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

Given, points are \(A(-2 \hat{i}+3 \hat{j}+5 \hat{k}), B(\hat{i}+2 \hat{j}+3 \hat{k})\) and \(C(7 \hat{i}-\hat{k})\).
Here, \(\overrightarrow{A B}=\vec{b}-\vec{a}=(\hat{i}+2 \hat{j}+3 \hat{k})-(-2 \hat{i}+3 \hat{j}+5 \hat{k})\)
\(=3 \hat{i}-\hat{j}-2 \hat{k}\)
and \(\begin{aligned} \overrightarrow{B C} & =\vec{c}-\vec{b}=(7 \hat{i}-\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k}) \end{aligned}\)
\(\begin{aligned} =6 \hat{i}-2 \hat{j}-4 \hat{k}=2(3 \hat{i}-\hat{j}-2 \hat{k}) \end{aligned}\)
Since, \(\overrightarrow{A B}=\lambda \overrightarrow{B C},\) where \(\lambda=2\)
So, the given points are collinear.

Let the position vectors of vertices A,B and C be \(\overset { \wedge }{ 2i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } ,\overset { \wedge }{ i } -\overset { \wedge }{ 3j } -\overset { \wedge }{ 5k } \quad and\quad \overset { \wedge }{ 3i } -\overset { \wedge }{ 4j } -\overset { \wedge }{ 4k } \) respectively
 \(\therefore \overset { \rightarrow }{ AB } =\overset { \rightarrow }{ OB } -\overset { \rightarrow }{ OA }\)
\(= \left( \overset { \wedge }{ i } -\overset { \wedge }{ 3j } -\overset { \wedge }{ 5k } \right) -\left( \overset { \wedge }{ 2i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right)\)
\(=\overset { \wedge }{ -i } -\overset { \wedge }{ 2j } -\overset { \wedge }{ 6k } \)
\( \therefore AB=\overset { \rightarrow }{ |AB| } \)
\(=\sqrt { { (-1) }^{ 2 }+{ (-2) }^{ 2 }+{ (-6) }^{ 2 } } \)
\(=\sqrt { 1+4+36 } =\sqrt { 41 } \)
\(\overset { \rightarrow }{ BC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OB } \) 
\(=\left( \overset { \wedge }{ 3i } -\overset { \wedge }{ 3j } -\overset { \wedge }{ 4k } \right) -\left( \overset { \wedge }{ i } -\overset { \wedge }{ 3j } -\overset { \wedge }{ 5k } \right) \)    
\(=\overset { \wedge }{ 2i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \) 
\(\therefore\)   \(BC=\overset { \rightarrow }{ |BC| } =\sqrt { { 2 }^{ 2 }+{ (-1) }^{ 2 }+{ 1 }^{ 2 } } \)
\(=\sqrt { 4+1+1 } \quad \sqrt { 6 } \) 
and  \(\overset { \rightarrow }{ AC } =\overset { \rightarrow }{ OC } -\overset { \rightarrow }{ OA } \)       
\(=\left( \overset { \wedge }{ 3i } -\overset { \wedge }{ 3j } -\overset { \wedge }{ 4k } \right) -\left( \overset { \wedge }{ 2i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) \)
\(=\overset { \wedge }{ i } -\overset { \wedge }{ 3j } -\overset { \wedge }{ 5k } \) 
\(\therefore \quad AC=\overset { \rightarrow }{ |AC| } =\sqrt { { 1 }^{ 2 }+{ (-3) }^{ 2 }+{ (-5) }^{ 2 } } \)        
\(=\sqrt { 1+9+25 } =\sqrt { 35 } \) 
Now \({ BC }^{ 2 }+{ AC }^{ 2 }=6+35=35=41=AB\) 
Hence \(\triangle\)ABC is a right angled triangle.

The feasible region determined by the system of constraints, 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, and y ≥ 0, are as follows.

The corner points of the feasible region are O (0, 0), A (2, 0), B (0, 3), and C\(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \)
The values of Z at these corner points are as follows.

Therefore, the maximum value of Z is \(\frac { 235 }{ 19 } \) at \(\left( \frac { 20 }{ 19 } ,\frac { 45 }{ 19 } \right) \).

First of all, let us graph the feasible region of the system of inequalities (2) to (5). The feasible region (shaded). Observe that the feasible region is unbounded.
We now evaluate Z at the corner points.

