\displaystyle \text{Vector Algebra}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{Solar panels have to be installed carefully so that the tilt of the roof and the direction of the sun}
\displaystyle \text{produce the largest possible electrical power in the solar panels. A surveyor determines the}
\displaystyle \text{coordinates of the four corners of the roof as }P_1(6,8,4),\ P_2(21,8,4),\ P_3(21,16,10)
\displaystyle \text{and }P_4(6,16,10),\text{ where all coordinates are measured in metres.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{What are the components of the two edge vectors }\overrightarrow{A}=\overrightarrow{P_1P_2}
\displaystyle \text{and }\overrightarrow{B}=\overrightarrow{P_1P_4}\text{?}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{A}=P_2-P_1.
\displaystyle =(21-6,\ 8-8,\ 4-4).
\displaystyle =(15,0,0).
\displaystyle \overrightarrow{B}=P_4-P_1.
\displaystyle =(6-6,\ 16-8,\ 10-4).
\displaystyle =(0,8,6).
\displaystyle \therefore \overrightarrow{A}=(15,0,0)\quad\text{and}\quad\overrightarrow{B}=(0,8,6).
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Write the vectors in standard notation using }\widehat{i},\widehat{j}\text{ and }\widehat{k}.
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{A}=15\widehat{i}+0\widehat{j}+0\widehat{k}.
\displaystyle \therefore \overrightarrow{A}=15\widehat{i}.
\displaystyle \overrightarrow{B}=0\widehat{i}+8\widehat{j}+6\widehat{k}.
\displaystyle \therefore \overrightarrow{B}=8\widehat{j}+6\widehat{k}.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What are the magnitudes of the vectors }\overrightarrow{A}\text{ and }\overrightarrow{B}\text{, and in what units?}
\displaystyle \text{Answer:}
\displaystyle |\overrightarrow{A}|=\sqrt{15^2+0^2+0^2}=15.
\displaystyle \therefore |\overrightarrow{A}|=15\text{ m}.
\displaystyle |\overrightarrow{B}|=\sqrt{0^2+8^2+6^2}.
\displaystyle =\sqrt{64+36}=\sqrt{100}=10.
\displaystyle \therefore |\overrightarrow{B}|=10\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What are the components of the vector }\overrightarrow{N}\text{ perpendicular to }\overrightarrow{A},
\displaystyle \overrightarrow{B}\text{ and the surface of the roof?}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{N}=\overrightarrow{A}\times\overrightarrow{B}.
\displaystyle =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\15&0&0\\0&8&6\end{vmatrix}.
\displaystyle =\widehat{i}(0)-\widehat{j}(15\times6)+\widehat{k}(15\times8).
\displaystyle =-90\widehat{j}+120\widehat{k}.
\displaystyle \therefore \overrightarrow{N}=0\widehat{i}-90\widehat{j}+120\widehat{k}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the magnitude of }\overrightarrow{N}\text{ and its units? The sun lies along the unit vector}
\displaystyle \overrightarrow{S}=\frac12\widehat{i}-\frac67\widehat{j}+\frac17\widehat{k}.\text{ If }\overrightarrow{F}=910\overrightarrow{S}\text{ W/m}^2,\text{ find }\overrightarrow{F}\cdot\overrightarrow{N}.
\displaystyle \text{Answer:}
\displaystyle |\overrightarrow{N}|=\sqrt{0^2+(-90)^2+120^2}.
\displaystyle =\sqrt{8100+14400}=\sqrt{22500}=150.
\displaystyle \therefore |\overrightarrow{N}|=150\text{ m}^2.
\displaystyle \overrightarrow{F}=910\left(\frac12\widehat{i}-\frac67\widehat{j}+\frac17\widehat{k}\right).
\displaystyle =455\widehat{i}-780\widehat{j}+130\widehat{k}.
\displaystyle \overrightarrow{F}\cdot\overrightarrow{N}=(455)(0)+(-780)(-90)+(130)(120).
\displaystyle =70200+15600=85800.
\displaystyle \therefore \overrightarrow{F}\cdot\overrightarrow{N}=85800\text{ W}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{What is the angle between }\overrightarrow{N}\text{ and }\overrightarrow{S}\text{? What is the elevation angle of the sun}
\displaystyle \text{above the plane of the roof? Given }\cos51^\circ=0.629.
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{N}\cdot\overrightarrow{S} =(0)\left(\frac12\right)+(-90)\left(-\frac67\right)+120\left(\frac17\right).
\displaystyle =\frac{540}{7}+\frac{120}{7}=\frac{660}{7}.
\displaystyle \cos\theta=\frac{\overrightarrow{N}\cdot\overrightarrow{S}} {|\overrightarrow{N}||\overrightarrow{S}|}.
\displaystyle \text{Taking }\overrightarrow{S}\text{ as the given unit vector,}
\displaystyle \cos\theta=\frac{\frac{660}{7}}{150}=\frac{22}{35}\approx0.629.
