\(\left( \hat { k } \times \hat { i } \right) .\hat { i } +\hat { j } .\hat { k } =\left( -\hat { j } \right) .\hat { i } +0\)
\(=-\hat { i } .\hat { i } +0\)
= -1 + 0 = -1

\((\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)=\overset\rightarrow a.\overset\rightarrow a+\overset\rightarrow a.\overset\rightarrow b-\overset\rightarrow b.\overset\rightarrow a-\overset\rightarrow b.\overset\rightarrow b\)
\(=\left| \overrightarrow {a } \right| ^{2 }-\left| \overrightarrow {b } \right| ^{ 2 }\)
\((\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)=\)\((8\left| \overrightarrow { b } \right| )^{ 2 }-\left| \overrightarrow {b } \right| ^{ 2}\)
\(=64(\left| \overrightarrow { b } \right| )^{ 2 }-\left| \overrightarrow {b } \right| ^{ 2}\)
\(=63\left| \overrightarrow { b } \right| ^{ 2 }\)
\((\overset\rightarrow a+\overset\rightarrow b).(\overset\rightarrow a-\overset\rightarrow b)=8\)
\(\Rightarrow \)\(63\left| \overrightarrow { b } \right| ^{ 2 }=8\)
\(\Rightarrow \left| \overrightarrow { b } \right|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}\)
\(\therefore \left| \overrightarrow { a } \right| =\frac{8\times 2\sqrt{7}}{3\sqrt{7}}=\frac{16\sqrt{2}}{3\sqrt{7}}\)

\(\overrightarrow a.\overrightarrow b =0\)
\(\left| \overrightarrow { a } \right| =0 or \left| \overrightarrow {b } \right| =0\)
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.

\((\overset\wedge k\times \overset\wedge j).\overset\wedge k i+\overset\wedge i+\overset\wedge j.\overset\wedge k=(-\overset\wedge j).\overset\wedge i+0\)
\(=-\overset\wedge i.\overset\wedge i+0\)
= -1 + 0 = -1

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

\(\overset\rightarrow c=\overset\rightarrow a-\overset\rightarrow b\)
\(=(3\overset\wedge i-2\overset\wedge j+3\overset\wedge k)-(\overset\wedge i-4\overset\wedge j-\overset\wedge k)\)
\(\Rightarrow \overset\rightarrow c=2 \overset\wedge i+2\overset\wedge j+4\overset\wedge k\)
\(\overset\wedge c=\frac{\overset\rightarrow c}{\left| c \right| }=\frac{2\overset\wedge i+2\overset\wedge j+4\overset\wedge k}{\sqrt{4+4+16}}\)
\(=\frac{2\overset\wedge i+2\overset\wedge j+4\overset\wedge k}{\sqrt{24}}\)
\(=\frac{2(\overset\wedge i+\overset\wedge j+2\overset\wedge k)}{2\sqrt6}\)
\(\overset\wedge c=\frac{\overset\wedge i+\overset\wedge j+2\overset\wedge k}{\sqrt{6}}\)
\(\Rightarrow\)Required vector =\(9\overset\wedge c=\frac{9\overset\wedge i+9\overset\wedge j+18\overset\wedge k}{\sqrt6}\)

\(\overset\rightarrow {AB}=\overset\rightarrow {OB}-\overset\rightarrow {OA}\).....(i)
and \(\overset\rightarrow {BC}=\overset\rightarrow {OC}-\overset\rightarrow {OB}\)....(ii)
\(\therefore \overset\rightarrow {CA}=\overset\rightarrow {OA}-\overset\rightarrow {OC}\)....(iii)
By adding eqn (i),(ii) and (iii)
\(\overset\rightarrow {AB}+\overset\rightarrow {BC}+\overset\rightarrow {CA}=0\)

\(\overset\rightarrow {AB} \) = Position vector of B-Position vector of A
= (−5\(\hat{i}\) +7\(\hat{j}\)) − (2\(\hat{i}\)+1\(\hat{j}\))
= (−5−2)\(\hat{i}\)+ (7−1)\(\hat{j}\)
= −7\(\hat{i}\)+ 6\(\hat{j}\)
∴ The scalar components are (-7,6,0).

Sum of given two vectors is given as
\( (\hat{i}+\hat{j}+\hat{k})+(2 \hat{i}-3 \hat{j}+5 \hat{k})=(1+2) \hat{i}+(1-3) \hat{j}+(1+5) \hat{k} \)
\( =3 \hat{i}-2 \hat{j}+6 \hat{k}=\vec{A}\)
A unit vector parallel to this vector
\( =\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{|\vec{A}|} \)
\( =\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{\sqrt{3^2+(-2)^2+6^2}}=\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{\sqrt{49}} \)
\( =\frac{3}{7} \hat{i}-\frac{2}{7} \hat{j}+\frac{6}{7} \hat{k}\)

\(10\over\sqrt6\)

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

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

\(\frac{3}{1}=\frac{2}{p}=\frac{9}{3} \Rightarrow p=\frac{2}{3}\)

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

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

Let \(\vec{c}=\hat{a}+\hat{b}\). 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.

Part (C) is the correct answer
here \(\overset { \rightarrow }{ AB } \)
\(\left( \overset { \wedge }{ i } +\frac { 1 }{ 2 } \overset { \wedge }{ j } +\overset { \wedge }{ 4k } \right) \quad -\quad \left( -\overset { \wedge }{ i } +\frac { 1 }{ 2 } \overset { \wedge }{ j } +\overset { \wedge }{ 4k } \right) \)
\(\therefore \overset { \rightarrow }{ |AB| } =2\)
and \(\overset { \rightarrow }{ AD } =\left( \overset { \wedge }{ -i } -\frac { 1 }{ 2 } \overset { \wedge }{ j } +\overset { \wedge }{ 4k } \right) \quad -\quad \left( -\overset { \wedge }{ i } +\frac { 1 }{ 2 } \overset { \wedge }{ j } +\overset { \wedge }{ 4k } \right) =-\overset { \wedge }{ j } \) 
\(\therefore \overset { \rightarrow }{ |AD| } = 1\)
\(\therefore \) Area of rect(ABCD) = \(\overset { \rightarrow }{ |AB } |\overset { \rightarrow }{ |AD } \) 
= (2)(1) = 2.

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, \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =0\text{ through}\ \overset { \rightarrow }{ a } \neq \overset { \rightarrow }{ 0 } ,\ \overset { \rightarrow }{ b } \neq \overset { \rightarrow }{ 0 } \).

\(\text { Let } \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \)
\(\therefore|\vec{a}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{1+4+9}=\sqrt{14} \)
\(\text { Hence, the direction cosines of }\vec{a} \text { are }\left(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right)\) 

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

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

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