From this table, we find that – 300 is the smallest value of Z at the corner point (6, 0). Can we say that minimum value of Z is – 300? Note that if the region would have been bounded, this smallest value of Z is the minimum value of Z (Theorem 2). But here we see that the feasible region is unbounded. Therefore, – 300 may or may not be the minimum value of Z. To decide this issue, we graph the inequality
– 50x + 20y < – 300 (see Step 3(ii) of corner Point Method.)
i.e., – 5x + 2y < – 30
and check whether the resulting open half plane has points in common with feasible region or not. If it has common points, then –300 will not be the minimum value of Z.
Otherwise, –300 will be the minimum value of Z. 
It has common points. Therefore, Z = –50 x + 20 y has no minimum value subject to the given constraints.
In the above example, can you say whether z = – 50 x + 20 y has the maximum value 100 at (0, 5)? For this, check whether the graph of – 50 x + 20 y > 100 has points in common with the feasible region.

\(\frac{3}{\sqrt{22}} \hat{i}-\frac{3}{\sqrt{22}} \hat{j}+\frac{2}{\sqrt{22}} \hat{k}\)

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 \(C=\frac{8\overset\rightarrow a-3\overset\rightarrow b}{5}\)

\(\left|\overset\rightarrow a+\overset\rightarrow b \right|^{ 2 } +\left|\overset\rightarrow a-\overset\rightarrow b \right|^{ 2 }=2{(\left| \overset\rightarrow a \right| ^{2 }+\left|\overset\rightarrow b \right| ^{ 2 })}\)
\(\Rightarrow\left| \overset\rightarrow b \right| ^{ 2 }=2,116\)
\(\Rightarrow\left| \overset\rightarrow b \right| =46\)

\((2\overset\rightarrow a-3\overset\rightarrow b).(3 \overset\rightarrow a+\overset\rightarrow b).=6\overset\rightarrow a.\overset\rightarrow a+2\overset\rightarrow a.\overset\rightarrow b-9\overset\rightarrow b.\overset\rightarrow a-3\overset\rightarrow b.\overset\rightarrow b\)
\(=6\left| \overset { \rightarrow }{ a } \right| ^{ 2}-(\overset\rightarrow a.\overset\rightarrow b)-3\left| \overset { \rightarrow }{ b } \right| ^{ 2}\)
\(=6(3)^{ 2}-7(2)-3(1)^{2 }\)
= 6(9) - 14 - 3
= 54 - 17 = 37

Let A(0, 3, 2), B(3, 5, 6)
Direction ratios of AB(a, b, c) are (3 - 0), (5 - 3), (6 - 2)
\((a_{ 1 },b_{ 1 },c_{ 1 })\) = (3, 2, 4)
Let C(1, - 1, 2), 0(3, 4, - 2)
Direction ratios of CD \((a_{ 2 },b_{ 2 },c_{ 2 })\) are (3 - 1), (4 + 1) (-2,2) \((a_{ 2 },b_{ 2 },c_{ 2 })\) are (2,56,-4)
When two lines are perpendicular if
\(a_{ 1 },a_{ 2 }+b_{ 1 },b_{ 2 },+c_{ 1 },c_{ 2 }=0\)
\(\Rightarrow 3\times 2+2\times 5+4\times -4=0\)
\(\Rightarrow 6+10-16=0\)
\(\Rightarrow 16-16=0\)
\(\therefore AB\bot to\quad CD\)

\(cos\theta =\left| \frac { a_{ 1 }a_{ 2 }+b_{ 1 }b_{ 2 }+c_{ 1 }c_{ 2 } }{ \sqrt { { a }_{ 1 }^{ 2 }+b_{ 1 }^{ 2 }+{ c }_{ 1 }^{ 2 } } \sqrt { { a }_{ 2 }^{ 2 }+b_{ 2 }^{ 2 }+c_{ 2 }^{ 2 } } } \right| \)
\(a_{ 1 }=7,b_{ 1 }=2,c_{ 1 }=6\)
\(a_{ 2 }=3,b_{ 2 }=-1,c_{ 2 }=-10\)
\(cos\theta =\left| \frac { 7\times 3+2\times -1+6\times -10 }{ \sqrt { 49+4+36 } \sqrt { 9+1+100 } } \right| \)
\(=\left| \frac { 21-2-60 }{ \sqrt { 89 } \sqrt { 110 } } \right| \)
\(cos\theta =\frac { 41 }{ \sqrt { 9790 } } \)
\(\Rightarrow \theta =cos^{ -1 }\left( \frac { 41 }{ \sqrt { 9790 } } \right) \)

