5}=6 \)

\( =\left|\begin{array}{rrr} 2 & -3 & 1 \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{array}\right| \)
\(=2(-1)+3(-5)+1(3) \)
\(=|-2-15+3|=|-14|=14 \mathrm{cu} \text { units } \)

\(\overrightarrow { AL } +\overrightarrow { BM } +\overrightarrow { CN } \)=\(\lambda \)\(\overrightarrow { CK } \)

Given, \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k} \text { and } \vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\)
Now, \(\vec{a}+\vec{b}=6 \hat{i}-2 \hat{j}+(7+\lambda) \hat{k}\)
and \(\vec{a}-\vec{b}=-4 \hat{i}+(7-\lambda) \hat{k}\)
\(\because(\vec{a}+\vec{b}) \text { and }(\vec{a}-\vec{b})\) are orthogonal.
\(\begin{aligned} \therefore & (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b}) =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & {[6 \hat{i}-2 \hat{j}+(7+\lambda) \hat{k}] \cdot[-4 \hat{i}+(7-\lambda) \hat{k}] } =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & -24+49-\lambda^2 =0 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \lambda^2 =25 \end{aligned}\)
\(\begin{aligned} \Rightarrow & \lambda = \pm 5 \end{aligned}\)

To prove, \((2 \vec{a}+\vec{b}) \perp \vec{b}\)
Given, \(|\vec{a}+\vec{b}|=|\vec{a}|\)
 On squaring both sides, we get
\(\begin{aligned} |\vec{a}+\vec{b}|^2=|\vec{a}|^2 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad & |\vec{a}|^2+2 \vec{a} \cdot \vec{b}+|\vec{b}|^2=|\vec{a}|^2 \end{aligned}\)
\(\Rightarrow \quad 2 \vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{b}=0 \quad\left[\because|\vec{x}|^2=\vec{x} \cdot \vec{x}\right]\)
\(\begin{array}{ll} \Rightarrow \quad(2 \vec{a}+\vec{b}) \cdot \vec{b}=0 \end{array}\)
\(\begin{array}{ll} \therefore \quad(2 \vec{a}+\vec{b}) \perp \vec{b} \end{array}\)
\(\begin{array}{ll} {[\because \text { if } \vec{a} \perp \vec{b} \Leftrightarrow \vec{a} \cdot \vec{b}=0]} \end{array}\)
Hence proved.

\(2\hat { i } +\hat { j } -4\hat { k } =(-\frac { 3 }{ 5 } \hat { i } -\frac { 4 }{ 5 } \hat { j } -\hat { k } )+(\frac { 13 }{ 5 } \hat { i } +\frac { 9 }{ 5 } \hat { j } -3\hat { k } )\)

\((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})=6 a^{2}+11 \vec{a} \cdot \vec{b}-35 \vec{b}^{2} \)
\(=24+11-35=0 \quad[\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}] \)

We have \(\overrightarrow { a } =3\hat { i } +2\hat { j } +2\hat { k } \quad and\quad \overrightarrow { b } =\hat { i } +2\hat { j } -2\hat { k } \)
\(\overrightarrow { a } +\overrightarrow { b } =\left( 3\hat { i } +2\hat { j } +2\hat { k } \right) +\left( \hat { i } +2\hat { j } -2\hat { k } \right) \)
\(=\overset { \wedge }{ 4i } +\overset { \wedge }{ 4j } \)
\(\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } =\left( \overset { \wedge }{ 3i } +\overset { \wedge }{ 2j } +\overset { \wedge }{ 2k } \right) -\left( \overset { \wedge }{ i } +\overset { \wedge }{ 2j } -\overset { \wedge }{ 2k } \right) \)
\(=\overset { \wedge }{ 2i } +k\)
\(\therefore \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) x\left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) =\left| \begin{matrix} \overset { \wedge }{ i } & \overset { \wedge }{ j } & \overset { \wedge }{ k } \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{matrix} \right| \)
\(=\left( 16-0 \right) \overset { \wedge }{ i } -\left( 16-0 \right) \overset { \wedge }{ j } +\left( 0-8 \right) \overset { \wedge }{ k } \)
\(=16\overset { \wedge }{ i } -16\overset { \wedge }{ j } -8\overset { \wedge }{ k } \)
\(\therefore \) the unit vector perpendicular to both \(\left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) \) and\(\left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) \) is given by:
\(\overset { \wedge }{ n } =\pm \frac { \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) x\left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) }{ \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) x\left( \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \right) } \)
\(=\pm \frac { \overset { \wedge }{ 16i } -\overset { \wedge }{ 16j } -\overset { \wedge }{ 16k } }{ \sqrt { { (16) }^{ 2 }+{ (-16) }^{ 2 }+{ (-8) }^{ 2 } } } \)
\(=\pm \frac { 8\left( \overset { \wedge }{ 2i } -\overset { \wedge }{ j } -\overset { \wedge }{ k } \right) }{ 8\sqrt { 4+4+1 } } =\pm \frac { \overset { \wedge }{ 2i } -\overset { \wedge }{ j } -\overset { \wedge }{ k } }{ 3 } \)
\(=\pm \frac { 2 }{ 3 } \overset { \wedge }{ i } \mp \frac { 1 }{ 3 } \overset { \wedge }{ j } \mp \frac { 1 }{ 3 } \overset { \wedge }{ k } \)