To find projection of \((\vec{b}+\vec{c}) \text { on } \vec{a}\)
Given, \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Consider, \(\begin{aligned} (\vec{b}+\vec{c}) & =(\hat{i}+2 \hat{j}-2 \hat{k})+(2 \hat{i}-\hat{j}+4 \hat{k}) \end{aligned}\)
\(\begin{aligned} =3 \hat{i}+\hat{j}+2 \hat{k} \end{aligned}\)
Now, the projection of \(\vec{b}+\vec{c} \text { on } \vec{a}\) is given by
\(\begin{aligned} \frac{(\vec{b}+\vec{c}) \vec{a}}{|\vec{a}|}=\frac{(3 \hat{i}+\hat{j}+2 \hat{k})(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{2^2+(-2)^2+1^2}} \end{aligned}\)
\(\begin{aligned} \frac{6-2+2}{\sqrt{4+4+1}}=\frac{6}{\sqrt{9}}=\frac{6}{3}=2 \end{aligned}\)

Given, \(\vec{a}, \vec{b}, \vec{c}\)are three non-zero unequal vectors such that \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\)
\(\Rightarrow \quad \vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0\)
                                    [subtracting \(\vec{a} \cdot \vec{c}\) from both sides]
\(\Rightarrow \quad \vec{a} \cdot(\vec{b}-\vec{c})=0\)                     ...(i)
Now, let \(\theta\) be the angle between \(\vec{a} \text { and } \vec{b}-\vec{c}\).
\(\therefore \quad \cos \theta=\frac{\vec{a} \cdot(\vec{b}-\vec{c})}{|\vec{a}||\vec{b}-\vec{c}|}\)
\(\Rightarrow \quad \cos \theta=\frac{0}{|\vec{a}||\vec{a}-\vec{c}|}\)        [using Eq. (i)]
\(\begin{array}{ll} \Rightarrow \quad & \cos \theta=0 \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2} \end{array}\)

Given, the adjacent sides of a parallelogram are \(\vec{a}=\hat{i}-\hat{j}+3 \hat{k} \text { and } \vec{b}=2 \hat{i}-7 \hat{j}+\hat{k}\)
\(\therefore \text { Area of parallelogram }=|\vec{a} \times \vec{b}|\)
Now, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{array}\right|\)
\(\begin{aligned} =\hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2) \end{aligned}\)
\(\begin{aligned} =20 \hat{i}+5 \hat{j}-5 \hat{k} \end{aligned}\)
\(\begin{aligned} \therefore|\vec{a} \times \vec{b}| & =\sqrt{(20)^2+(5)^2+(-5)^2} \end{aligned}\)
\(\begin{aligned} =\sqrt{400+25+25}=\sqrt{450}=15 \sqrt{2} \end{aligned}\)
Hence, the area of parallelogram is \(15 \sqrt{2}\) sq units.

Let \(\theta\) be the angle between \(\vec{a} \text { and } \vec{b}\).
Given, \(|\vec{a}|=8,|\vec{b}|=3 \text { and }|\vec{a} \times \vec{b}|=12\)
We know that \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta\)
\(\begin{aligned} \therefore \quad |\vec{a}||\vec{b}| \sin \theta =12 \end{aligned}\)
\(\begin{array}{ll} \Rightarrow & \sin \theta=\frac{12}{|\vec{a}||\vec{b}|} \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & \sin \theta=\frac{12}{8 \times 3} \end{array}\)
\(\begin{array}{ll} \Rightarrow \quad & \sin \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{6} \end{array}\)
Hence, the required angle between \(\vec{a} \text { and } \vec{b} \text { is } \frac{\pi}{6}\)

We have, \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}  \)
\(\Rightarrow\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{array}\right|=\overrightarrow{0}  \)
\(\Rightarrow(6 \mu-27 \lambda) \hat{i}-(2 \mu-27) \hat{j}+(2 \lambda-6) \hat{k}=0 \hat{i}+0 \hat{j}+0 \hat{k} \)
On equating the coefficients of \(\hat{i}, \hat{j} \) and \(\hat{k}\)both sides we get
\(6 \mu-27 \lambda=0,2 \mu-27=0 \text { and } 2 \lambda-6=0  \)
\(\Rightarrow \lambda=3 \text { and } \mu=\frac{27}{2}\)

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

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

\(-\vec{a}+4 \vec{b}\)

a = -15
a = 2/3

\(\text { Consider, }(2 \vec{x}-3 \vec{a}) \cdot(2 \vec{x}+3 \vec{a})=91 \)
\(\Rightarrow 4 \overrightarrow{x^{2}}-9 \overrightarrow{a^{2}}=91 \Rightarrow 4|\vec{x}|^{2}-9|\vec{a}|^{2}=91 \)
\(\Rightarrow 4|\vec{x}|^{2}-9(1)^{2}=91 \Rightarrow 4|\vec{x}|^{2}=100 \)
\(\Rightarrow|\vec{x}|^{2}=25 \Rightarrow|\vec{x}|=5 \)