\displaystyle \therefore \theta\approx51^\circ.
\displaystyle \text{The angle }\theta\text{ is measured from the normal to the roof.}
\displaystyle \therefore \text{Elevation angle above the roof}=90^\circ-51^\circ=39^\circ.
\displaystyle \therefore \text{The elevation angle of the sun above the roof is }39^\circ.
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{A Class XII student appearing for a competitive examination was asked to attempt the following}
\displaystyle \text{questions. Let }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ be three non-zero vectors.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are such that } |\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|,\text{ then}
\displaystyle \text{(a) }\overrightarrow{a}\perp\overrightarrow{b}\qquad \text{(b) }\overrightarrow{a}\parallel\overrightarrow{b}\qquad \text{(c) }\overrightarrow{a}=\overrightarrow{b}\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle |\overrightarrow{a}+\overrightarrow{b}|= |\overrightarrow{a}-\overrightarrow{b}|.
\displaystyle \text{Squaring both sides,}
\displaystyle |\overrightarrow{a}|^2+|\overrightarrow{b}|^2 +2\overrightarrow{a}\cdot\overrightarrow{b} =|\overrightarrow{a}|^2+|\overrightarrow{b}|^2 -2\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle 4\overrightarrow{a}\cdot\overrightarrow{b}=0.
\displaystyle \therefore \overrightarrow{a}\cdot\overrightarrow{b}=0.
\displaystyle \therefore \overrightarrow{a}\perp\overrightarrow{b}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\overrightarrow{a}=\widehat{i}-2\widehat{j}\text{ and } \overrightarrow{b}=2\widehat{i}+\widehat{j}+3\widehat{k},\text{ evaluate}
\displaystyle (2\overrightarrow{a}+\overrightarrow{b})\cdot [(\overrightarrow{a}+\overrightarrow{b})\times(\overrightarrow{a}-2\overrightarrow{b})].
\displaystyle \text{(a) }0\qquad\text{(b) }4\qquad\text{(c) }3\qquad\text{(d) }2
\displaystyle \text{Answer:}
\displaystyle (\overrightarrow{a}+\overrightarrow{b})\times (\overrightarrow{a}-2\overrightarrow{b})
\displaystyle =\overrightarrow{a}\times\overrightarrow{a} -2\overrightarrow{a}\times\overrightarrow{b} +\overrightarrow{b}\times\overrightarrow{a} -2\overrightarrow{b}\times\overrightarrow{b}.
\displaystyle =-2\overrightarrow{a}\times\overrightarrow{b} -\overrightarrow{a}\times\overrightarrow{b}.
\displaystyle =-3(\overrightarrow{a}\times\overrightarrow{b}).
\displaystyle \therefore (2\overrightarrow{a}+\overrightarrow{b})\cdot [-3(\overrightarrow{a}\times\overrightarrow{b})]=0,
\displaystyle \text{since }\overrightarrow{a}\times\overrightarrow{b}\text{ is perpendicular to both } \overrightarrow{a}\text{ and }\overrightarrow{b}.
\displaystyle \therefore \text{The required value is }0.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are unit vectors and }\theta \text{ is the angle between them, then }|\overrightarrow{a}-\overrightarrow{b}|\text{ is}
\displaystyle \text{(a) }\sin\frac{\theta}{2}\qquad \text{(b) }2\sin\frac{\theta}{2}\qquad \text{(c) }2\cos\frac{\theta}{2}\qquad \text{(d) }\cos\frac{\theta}{2}
\displaystyle \text{Answer:}
\displaystyle |\overrightarrow{a}-\overrightarrow{b}|^2 =|\overrightarrow{a}|^2+|\overrightarrow{b}|^2 -2\overrightarrow{a}\cdot\overrightarrow{b}.
\displaystyle =1+1-2\cos\theta.
\displaystyle =2(1-\cos\theta).
\displaystyle =4\sin^2\frac{\theta}{2}.
\displaystyle \therefore |\overrightarrow{a}-\overrightarrow{b}| =2\sin\frac{\theta}{2}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Let }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ be unit vectors such that } \overrightarrow{a}\cdot\overrightarrow{b}=\overrightarrow{a}\cdot\overrightarrow{c}=0
\displaystyle \text{and the angle between }\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ is }\frac{\pi}{6}. \text{ Then }\overrightarrow{a}\text{ is}
\displaystyle \text{(a) }2(\overrightarrow{b}\times\overrightarrow{c})\qquad \text{(b) }-2(\overrightarrow{b}\times\overrightarrow{c})
\displaystyle \text{(c) }\pm2(\overrightarrow{b}\times\overrightarrow{c})\qquad \text{(d) }2(\overrightarrow{b}\pm\overrightarrow{c})
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=0 \quad\text{and}\quad\overrightarrow{a}\cdot\overrightarrow{c}=0.
\displaystyle \therefore \overrightarrow{a}\text{ is perpendicular to both } \overrightarrow{b}\text{ and }\overrightarrow{c}.