Let the number of large vans = x
and number of small vans = y
Minimise cost z = 400x + 200y
subject to constraints
\(200 x+80 y \geq 1200 \text { or } 5 x+2 y \geq 30\) ,
\(400 x+200 y \leq 3000 \text { or } 2 x+y \leq 15\)
\(x 

The equation of the line passing through point (3, -2, 1) and parallel to the given line is \(\frac { x-3 }{ 2 } =\frac { y+2 }{ -3 } =\frac { z-1 }{ 1 } =\lambda\)
Any point on this line, we have (2\(\lambda \)+3, -3\(\lambda \)-2, \(\lambda \)+1)
If it lies on the plane, we have 3(2\(\lambda \)+3)-3\(\lambda \)-2-\(\lambda \)-1+2 = 0
\(\Rightarrow \lambda =-4\)
Hence, the point common to the plane and the line is (-5, 10, -3).
Hence the required distance = \(\sqrt { { \left( 3+5 \right) }^{ 2 }+{ \left( -2-10 \right) }^{ 2 }+{ { \left( 1+3 \right) } }^{ 2 } } =4\sqrt { 14 } units\)
The equation of the line passing through (3, -2, 1) and perpendicular to the plane is \(\frac { x-3 }{ 3 } =\frac { y+2 }{ 1 } =\frac { z-1 }{ -1 } =\mu\)
Any point on it is (3\(\mu\)+3, \(\mu\)-2, -\(\mu\) +1)
If it lies on the plane, we get 3(3\(\mu\)+3)+\(\mu\) -2+ \(\mu\)-1+2 = 0
\(\Rightarrow\) \(\mu=\frac{8}{11}\)
The required foot of the perpendicular is \(\left( \frac { 9 }{ 11 } ,\ \frac { -30 }{ 11 } ,\ \frac { 19 }{ 11 } \right) \).

Given inequations are
\(2 x+4 y \leq 8 \text { or } \ x+2 y \leq 4 \)
\(3 x+y \leq 6, x+y \leq 4, \ x \geq 0, \ y \geq 0 \)
Maximise Z = 2x + 5y on plotting the graph of the inequations we notice shaded portion as feasible solution

Possible points for maximum Z are A(2, 0),\(B\left(\frac{8}{5}, \frac{6}{5}\right)\) C (0,2)

Z is maximum at qo, 2), i.e. x = 0, y = 2, maximum value = 10

Let land allocated to crop a = x hectare
Let land allocated to crop b = y hectare
Obviously, x \(\ge \) 0, y \(\ge \) 0
Profit per hectare on a = Rs. 10,000
Profit per hectare on b = Rs. 9,500
Total profit = 10,00x + 9,500y
To maximise: Z=10,000x + 9,500y

Given: x + y \(\le \) 50   (related to land)   ...(i)
20x + 10y \(\le \)  800   (related to herbicide)
\(\Rightarrow\) 2x +  y\(\le \)  80..(ii)
x \(\ge \) 0, y\(\ge \)  0
Calculate value on O, A, C, D and note,
At point O(0, 0),
Z = 0
At point A(00, 50),
Z = 0 + 9500 \(\times\)50
= Rs. 4,75,000
At point D(40, 0),
Z = Rs. 4,00,000
At point C(30, 20),
Z = Rs. 4,90,000,
Which is maximum i.e., society will get maximum profit of Rs. 4,90,000 by allocationg 30 hectare to crop a and 20 hectare to crop b.
Value: Awareness would increase the profit of farmers.