The projection of \(\vec{a}\) on \(\vec{b} \vec{b}=\frac{\vec{a} \cdot \vec{b}}{\overrightarrow{|b|}}\) by the question \(\frac{(\lambda \hat{i}+\hat{j}+4 \hat{k}) \cdot(2 \hat{i}+6 \hat{j}+3 \hat{k})}{|2 \hat{i}+6 \hat{j}+3 \hat{k}|}=4\)
\( \Rightarrow \frac{(\lambda)(2)+(1)(6)+(4)(3)}{\sqrt{4+36+9}}=4 \)
\( \Rightarrow \frac{2 \lambda+6+12}{7}=4 \)
\( \Rightarrow 2 \lambda+18=28 \)
\( \Rightarrow 2 \lambda=10 .\)
Hence \(\lambda=5\)

For vectors to be collinear \(\frac{2}{a}=\frac{-3}{6}=\frac{4}{-8}\)
a = -4

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

\(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a} \| \vec{b}|}=\frac{\sqrt{3}}{2 \sqrt{3}}=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}\)

3,-4 and -6

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

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

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

For points A,B,C to be collinear,show that \(\overrightarrow{A B}=\lambda \overrightarrow{B C}\) \(\Rightarrow A B \| B C, B\) is common.
Hence, points are collinear

Let \(\vec{a}, \vec{b} \text { and } \vec{c}\) be the position vectors of points \(P, Q \text { and } R\) . respectively. Then, \(\vec{a}=-\hat{i}+2 \hat{j}-\hat{k}, \vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{c}=2 \hat{i}+3 \hat{j}-\hat{k}\) .
Clearly, the area of \(\Delta P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|\) 
Now,\(\overrightarrow{P Q}=\text { Position vector of } Q-\text { Position vector of } P\) 
\(=\vec{b}-\vec{a}=(3 \hat{i}-\hat{j}+2 \hat{k})-(-\hat{i}+2 \hat{j}-\hat{k})\) 
\(=4 \hat{i}-3 \hat{j}+3 \hat{k}\) 
\(\overrightarrow{P R}=\text { Position vector of } R \text { - Position vector of } P\) 
\(=\vec{c}-\vec{a}=(2 \hat{i}+3 \hat{j}-\hat{k})-(-\hat{i}+2 \hat{j}-\hat{k})=3 \hat{i}+\hat{j}\) 
\(\therefore \overrightarrow{P Q} \times \overrightarrow{P R}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & -3 & 3 \\ 3 & 1 & 0 \end{array}\right|=(0-3) \hat{i}-(0-9) \hat{j}+(4+9) \hat{k}\) 
\(=-3 \hat{i}+9 \hat{j}+13 \hat{k}\) 
and \(|\overrightarrow{P Q} \times \overrightarrow{P R}|=\sqrt{(-3)^{2}+(9)^{2}+(13)^{2}}\) 
\(=\sqrt{9+81+169}=\sqrt{259}\) 
So, area of \(\Delta P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|=\frac{1}{2} \sqrt{259}\)

\(\text {Let vector be } \vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)
According to given condition \( x-y+z=4 \text { ; }\)
\(2 x+y-3 z=0 ; x+y+z=2 \)
\(x-y+z=4 \)
\(2 x+y-3 z=0 \)
\(x+y+z=2 \)
From (ii), y = 3z − 2x
Substituting in (i) and (iii), we get
\(x-3 z+2 x+z=4 \Rightarrow 3 x-2 z=4 \)
\(x+3 z-2 x+z=2 \Rightarrow-x+4 z=2 \)
Solving (v) and (vi), we get
\(x=2, z=1\)
Substituting in (iv), we get
\(y=-1\)
\(\therefore \text { Vector is } 2 \hat{i}-\hat{j}+\hat{k} \)