\displaystyle \therefore \overrightarrow{a}\parallel (\overrightarrow{b}\times\overrightarrow{c}).
\displaystyle |\overrightarrow{b}\times\overrightarrow{c}| =|\overrightarrow{b}||\overrightarrow{c}|\sin\frac{\pi}{6}.
\displaystyle =1\times1\times\frac12=\frac12.
\displaystyle \text{Since }|\overrightarrow{a}|=1,\text{ we require a multiplier of }2.
\displaystyle \text{The direction may be along or opposite to } \overrightarrow{b}\times\overrightarrow{c}.
\displaystyle \therefore \overrightarrow{a} =\pm2(\overrightarrow{b}\times\overrightarrow{c}).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The area of the parallelogram formed by }\overrightarrow{a}\text{ and } \overrightarrow{b}\text{ as diagonals is}
\displaystyle \text{(a) }70\qquad\text{(b) }35\qquad \text{(c) }\frac{\sqrt{70}}{2}\qquad\text{(d) }\sqrt{70}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{a}=\widehat{i}-2\widehat{j} =(1,-2,0),
\displaystyle \overrightarrow{b}=2\widehat{i}+\widehat{j}+3\widehat{k} =(2,1,3).
\displaystyle \overrightarrow{a}\times\overrightarrow{b} =\begin{vmatrix} \widehat{i}&\widehat{j}&\widehat{k}\\ 1&-2&0\\ 2&1&3 \end{vmatrix}.
\displaystyle =-6\widehat{i}-3\widehat{j}+5\widehat{k}.
\displaystyle |\overrightarrow{a}\times\overrightarrow{b}| =\sqrt{(-6)^2+(-3)^2+5^2}.
\displaystyle =\sqrt{36+9+25}=\sqrt{70}.
\displaystyle \text{If }\overrightarrow{a}\text{ and }\overrightarrow{b}\text{ are diagonals, area of the parallelogram} =\frac12|\overrightarrow{a}\times\overrightarrow{b}|.
\displaystyle \therefore \text{Area}=\frac{\sqrt{70}}{2}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 3}

\displaystyle \text{A cricket match is organized between Clubs A and B, for which a team from each club is chosen.}
\displaystyle \text{The remaining players of Clubs A and B are respectively seated on the planes represented by}
\displaystyle \overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=3\text{ and }\overrightarrow{r}\cdot(\widehat{i}+3\widehat{j}+2\widehat{k})=8. \displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{The Cartesian equation of the plane on which the players of Club A are seated is}
\displaystyle \text{(a) }2x-y+z=3\qquad\text{(b) }2x-y+2z=3
\displaystyle \text{(c) }2x-y+z=-3\qquad\text{(d) }x-y+z=3
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}.
\displaystyle \overrightarrow{r}\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=3.
\displaystyle (x\widehat{i}+y\widehat{j}+z\widehat{k})\cdot(2\widehat{i}-\widehat{j}+\widehat{k})=3.
\displaystyle 2x-y+z=3.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The magnitude of the normal to the plane on which the players of Club B are seated is}
\displaystyle \text{(a) }\sqrt{15}\qquad\text{(b) }\sqrt{14}\qquad\text{(c) }\sqrt{17}\qquad\text{(d) }\sqrt{20}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+3\widehat{j}+2\widehat{k})=8.
\displaystyle \therefore \text{A normal vector to the plane is }
\displaystyle \overrightarrow{n}=\widehat{i}+3\widehat{j}+2\widehat{k}.
\displaystyle |\overrightarrow{n}|=\sqrt{1^2+3^2+2^2}.
\displaystyle =\sqrt{1+9+4}=\sqrt{14}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The intercept form of the equation of the plane on which the players of Club B are}
\displaystyle \text{seated is}
\displaystyle \text{(a) }\frac{x}{8}+\frac{y}{8/3}+\frac{z}{2}=1
\displaystyle \text{(b) }\frac{x}{5}+\frac{y}{8/3}+\frac{z}{3}=1
\displaystyle \text{(c) }\frac{x}{8}+\frac{y}{8/3}+\frac{z}{4}=1
\displaystyle \text{(d) }\frac{x}{8}+\frac{y}{7}+\frac{z}{2}=1
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{r}\cdot(\widehat{i}+3\widehat{j}+2\widehat{k})=8.
\displaystyle \therefore x+3y+2z=8.
\displaystyle \text{Dividing throughout by }8,
\displaystyle \frac{x}{8}+\frac{3y}{8}+\frac{2z}{8}=1.
\displaystyle \therefore \frac{x}{8}+\frac{y}{8/3}+\frac{z}{4}=1.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Which of the following is a player of Club B?}
\displaystyle \text{(a) Player sitting at }(1,2,1)\qquad\text{(b) Player sitting at }(0,1,2)
\displaystyle \text{(c) Player sitting at }(1,4,1)\qquad\text{(d) Player sitting at }(1,1,2)
\displaystyle \text{Answer:}
\displaystyle \text{The plane for Club B is }x+3y+2z=8.