The vector equations of given lines are
\(\begin{aligned} \vec{r}=3 \hat{i}+2 \hat{j}-4 \hat{k}+\lambda(\hat{i}+2 \hat{j}+2 \hat{k}) \end{aligned}\)
and \(\begin{aligned} \vec{r}=5 \hat{i}-2 \hat{j}+\mu(3 \hat{i}+2 \hat{j}+6 \hat{k}) \end{aligned}\)
On comparing them with \(\vec{r}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1}\) and \(\vec{r}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}\), we get
\(\overrightarrow{a_1}=3 \hat{i}+2 \hat{j}-4 \hat{k}, \overrightarrow{a_2}=5 \hat{i}-2 \hat{j}, \vec{b}_1=\hat{i}+2 \hat{j}+2 \hat{k}\)
and \(\vec{b}_2=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
\(\therefore \quad \overrightarrow{a_2}-\overrightarrow{a_1}=(5 \hat{i}-2 \hat{j})-(3 \hat{i}+2 \hat{j}-4 \hat{k})\)
\(=2 \hat{i}-4 \hat{j}+4 \hat{k}\)
\(\begin{aligned} \therefore \quad \vec{b}_1 \times \vec{b}_2 & =\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 2 \\ 3 & 2 & 6 \end{array}\right| \end{aligned}\)
\(\begin{aligned} =\hat{i}(12-4)-\hat{j}(6-6)+\hat{k}(2-6) \end{aligned}\)
\(\begin{aligned} =8 \hat{i}-4 \hat{k} \end{aligned}\)
\(\therefore\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)=(2 \hat{i}-4 \hat{j}+4 \hat{k}) \cdot(8 \hat{i}-4 \hat{k})\)
= 16 + 0 - 16 = 0
\(\therefore\) The lines are intersecting and the shortest distance between the lines is 0.
Now, the position vectors of arbitrary points on the given lines are \((3+\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(-4+2 \lambda) \hat{k}\) and \((5+3 \mu) \hat{i}+(-2+2 \mu) \hat{j}+6 \mu \hat{k}\), respectively.
Since, lines intersect then they have a common point.
\(\therefore \begin{aligned} 3+\lambda & =5+3 \mu \end{aligned}\)         ...(i)
\(\begin{aligned} 2+2 \lambda & =-2+2 \mu \end{aligned}\)        ...(ii)
\(\begin{aligned} -4+2 \lambda & =6 \mu \end{aligned}\)               ...(iii)
On solving Eqs. (i) and (ii), we get
\(\lambda=-4 \text { and } \mu=-2\)
\(\therefore\) Point of intersection is (3 - 4, 2 - 8, -4 - 8)
i.e. (-1, -6, -12). 

We have, \(\vec{a}=\hat{i}+2 \hat{\jmath}+3 \hat{k} \text { and } \vec{b}=2 \hat{i}+4 \hat{\jmath}-5 \hat{k}\)
So, the diagonals of the parallelogram whose adjacent sides are \(\vec{a} \text { and } \vec{b}\) are given by
\(\begin{aligned} \vec{p} & =\vec{a}+\vec{b} \text { and } \vec{q}=\vec{a}-\vec{b} \end{aligned}\)
Now, \(\begin{aligned} \vec{p} & =(\hat{i}+2 \hat{j}+3 \hat{k})+(2 \hat{i}+4 \hat{j}-5 \hat{k}) \end{aligned}\)
\(\begin{aligned} =3 \hat{i}+6 \hat{j}-2 \hat{k} \end{aligned}\)
and \(\begin{aligned} \vec{q} & =(\hat{i}+2 \hat{j}+3 \hat{k})-(2 \hat{i}+4 \hat{j}-5 \hat{k}) \end{aligned}\)
\(\begin{aligned} =-\hat{i}-2 \hat{j}+8 \hat{k} \end{aligned}\)
\(\begin{aligned} \therefore \quad \hat{p} & =\frac{\vec{p}}{|\vec{p}|}=\frac{3 \hat{i}+6 \hat{j}-2 \hat{k}}{\sqrt{9+36+4}} \end{aligned}\)
\(\begin{aligned} =\frac{3 \hat{i}+6 \hat{j}-2 \hat{k}}{7}=\frac{3}{7} \hat{i}+\frac{6}{7} \hat{j}-\frac{2}{7} \hat{k} \end{aligned}\)
and \(\begin{aligned} \hat{q} & =\frac{\vec{q}}{|\vec{q}|}=\frac{-\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{1+4+64}}=\frac{-\hat{i}-2 \hat{j}+8 \hat{k}}{\sqrt{69}} \end{aligned}\)
\(\begin{aligned} =\frac{-1}{\sqrt{69}} \hat{i}-\frac{2}{\sqrt{69}} \hat{j}+\frac{8}{\sqrt{69}} \hat{k} \end{aligned}\)

As \({ \left| \sqrt { 3 } \vec { a } -\vec { b } \right| }^{ 2 }=({ \sqrt { 3 } \vec { a } -\vec { b } ) }^{ 2 }\)
\(=3\vec { { a }^{ 2 } } +\vec { { b }^{ 2 } } -2\sqrt { 3 } \vec { a } .\vec { b } \)
\(1=3+1-2\sqrt { 3 } \vec { a } .\vec { b } \)
\(\Rightarrow \vec { a } .\vec { b } =\frac { \sqrt { 3 } }{ 2 } \)
\(\therefore cos\theta =\frac { \vec { a } .\vec { b } }{ |\vec { b } ||\vec { b } | } =\frac { \sqrt { 3 } }{ 2 } \)\(\Rightarrow \theta ={ 30 }^{ 0 }\)