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

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

(i) (c): Position vector of \(\overrightarrow{A B} \) 
\(=(2-1) \hat{i} \dot{+}(1-1) \hat{j}+(3-1) \hat{k}=\hat{i}+2 \hat{k}\)
(ii) (b): Position vector of  \(\overrightarrow{A C} \)
\(=(3-1) \hat{i}+(2-1) \hat{j}+(2-1) \hat{k}=2 \hat{i}+\hat{j}+\hat{k}\)
(iii) (d): Position vector of \(\overrightarrow{AD} \)
\(=(3-1) \hat{i}+(3-1) \hat{j}+(4-1) \hat{k}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)
(iv) (b): Area of \(\Delta A B C\) = \(\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{A C}|\)
\(\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 2 \\ 2 & 1 & 1 \end{array}\right|=\hat{i}(0-2)-\hat{j}(1-4)+\hat{k}(1-0)\)
\(=-2 \hat{i}+3 \hat{j}+\hat{k}\)
\( \Rightarrow |\overrightarrow{A B} \times \overrightarrow{A C}| =\sqrt{(-2)^{2}+3^{2}+1^{2}} \)
\(=\sqrt{4+9+1}=\sqrt{14}\)
Area of \(\Delta A B C\) \(=\frac{1}{2} \sqrt{14} \text { sq. units }\)
(v) (a): Unit vector along \(\overrightarrow{A D}=\frac{\overrightarrow{A D}}{|\overrightarrow{A D}|}\)
\(=\frac{2 \hat{i}+2 \hat{j}+3 k}{\sqrt{2^{2}+2^{2}+3^{2}}}=\frac{2 \hat{i}+2 \hat{j}+3 \hat{k}}{\sqrt{4+4+9}}=\frac{1}{\sqrt{17}}(2 \hat{i}+2 \hat{j}+3 \hat{k})\)

Given \(\vec{a} \perp(\vec{b}+\vec{c}), \vec{b} \perp(\vec{c}+\vec{a}), \vec{c} \perp(\vec{a}+\vec{b})\)
and \(|\vec{a}|=3,|\vec{b}|=4,|\vec{c}|=5\)
To prove \(|\vec{a}+\vec{b}+\vec{c}|=5 \sqrt{2}\)
Consider, \(\begin{array}{r} |\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c}) \end{array}\)
\(\begin{array}{r} {\left[\because|\vec{x}|^2=\vec{x} \cdot \vec{x}\right]} \end{array}\)
\(\begin{aligned} &=\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+\vec{c} \cdot \vec{c} \end{aligned}\)
\(\begin{aligned} &=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+\vec{a} \cdot(\vec{b}+\vec{c})+\vec{b} \cdot(\vec{a}+\vec{c})+\vec{c} \cdot(\vec{a}+\vec{b}) \end{aligned}\)
\(=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+0+0+0\)
                                                \(\left[\begin{array}{l} \because \vec{a} \perp(\vec{b}+\vec{c}) \text {, therefore } \\ \vec{a} \cdot(\vec{b}+\vec{c})=0 \\ \text { Similarly, } \vec{b} \cdot(\vec{a}+\vec{c})=0 \\ \text { and } \vec{c} \cdot(\vec{a}+\vec{b})=0 \end{array}\right]\)
= 3² + 4² + 5² = 9 + 16 + 25          [given]
\(\Rightarrow|\vec{a}+\vec{b}+\vec{c}|^2=50 \Rightarrow|\vec{a}+\vec{b}+\vec{c}|=5 \sqrt{2}\)
                                                       [length cannot be '-' ve]

We have,
\(\overrightarrow{A B}\) = (position vector of B) - (position vector of A)
\(=(\hat{i}-3 \hat{j}-5 \hat{k})-(2 \hat{i}-\hat{j}+\hat{k})=-\hat{i}-2 \hat{j}-6 \hat{k}\)
\(\begin{aligned} \overrightarrow{B C}=(3 \hat{i}-4 \hat{j}-4 \hat{k})-(\hat{i}-3 \hat{j}-5 \hat{k})=2 \hat{i}-\hat{j}+\hat{k} \end{aligned}\)
and \(\begin{aligned} \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k})-(3 \hat{i}-4 \hat{j}-4 \hat{k}) \end{aligned}\)
\(=-\hat{i}+3 \hat{j}+5 \hat{k}\)
Here, \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}=0\)
\(\Rightarrow\) A, B and C are the vertices of a triangle.
Now, \(\overrightarrow{B C} \cdot \overrightarrow{C A}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(-\hat{i}+3 \hat{j}+5 \hat{k})\)
= - 2 - 3 + 5 = 0
\(\Rightarrow \overrightarrow{B C} \perp \overrightarrow{C A} \Rightarrow \angle C=90^{\circ}\)

Now, area of \(\Delta A B C=\frac{1}{2}|\overrightarrow{C A} \times \overrightarrow{B C}|\)
\(=\frac{1}{2}\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ -1 & 3 & 5 \\ 2 & -1 & 1 \end{array}\right|=\frac{1}{2}|(8 \hat{i}-11 \hat{j}-5 \hat{k})|\)
\(=\frac{1}{2} \sqrt{64+121+25}\)
\(=\frac{1}{2} \sqrt{210}\) sq units