\displaystyle \text{For }(1,1,2),
\displaystyle x+3y+2z=1+3(1)+2(2).
\displaystyle =1+3+4=8.
\displaystyle \therefore (1,1,2)\text{ lies on the plane of Club B.}
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The distance of the plane on which the players of Club B are seated from the origin is}
\displaystyle \text{(a) }\frac{8}{\sqrt{14}}\text{ units}\qquad\text{(b) }\frac{6}{\sqrt{14}}\text{ units}
\displaystyle \text{(c) }\frac{7}{\sqrt{14}}\text{ units}\qquad\text{(d) }\frac{9}{\sqrt{14}}\text{ units}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane is }x+3y+2z-8=0.
\displaystyle \text{Distance of }(0,0,0)\text{ from the plane}
\displaystyle =\frac{|1(0)+3(0)+2(0)-8|}{\sqrt{1^2+3^2+2^2}}.
\displaystyle =\frac{8}{\sqrt{14}}.
\displaystyle \therefore \text{The distance from the origin is }\frac{8}{\sqrt{14}}\text{ units.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 5}

\displaystyle \text{The Indian Coast Guard, while patrolling, saw a suspicious boat with people who did not appear to be}
\displaystyle \text{fishermen. The Coast Guard closely observed the movement of the boat for an opportunity to seize it.}
\displaystyle \text{They observed that the boat was moving along a planar surface. At an instant of time, the positions}
\displaystyle \text{of the Coast Guard helicopter and the boat were }(1,3,5)\text{ and }(2,5,3)\text{ respectively.} \displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{If the line joining the positions of the helicopter and the boat is perpendicular}
\displaystyle \text{to the plane in which the boat moves, then the equation of the plane is}
\displaystyle \text{(a) }-x+2y-2z=6\qquad\text{(b) }x+2y+2z=6
\displaystyle \text{(c) }x+2y-2z=6\qquad\text{(d) }x-2y-2z=6
\displaystyle \text{Answer:}
\displaystyle \text{Let }H(1,3,5)\text{ and }B(2,5,3)\text{ denote the positions of the helicopter and boat.}
\displaystyle \overrightarrow{HB}=(2-1)\widehat{i}+(5-3)\widehat{j}+(3-5)\widehat{k}.
\displaystyle \therefore \overrightarrow{HB}=\widehat{i}+2\widehat{j}-2\widehat{k}.
\displaystyle \text{Since }\overrightarrow{HB}\text{ is perpendicular to the plane, it is a normal vector to the plane.}
\displaystyle \text{The plane passes through the position of the boat }B(2,5,3).
\displaystyle 1(x-2)+2(y-5)-2(z-3)=0.
\displaystyle x-2+2y-10-2z+6=0.
\displaystyle \therefore x+2y-2z=6.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If the Coast Guard decides to shoot the boat at that instant, what is the distance}
\displaystyle \text{in metres that the bullet has to travel?}
\displaystyle \text{(a) }5\text{ m}\qquad\text{(b) }3\text{ m}\qquad\text{(c) }6\text{ m}\qquad\text{(d) }4\text{ m}
\displaystyle \text{Answer:}
\displaystyle \text{Distance between }H(1,3,5)\text{ and }B(2,5,3)
\displaystyle =\sqrt{(2-1)^2+(5-3)^2+(3-5)^2}.
\displaystyle =\sqrt{1+4+4}=\sqrt{9}=3\text{ m}.
\displaystyle \therefore \text{The bullet has to travel }3\text{ m.}
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the Coast Guard decides to shoot the boat at that instant and the speed of the}
\displaystyle \text{bullet is }36\text{ m/sec, then what is the time taken for the bullet to travel and hit the boat?}
\displaystyle \text{(a) }\frac18\text{ seconds}\qquad\text{(b) }\frac1{14}\text{ seconds}\qquad \text{(c) }\frac1{10}\text{ seconds}\qquad\text{(d) }\frac1{12}\text{ seconds}
\displaystyle \text{Answer:}
\displaystyle \text{Distance travelled by the bullet}=3\text{ m}.
\displaystyle \text{Speed of the bullet}=36\text{ m/sec}.
\displaystyle \text{Time}=\frac{\text{Distance}}{\text{Speed}}=\frac{3}{36}=\frac1{12}\text{ seconds}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{At that instant, the equation of the line passing through the positions of}
\displaystyle \text{the helicopter and the boat is}
\displaystyle \text{(a) }\frac{x-1}{1}=\frac{y-3}{2}=\frac{z-5}{-2}\qquad \text{(b) }\frac{x-1}{2}=\frac{y+3}{1}=\frac{z-5}{-2}
\displaystyle \text{(c) }\frac{x+1}{-2}=\frac{y-3}{1}=\frac{z-5}{-2}\qquad \text{(d) }\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+5}{2}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{HB}=\widehat{i}+2\widehat{j}-2\widehat{k}.