As normal form of plane is
\(\overrightarrow { r } .(-\frac { 2 }{ 7 } \widehat { i } -\frac { 3 }{ 7 } \widehat { j } +\frac { 9 }{ 7 } \widehat { k } )\) = \(\frac27\)
∴ distance = \(\frac27\)
∴ p = \(\frac27\)

(i) (b) : We have, \(\overrightarrow{O A}=8 \hat{i} \text { and } \overrightarrow{A B}=6 \hat{j}\)
\(\therefore \overrightarrow{O B}=\overrightarrow{O A}+\overrightarrow{A B}=8 \hat{i}+6 \hat{j}\)
(ii) (a) : To reach school Geetika travels
=(8+6) km=14km
(iii) (b): Vector distance from school to Alok's house
\( =6 \cos 30^{\circ} \hat{i}+6 \sin 30^{\circ} \hat{j} \)
\(=6 \times \frac{\sqrt{3}}{2} \hat{i}+6 \times \frac{1}{2} \hat{j}=3 \sqrt{3} \hat{i}+3 \hat{j}\)
(iv) (a): Vector distance from Geetika's house to
Alok'shouse =\(8 \hat{i}+6 \hat{j}+3 \sqrt{3} \hat{i}+3 \hat{j}=(8+3 \sqrt{3}) \hat{i}+9 \hat{j}\)
(v) (b): Total distance travelled by Geetika from her house to Alok's house = (8 + 6 + 6) km = 20 km.

(i) (b): Clearly, the coordinates of A are (8, -6, 0) and that of E are (0, 0, 24).
Also, cartesian equation of line along EA is given by
\(\frac{x-0}{8-0}=\frac{y-0}{-6-0}=\frac{z-24}{0-24}\)
\(\Rightarrow \frac{x}{8}=\frac{y}{-6}=\frac{z-24}{-24} \Rightarrow \frac{x}{-4}=\frac{y}{3}=\frac{z-24}{12}\)
(ii) (c): Clearly, the coordinates of Dare (-8, -6, 0) and that of E are (0, 0, 24)
\(\therefore \text { Vector } \overline{E D} \text { is }(-8-0) \hat{i}+(-6-0) \hat{j}+(0-24) \hat{k} \ \text { i.e., }-8 \hat{i}-6 \hat{i}-24 \hat{k} \)
(iii) (b): Since, the coordinates of Bare (8, 6, 0) and that of E are (0, 0, 24), therefore length of cable
\(E B=\sqrt{(8-0)^{2}+(6-0)^{2}+(0-24)^{2}} \)
\(=\sqrt{64+36+576}=\sqrt{676}=26 \text { units }\)
(iv) (d): Since, the coordinates of Care (-8,6,0) therefore length of cable EC = \(\sqrt{(-8-0)^{2}+(6-0)^{2}+(0-24)^{2}}\)
\(=\sqrt{64+36+576}=\sqrt{676}=26 \mathrm{units}\)
Similarly, length of cable EA = ED = 26 units.
(v) (c): Sum of all vectors along the cables
\(=\overrightarrow{E A}+\overrightarrow{E B}+\overrightarrow{E C}+\overline{E D}\)
\( =(8 \hat{i}-6 \hat{j}-24 \hat{k})+(8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}+6 \hat{j}-24 \hat{k})+(-8 \hat{i}-6 \hat{j}-24 \hat{k}) \)
\(=-96 \hat{k}\)

(b) Assertion is correct, Reason is correct; Reason is not a correct explanation for Assertion

(a) Assertion The given lines are \(\vec{r}=\vec{a}_1+\lambda \overrightarrow{b_1}\) and \(\vec{r}=\vec{a}_2+\lambda \vec{b}_2\)
Let \(\theta\) be the angle between these lines
\(\therefore \cos \theta=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|| \vec{b}_2 \mid}\)
When \(\overrightarrow{b_1} \cdot \overrightarrow{b_2}=0\)
Then, \(\cos \theta=\frac{0}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}=0\)
\(\Rightarrow \quad \cos \theta=\cos \frac{\pi}{2} \Rightarrow \theta=\frac{\pi}{2}\)
Hence, the given lines are perpendicular.
Both Assertion and Reason are true and Reason is a correct explanation of Assertion.