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

Given equation of line is
\(a\vec{r}=(-\hat{i}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})\) 
and equation of plane is \(\vec{r} \cdot(10 \hat{i}+2 \hat{j}-11 \hat{k})=3\) .
On comparing with \(\vec{r}=\vec{a}+\lambda \vec{b} \text { and } \vec{r} \cdot \vec{n}=d\) ,we get
\(\vec{b}=(2 \hat{i}+3 \hat{j}+6 \hat{k}) \text { and } \vec{n}=10 \hat{i}+2 \hat{j}-11 \hat{k}\) 
The angle \(\phi\) between the line and the plane is
\(\sin \phi=\left|\frac{\vec{b} \cdot \vec{n}}{|\vec{b}||\vec{n}|}\right|\) 
\(=\mid \frac{(2 \hat{i}+3 \hat{j}+6 \hat{k}) \cdot(10 \hat{i}+2 \hat{j}-11 \hat{k})}{\sqrt{(2)^{2}+(3)^{2}+(6)^{2}} \sqrt{(10)^{2}+(2)^{2}+(-11)^{2}}}\) 
\(=\mid \frac{(2)(10)+(3)(2)+(6)(-11)}{\sqrt{4+9+36} \sqrt{100+4+121}}\) 
\(=\left|\frac{20+6-66}{\sqrt{49} \sqrt{225}}\right|=\left|\frac{-40}{7 \times 15}\right|=\frac{8}{21}\) 
\(\therefore \phi=\sin ^{-1}\left(\frac{8}{21}\right)\)

Given, Z = -3x + 4y
Subject to the constraints
x + 2y \( \leq\) 8; 3x + 2y \( \leq\) 12 and x \(\geq\)0, y \(\geq\) 0
Now, considering the inequations as equations, we get
  x + 2y = 8                  ...(i)
3x + 2y = 12                ...(ii)
Table for line x + 2y = 8 is

On putting (0, 0) in the inequality x + 2y \( \leq\) 8
0 \( \leq\) 8  (which is true)
So, half plane is towards the origin.
Table for line 3x + 2y = 12

On putting (0, 0) in the inequality 3x + 2y \( \leq\)12
0 \( \leq\) 12 (which is true)
So, half plane is towards the origin.
Also, x \(\geq\) 0 and y \(\geq\) 0, so the feasible region lies in the Ist quadrant.
The point of intersection of Eqs. (i) and (ii) is (2, 3).
The graphical representation of the above system of inequations is given below.

Hence, Z = -12 is minimum at (4, 0).

The angle between the lines whose direction ratios are a, b, c and b-c,c-a, a-b  \(\)
\(\cos Q=\left|\frac{a(b-c)+b(c-a)+c(a-b)}{\sqrt{a^{2}+b^{2}+c^{2}}+\sqrt{(b-c)^{2}+(c-a)^{2}+(a-b)^{2}}}\right| \)
\(\Rightarrow \cos Q=0 \)
\(\Rightarrow Q=\cos ^{-1} 0 \)
\(\Rightarrow Q=90^{\circ} \)
Thus, the angle between the lines is \( 90^{\circ} \text {. }\)

Let E denote the event that a student chosen randomly studies in Class XII and F be the event that the randomly chosen student is a girl. We have to find P (E|F).
\(\text { Now } \mathrm{P}(\mathrm{F}) =\frac{430}{1000}=0.43 \text { and } \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{43}{1000}=0.043 \text { (Why?) }\)
\(\text { Then } \mathrm{P}(\mathrm{E} \mid \mathrm{F}) =\frac{\mathrm{P}(\mathrm{E} \cap \mathrm{F})}{\mathrm{P}(\mathrm{F})}=\frac{0.043}{0.43}=0.1 \)

Total cards are 12
A : number drawn is more than 3, i.e. 4, 5, 6, ..., 12.
B : getting an even number, i.e. 2, 4, 6, 8, 10, 12.
\(A \cap B: 4,6,8,10,12 \)
\(P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{5 / 12}{9 / 12}=\frac{5}{9} . \)