\displaystyle \text{Thus, the direction ratios of the line are }1,2,-2.
\displaystyle \text{The line passes through }H(1,3,5).
\displaystyle \therefore \frac{x-1}{1}=\frac{y-3}{2}=\frac{z-5}{-2}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{At a different instant, the boat moves to another position along the planar}
\displaystyle \text{surface. What should be the coordinates of the boat if the Coast Guard shoots along the line}
\displaystyle \frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{1}\text{ for the bullet to hit the boat?}
\displaystyle \text{(a) }\left(-\frac83,\frac{19}{3},-\frac{14}{3}\right)\qquad \text{(b) }\left(\frac83,-\frac{19}{3},-\frac{14}{3}\right)
\displaystyle \text{(c) }\left(\frac83,-\frac{19}{3},\frac{14}{3}\right)\qquad \text{(d) None of the above}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{1}=\lambda.
\displaystyle \therefore x=\lambda,\qquad y=1+2\lambda,\qquad z=2+\lambda.
\displaystyle \text{The boat moves on the plane }x+2y-2z=6.
\displaystyle \therefore \lambda+2(1+2\lambda)-2(2+\lambda)=6.
\displaystyle \lambda+2+4\lambda-4-2\lambda=6.
\displaystyle 3\lambda-2=6.
\displaystyle 3\lambda=8\qquad\therefore\lambda=\frac83.
\displaystyle \therefore x=\frac83.
\displaystyle y=1+\frac{16}{3}=\frac{19}{3},\qquad z=2+\frac83=\frac{14}{3}.
\displaystyle \therefore \text{The position of the boat is } \left(\frac83,\frac{19}{3},\frac{14}{3}\right).
\displaystyle \text{This point is not among options (a), (b) or (c).}
\displaystyle \therefore \text{Option (d), None of the above, is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 6}

\displaystyle \text{The equations of motion of a missile are }x=3t,\ y=-4t,\ z=t,\text{ where time }t\text{ is given in}
\displaystyle \text{seconds and distance is measured in kilometres. Based on the above information, answer the following.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{What is the path of the missile?}
\displaystyle \text{(a) Straight line}\qquad\text{(b) Parabola}\qquad\text{(c) Circle}\qquad\text{(d) Ellipse}
\displaystyle \text{Answer:}
\displaystyle x=3t,\qquad y=-4t,\qquad z=t.
\displaystyle \therefore \frac{x}{3}=\frac{y}{-4}=\frac{z}{1}=t.
\displaystyle \text{This is the symmetric equation of a straight line through the origin.}
\displaystyle \therefore \text{The path of the missile is a straight line.}
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Which of the following points lies on the path of the missile?}
\displaystyle \text{(a) }(6,8,2)\qquad\text{(b) }(6,-8,-2)\qquad \text{(c) }(6,-8,2)\qquad\text{(d) }(-6,-8,2)
\displaystyle \text{Answer:}
\displaystyle \text{Since }x=3t,\text{ for }x=6,\quad 3t=6.
\displaystyle \therefore t=2.
\displaystyle \text{At }t=2,\qquad y=-4(2)=-8,\qquad z=2.
\displaystyle \therefore \text{The point on the path is }(6,-8,2).
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{At what distance will the missile be from the starting point }(0,0,0)\text{ in }5\text{ seconds?}
\displaystyle \text{(a) }\sqrt{550}\text{ km}\qquad\text{(b) }\sqrt{650}\text{ km}\qquad \text{(c) }\sqrt{450}\text{ km}\qquad\text{(d) }\sqrt{750}\text{ km}
\displaystyle \text{Answer:}
\displaystyle \text{At }t=5,\qquad x=3(5)=15,\qquad y=-4(5)=-20,\qquad z=5.
\displaystyle \therefore \text{The position of the missile is }(15,-20,5).
\displaystyle \text{Distance from }(0,0,0)=\sqrt{15^2+(-20)^2+5^2}.
\displaystyle =\sqrt{225+400+25}=\sqrt{650}\text{ km}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If the position of a rocket at a certain instant is }(5,-8,10),\text{ what will be its height}
\displaystyle \text{from the ground, where the ground is considered as the }xy\text{-plane?}
\displaystyle \text{(a) }12\text{ km}\qquad\text{(b) }11\text{ km}\qquad \text{(c) }20\text{ km}\qquad\text{(d) }10\text{ km}
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the }xy\text{-plane is }z=0.
\displaystyle \text{The perpendicular distance of }(x_1,y_1,z_1)\text{ from the }xy\text{-plane is }|z_1|.
\displaystyle \therefore \text{Height of the rocket}=|10|=10\text{ km}.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{At a certain instant, the missile is above sea level, where the equation of the sea surface}
\displaystyle \text{is }2x+y+3z=1.\text{ If the position of the missile is }(1,1,2),\text{ then its image in the sea is}
\displaystyle \text{(a) }\left(-\frac97,-\frac17,-\frac{10}{7}\right)\qquad \text{(b) }\left(\frac97,-\frac17,-\frac{10}{7}\right)
\displaystyle \text{(c) }\left(-\frac97,\frac17,-\frac{10}{7}\right)\qquad \text{(d) }\left(-\frac97,-\frac17,\frac{10}{7}\right)
\displaystyle \text{Answer:}
\displaystyle \text{The plane is }2x+y+3z-1=0.
\displaystyle \text{For a point }(x_1,y_1,z_1),\text{ its image in the plane }ax+by+cz+d=0\text{ is}
\displaystyle \left(x_1-\frac{2aD}{a^2+b^2+c^2},\ y_1-\frac{2bD}{a^2+b^2+c^2},\ z_1-\frac{2cD}{a^2+b^2+c^2}\right),
\displaystyle \text{where }D=ax_1+by_1+cz_1+d.
\displaystyle D=2(1)+1(1)+3(2)-1=8.
\displaystyle a^2+b^2+c^2=2^2+1^2+3^2=14.
\displaystyle x'=1-\frac{2(2)(8)}{14}=1-\frac{16}{7}=-\frac97.
\displaystyle y'=1-\frac{2(1)(8)}{14}=1-\frac87=-\frac17.
\displaystyle z'=2-\frac{2(3)(8)}{14}=2-\frac{24}{7}=-\frac{10}{7}.
\displaystyle \therefore \text{The image is }\left(-\frac97,-\frac17,-\frac{10}{7}\right).
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 7}

\displaystyle \text{Suppose the floor of a hotel is made of mirror-polished Salvatore stone. A large crystal chandelier}
\displaystyle \text{is attached to the ceiling of the hotel room. Consider the floor as a plane having the equation}
\displaystyle x-y+z=4,\text{ and suppose the crystal chandelier is suspended at the point }(1,0,1).
\displaystyle \text{Based on the above information, answer the following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{Find the direction ratios of the perpendicular from the point }(1,0,1)\text{ to the plane}
\displaystyle x-y+z=4.
\displaystyle \text{(a) }(-1,-1,1)\qquad\text{(b) }(1,-1,-1)
\displaystyle \text{(c) }(-1,-1,-1)\qquad\text{(d) }(1,-1,1)
\displaystyle \text{Answer:}
\displaystyle \text{The equation of the plane is }x-y+z-4=0.
\displaystyle \text{A normal vector to the plane is }
\displaystyle \overrightarrow{n}=\widehat{i}-\widehat{j}+\widehat{k}.
\displaystyle \text{The perpendicular to the plane is parallel to its normal.}
\displaystyle \therefore \text{The required direction ratios are }(1,-1,1).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find the length of the perpendicular from the point }(1,0,1)\text{ to the plane}
\displaystyle x-y+z=4.
\displaystyle \text{(a) }\frac{2}{\sqrt3}\text{ units}\qquad\text{(b) }\frac{4}{\sqrt3}\text{ units}
\displaystyle \text{(c) }\frac{6}{\sqrt3}\text{ units}\qquad\text{(d) }\frac{8}{\sqrt3}\text{ units}
\displaystyle \text{Answer:}
\displaystyle \text{The plane is }x-y+z-4=0.
\displaystyle \text{Distance of }(1,0,1)\text{ from the plane}
\displaystyle =\frac{|1(1)-1(0)+1(1)-4|}{\sqrt{1^2+(-1)^2+1^2}}.
\displaystyle =\frac{|-2|}{\sqrt3}=\frac{2}{\sqrt3}\text{ units}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The equation of the perpendicular from the point }(1,0,1)\text{ to the plane }x-y+z=4
\displaystyle \text{is}
\displaystyle \text{(a) }\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+5}{2}
\displaystyle \text{(b) }\frac{x-1}{-2}=\frac{y+3}{-1}=\frac{z-5}{2}
\displaystyle \text{(c) }\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{1}
\displaystyle \text{(d) }\frac{x-1}{2}=\frac{y}{-2}=\frac{z-1}{1}
\displaystyle \text{Answer:}
\displaystyle \text{The normal to the plane has direction ratios }1,-1,1.
\displaystyle \text{The required perpendicular passes through }(1,0,1).
\displaystyle \therefore \frac{x-1}{1}=\frac{y-0}{-1}=\frac{z-1}{1}.
\displaystyle \therefore \frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{1}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The equation of a plane parallel to }x-y+z=4\text{ and at a unit distance from}
\displaystyle \text{the point }(1,0,1)\text{ is}
\displaystyle \text{(a) }x-y+z+(2-\sqrt3)=0
\displaystyle \text{(b) }x-y+z-(2+\sqrt3)=0
\displaystyle \text{(c) }x-y+z+(2+\sqrt3)=0\qquad\text{(d) Both (a) and (c)}
\displaystyle \text{Answer:}
\displaystyle \text{Let the required plane be }x-y+z+d=0.