As area of parallelogram
= \(\left| \begin{matrix} \widehat { i } & \widehat { j } & \widehat { k } \\ 2 & 1 & -2 \\ 3 & 1 & -1 \end{matrix} \right| \)
= \(\left| \widehat { i } -4\widehat { j } -\widehat { k } \right| \)
= \(\sqrt { 1+16+1 } \)
= \(3\sqrt { 2 } \) sq units

Given, lines are
\(\begin{aligned} \frac{x-4}{3}=\frac{y+3}{4}=\frac{z+1}{5} \end{aligned}\)           ...(i)
and \(\begin{aligned} \frac{x-1}{4}=\frac{y+1}{-3}=\frac{z+10}{5} \end{aligned}\)   ...(ii)
DR's of line (i) is 3, 4, 5 and DR's of line (ii) is 4, -3, 5.
Vector in the direction of first line \(\vec{b}=3 \hat{i}+4 \hat{j}+5 \hat{k}\)
Vector in the direction of second line
\(\vec{d}=4 \hat{i}-3 \hat{j}+5 \hat{k}\)
Acute angle \(\theta\) between two lines is given by
\(\cos \theta=\frac{\overrightarrow{\mid b} \cdot \vec{d} \mid}{|\vec{b}||\vec{d}|}\)
\(\begin{aligned} \cos \theta & =\frac{|(3 \hat{i}+4 \hat{j}+5 \hat{k}) \cdot(4 \hat{i}-3 \hat{j}+5 \hat{k})|}{|(3 \hat{i}+4 \hat{j}+5 \hat{k})||(4 \hat{i}-3 \hat{j}+5 \hat{k})|} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \cos \theta & =\frac{|12-12+25|}{\sqrt{9+16+25} \sqrt{16+9+25}} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \cos \theta=\frac{25}{\sqrt{50} \sqrt{50}} \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \end{aligned}\)

(i) Collinear vectors : \(\vec{a}, \vec{c} \text { and } \vec{d}\)
(ii) Equal vectors : \(\vec{a} \text { and } \vec{c} \text { . }\)
(iii) Coinitial vectors : \(\vec{b}, \vec{c} \text { and } \vec{d}\)

Let x necklaces and y bracelets manufactured by small firm.
\(\begin{array}{c|c|c|c} \hline \text { Item } & \text { Number } & \begin{array}{c} \text { Manufactures } \\ \text { times (in hours) } \end{array} & \text { Profit (in Rs.) } \\ \hline \text { Necklaces } & x & x / 2 & 100 x \\ \text { Bracelets } & y & y & 300 y \\ \text { Total } & x+y & \frac{x}{2}+y & 100 x+300 y \\ \hline \text { Availability } & 24 & 16 & \\ \hline \end{array}\) 
Our problem is to maximise Z = 100x + 300y subject to constraints are 
\(x \geq 1, y \geq 1 \)
\(x+y \leq 24 ; \frac{1}{2} x+y \leq 16\)

Let the d.c. 's of the lines be l, m, n.
Then \(\mathrm{I}=\cos 90^{\circ}=0, \mathrm{~m}=\cos 60^{\circ}=\frac{1}{2} n=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)

Since E and F are independent, we have
P(E ∩ F) = P(E) . P(F) ....(1)
From the venn diagram in Fig 13.3, it is clear that E ∩ F and E ∩ F′ are mutually exclusive events and also E =(E ∩ F) ∪ (E ∩ F′). Therefore P(E) = P(E ∩ F) + P(E ∩ F′)
or P(E ∩ F′) = P(E) − P(E ∩ F)
= P(E) − P(E) . P(F) (by (1))
= P(E) (1−P(F))
= P(E). P(F′)
Hence, E and F′ are independent