\displaystyle \text{Its distance from }(1,0,1)\text{ is }1.
\displaystyle \frac{|1-0+1+d|}{\sqrt3}=1.
\displaystyle |2+d|=\sqrt3.
\displaystyle 2+d=\pm\sqrt3.
\displaystyle \therefore d=-2+\sqrt3\quad\text{or}\quad d=-2-\sqrt3.
\displaystyle \therefore x-y+z-(2-\sqrt3)=0
\displaystyle \text{or}\quad x-y+z-(2+\sqrt3)=0.
\displaystyle \text{The second equation is given as option (b).}
\displaystyle \therefore \text{Among the given options, option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The direction cosines of the normal to the plane }x-y+z=4\text{ are}
\displaystyle \text{(a) }\left(\frac1{\sqrt3},-\frac1{\sqrt3},-\frac1{\sqrt3}\right)
\displaystyle \text{(b) }\left(\frac1{\sqrt3},-\frac1{\sqrt3},\frac1{\sqrt3}\right)
\displaystyle \text{(c) }\left(-\frac1{\sqrt3},-\frac1{\sqrt3},\frac1{\sqrt3}\right)
\displaystyle \text{(d) }\left(-\frac1{\sqrt3},-\frac1{\sqrt3},-\frac1{\sqrt3}\right)
\displaystyle \text{Answer:}
\displaystyle \text{A normal vector to the plane is }\overrightarrow{n}=\widehat{i}-\widehat{j}+\widehat{k}.
\displaystyle |\overrightarrow{n}|=\sqrt{1^2+(-1)^2+1^2}=\sqrt3.
\displaystyle \text{Therefore, the corresponding unit normal vector is}
\displaystyle \frac{\overrightarrow{n}}{|\overrightarrow{n}|} =\frac1{\sqrt3}\widehat{i}-\frac1{\sqrt3}\widehat{j}+\frac1{\sqrt3}\widehat{k}.
\displaystyle \therefore \text{The direction cosines are } \left(\frac1{\sqrt3},-\frac1{\sqrt3},\frac1{\sqrt3}\right).
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 8}

\displaystyle \text{A mobile tower stands at the top of a hill. Consider the surface on which the tower stands as a}
\displaystyle \text{plane containing the points }A(1,0,2),\ B(3,-1,1)\text{ and }C(1,2,1).\text{ The mobile tower is tied}
\displaystyle \text{with three cables from the points }A,\ B\text{ and }C\text{ such that it stands vertically on the ground.}
\displaystyle \text{The top of the tower is at the point }(2,3,1).\text{ Based on the above information, answer the}
\displaystyle \text{following questions.} \displaystyle \\

\displaystyle \textbf{Question 1: }\text{The equation of the plane passing through the points }A,\ B\text{ and }C\text{ is}
\displaystyle \text{(a) }3x-2y+4z=-11\qquad\text{(b) }3x+2y+4z=11
\displaystyle \text{(c) }3x-2y-4z=11\qquad\text{(d) }-3x+2y+4z=-11
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{AB}=(3-1)\widehat{i}+(-1-0)\widehat{j}+(1-2)\widehat{k}.
\displaystyle =2\widehat{i}-\widehat{j}-\widehat{k}.
\displaystyle \overrightarrow{AC}=(1-1)\widehat{i}+(2-0)\widehat{j}+(1-2)\widehat{k}.
\displaystyle =2\widehat{j}-\widehat{k}.
\displaystyle \overrightarrow{AB}\times\overrightarrow{AC} =\begin{vmatrix}\widehat{i}&\widehat{j}&\widehat{k}\\2&-1&-1\\0&2&-1\end{vmatrix}.
\displaystyle =3\widehat{i}+2\widehat{j}+4\widehat{k}.
\displaystyle \therefore \text{A normal vector to the plane is }3\widehat{i}+2\widehat{j}+4\widehat{k}.
\displaystyle \text{Using the point }A(1,0,2),
\displaystyle 3(x-1)+2(y-0)+4(z-2)=0.
\displaystyle 3x+2y+4z=11.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The height of the tower from the ground is}
\displaystyle \text{(a) }\frac{5}{\sqrt{29}}\text{ units}\qquad\text{(b) }\frac{7}{\sqrt{29}}\text{ units}
\displaystyle \text{(c) }\frac{6}{\sqrt{29}}\text{ units}\qquad\text{(d) }\frac{8}{\sqrt{29}}\text{ units}
\displaystyle \text{Answer:}
\displaystyle \text{The ground plane is }3x+2y+4z-11=0.
\displaystyle \text{The top of the tower is }P(2,3,1).
\displaystyle \text{Height}=\frac{|3(2)+2(3)+4(1)-11|}{\sqrt{3^2+2^2+4^2}}.