Let us define the events as|
E₁ : Insured person is a scooter driver
E₂ : Insured person is a car driver
E₃ : Insured person is a truck driver
A : Insured person meets with an accident
Then, n(E₁) = 2000, n(E₂) = 4000 and n(E₃) = 6000
Here, total insured person, n(S) = 12000
Now, P(E₁) = Probability that the insured person is a scooter driver
\(=\frac{n\left(E_1\right)}{n(S)}=\frac{2000}{12000}=\frac{1}{6}\)
P(E₂) = Probability that the insured person is a car driver
\(=\frac{n\left(E_2\right)}{n(S)}=\frac{4000}{12000}=\frac{1}{3}\)
and P(E₃) = Probability that the insured person is a truck driver
\(=\frac{n\left(E_3\right)}{n(S)}=\frac{6000}{12000}=\frac{1}{2}\)
Also, P(A/E₁)= Probability that scooter driver meets with an accident
= 0.01
P(A/E₂) = Probability that car driver meets with an accident
= 0.03
and P(A/E₃) = Probability that truck driver meets with an accident
= 0.15
The probability that the person met with an accident was a scooter driver,
\(P\left(E_1 / A\right)=\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{\left[\begin{array}{r} P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right) \\ +P\left(E_3\right) \cdot P\left(A / E_3\right) \end{array}\right]}\)
                                                           [by Baye's theorem]
\(=\frac{\frac{1}{6} \times 0.01}{\left[\left(\frac{1}{6} \times 0.01\right)+\left(\frac{1}{3} \times 0.03\right)+\left(\frac{1}{2} \times 0.15\right)\right]}\)
\(=\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}}=\frac{1}{6} \times \frac{6}{1+6+45}=\frac{1}{52}\)
The probability that the person met with an accident was a car driver, P (E₂ /A)
\(=\frac{P\left(E_2\right) \cdot P\left(A / E_2\right)}{P\left(E_1\right) P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)+P\left(E_3\right) \cdot P\left(A / E_3\right)}\)
\(=\frac{\frac{1}{3} \times 0.03}{\left(\frac{1}{6} \times 0.01\right)+\left(\frac{1}{3} \times 0.03\right)+\left(\frac{1}{2} \times 0.15\right)}\)
\(=\frac{\frac{1}{100}}{\frac{1}{600}+\frac{1}{100}+\frac{15}{200}}=\frac{\frac{1}{100}}{\frac{1+6+45}{600}}=\frac{6}{52}\)
Hence, the required probability is P(E₁ U E₂ / A) = P(E₁/A) + P(E₂ / A)
\(=\frac{1}{52}+\frac{6}{52}=\frac{7}{52}\)

Equation of plane passing through the intersection of given planes is
\(\vec{r} \cdot[(\hat{i}+3 \hat{j})+\lambda(3 \hat{i}-\hat{j}-4 \hat{k})]=6+0 \cdot \lambda\) 
\(\Rightarrow \vec{r} \cdot[(1+3 \lambda) \hat{i}+(3-\lambda) \hat{j}+\hat{k}(-4 \lambda)]=6\)  ...(i)
On dividing both sides by
\(\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}\) ,We get 
\(\frac{\vec{r} \cdot[(1+3 \lambda) \hat{i}+(3-\lambda) \hat{j}+\hat{k}(-4 \lambda)]}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}\) 
\(=\frac{6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}\) 
Given the perpendicular distance from origin is unity.
\(\therefore \frac{6}{\sqrt{(1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}}}=1\) 
\(\Rightarrow (1+3 \lambda)^{2}+(3-\lambda)^{2}+(-4 \lambda)^{2}=36\) 
\(\Rightarrow 1+9 \lambda^{2}+6 \lambda+9+\lambda^{2}-6 \lambda+16 \lambda^{2}=36\) 
\(\Rightarrow 26 \lambda^{2}+10=36 \Rightarrow \lambda^{2}=1\) 
\(\Rightarrow \lambda=\pm 1\) 
On putting the value of \(\lambda\) in Eq. (i), the required equation of plane are
\(\vec{r} \cdot[(1 \pm 3) \hat{i}+(3 \mp 1) \hat{j}+(\mp 4) \hat{k}]=6\) 
\(\Rightarrow \vec{r} \cdot[(1+3) \hat{i}+(3-1) \hat{j}+(-4) \hat{k}]=6\) 
and \(\vec{r} \cdot[(1-3) \hat{i}+(3+1) \hat{j}+4 \hat{k}]=6\) 
\(\Rightarrow \vec{r} \cdot(4 \hat{i}+2 \hat{j}-4 \hat{k})=6\) 
and \(\vec{r} \cdot(-2 \hat{i}+4 \hat{j}+4 \hat{k})=6\) 
and \(-2 x+4 y+4 z-6=0\) 

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.

(i) (b) : D.R:s of OA are < 1-0, 0-0, 0-0 >, i.e., < 1, 0, 0 >.
(ii) (a) : Equation of diagonal OB' is \(\frac{x-0}{1}=\frac{y-0}{2}=\frac{z-0}{3} \text { i.e., } \frac{x}{1}=\frac{y}{2}=\frac{z}{3}\)
(iii) (c) : OABC is xy-plane, therefore its equation is z = 0.
(iv) (c) : Plane O'A'B'C is parallel to xy-plane passing through (0, 0, 3), therefore its equation is z = 3.
(v) (a) : Plane ABB' A' is parallel to yz-plane passing through (1, 0, 0), therefore its equation is x = 1.

Let E be the event that the doctor visit the patient late and let A₁, A₂, A₃, A₄ be the events that the doctor comes by cab, metro, bike and other means of transport respectively.
\(P\left(A_{1}\right)=0.3, P\left(A_{2}\right)=0.2, P\left(A_{3}\right)=0.1, P\left(A_{4}\right)=0.4\)
P(E I A₁) = Probability that the doctor arriving late when he comes by cab = 0.25
Similarly, P ( E I A₂) = 0.3, P (E I A₃) = 0.35 and P ( E I A₃) = 0.1
(i) (b): P(A₂ | E) = Probability that the doctor arriving late and he comes by metro
\(=\frac{P\left(A_{2}\right) P\left(E \mid A_{2}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.2)(0.3)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.06}{0.21}=\frac{2}{7}\)
(ii) (c): P(A₁ | E) = Probability that the doctor arriving late and he comes by cab
\(=\frac{P\left(A_{1}\right) P\left(E \mid A_{1}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.3)(0,25)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.075}{0.21}=\frac{5}{14}\)
(iii) (d): P(A₃| E) = Probability that the doctor arriving late and he comes by bike
\(=\frac{P\left(A_{3}\right) P\left(E \mid A_{3}\right)}{\sum P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.1)(0.35)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.035}{0.21}=\frac{1}{6}\)
(iv) (c): P(A₄ | E) = Probability that the doctor arriving late and he comes by other means of transport
\(=\frac{P\left(A_{4}\right) P\left(E \mid A_{4}\right)}{\Sigma P\left(A_{i}\right) P\left(E \mid A_{i}\right)}\)
\(=\frac{(0.4)(0.1)}{(0.3)(0.25)+(0.2)(0.3)+(0.1)(0.35)+(0.4)(0.1)}\)
\(=\frac{0.04}{0.21}=\frac{4}{21}\)
(v) (a): Probability that the doctor is late by any means
\(=\frac{2}{7}+\frac{5}{14}+\frac{1}{6}+\frac{4}{21}=1\)

Construct the following table of values of objective function

(i) (d): Minimum value of Z is -48 which occurs at (0,8).
(ii) (a): Maximumvalue of Z is 20, which occurs at (5,0).
(iii) (b): Maximum of Z - Minimum of Z
= 20 - (-48) = 20 + 48 = 68
(iv) (c): The corner points of the feasible region are O(0,0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).
(v) (d)