\displaystyle =\frac{|6+6+4-11|}{\sqrt{29}}=\frac{5}{\sqrt{29}}\text{ units}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The equation of the perpendicular line drawn from the top of the tower to the ground is}
\displaystyle \text{(a) }\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z-5}{-2}
\displaystyle \text{(b) }\frac{x-2}{-3}=\frac{y-3}{-2}=\frac{z-1}{-4}
\displaystyle \text{(c) }\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}
\displaystyle \text{(d) }\frac{x+1}{-2}=\frac{y+3}{-1}=\frac{z-5}{2}
\displaystyle \text{Answer:}
\displaystyle \text{A normal vector to the ground plane has direction ratios }3,2,4.
\displaystyle \text{The perpendicular passes through the top of the tower }(2,3,1).
\displaystyle \therefore \frac{x-2}{3}=\frac{y-3}{2}=\frac{z-1}{4}.
\displaystyle \text{Multiplying all direction ratios by }-1\text{ gives the same line:}
\displaystyle \frac{x-2}{-3}=\frac{y-3}{-2}=\frac{z-1}{-4}.
\displaystyle \therefore \text{Options (b) and (c) represent the same line.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The coordinates of the foot of the perpendicular drawn from the top of the tower to}
\displaystyle \text{the ground are}
\displaystyle \text{(a) }\left(\frac{43}{29},-\frac{77}{29},-\frac{9}{29}\right)
\displaystyle \text{(b) }\left(\frac97,-\frac{11}{7},-\frac{10}{7}\right)
\displaystyle \text{(c) }\left(-\frac{43}{29},\frac{77}{29},-\frac{9}{29}\right)
\displaystyle \text{(d) }\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right)
\displaystyle \text{Answer:}
\displaystyle \text{Let the foot of the perpendicular be }Q.
\displaystyle \text{The line through }P(2,3,1)\text{ perpendicular to the plane is}
\displaystyle x=2+3\lambda,\qquad y=3+2\lambda,\qquad z=1+4\lambda.
\displaystyle \text{Since }Q\text{ lies on }3x+2y+4z=11,
\displaystyle 3(2+3\lambda)+2(3+2\lambda)+4(1+4\lambda)=11.
\displaystyle 16+29\lambda=11.
\displaystyle \therefore \lambda=-\frac{5}{29}.
\displaystyle x=2-\frac{15}{29}=\frac{43}{29}.
\displaystyle y=3-\frac{10}{29}=\frac{77}{29}.
\displaystyle z=1-\frac{20}{29}=\frac{9}{29}.
\displaystyle \therefore Q=\left(\frac{43}{29},\frac{77}{29},\frac{9}{29}\right).
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The area of }\triangle ABC\text{ is}
\displaystyle \text{(a) }\frac{\sqrt{29}}{4}\text{ sq. units}\qquad \text{(b) }\frac{\sqrt{29}}{2}\text{ sq. units}
\displaystyle \text{(c) }\frac{\sqrt{39}}{2}\text{ sq. units}\qquad \text{(d) }\frac{\sqrt{39}}{4}\text{ sq. units}
\displaystyle \text{Answer:}
\displaystyle \overrightarrow{AB}\times\overrightarrow{AC} =3\widehat{i}+2\widehat{j}+4\widehat{k}.
\displaystyle |\overrightarrow{AB}\times\overrightarrow{AC}| =\sqrt{3^2+2^2+4^2}=\sqrt{29}.
\displaystyle \text{Area of }\triangle ABC =\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|.
\displaystyle =\frac{\sqrt{29}}{2}\text{ sq. units}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

 

\displaystyle \text{PROBABILITY}

\displaystyle \textbf{Case Study - 1}

\displaystyle \text{A coach is training three players. He observes that player A can hit a target }4\text{ times in }5
\displaystyle \text{shots, player B can hit }3\text{ times in }4\text{ shots and player C can hit }2\text{ times in }3\text{ shots.} \displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{What is the probability that A, B and C will all hit the target?}
\displaystyle \text{(a) }\frac45\qquad\text{(b) }\frac35\qquad\text{(c) }\frac25\qquad\text{(d) }\frac15
\displaystyle \text{Answer:}
\displaystyle P(A)=\frac45,\qquad P(B)=\frac34,\qquad P(C)=\frac23.
\displaystyle \text{Since the shots are independent,}
\displaystyle P(A\cap B\cap C)=P(A)P(B)P(C).
\displaystyle =\frac45\times\frac34\times\frac23.
\displaystyle =\frac25.
\displaystyle \therefore \text{The probability that all three hit the target is }\frac25.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the probability that B and C hit the target and A misses it?}
\displaystyle \text{(a) }\frac1{10}\qquad\text{(b) }\frac3{10}\qquad\text{(c) }\frac7{10}\qquad\text{(d) }\frac4{10}
\displaystyle \text{Answer:}
\displaystyle P(A')=1-\frac45=\frac15.
\displaystyle P(A'\cap B\cap C)=P(A')P(B)P(C).
\displaystyle =\frac15\times\frac34\times\frac23.
\displaystyle =\frac1{10}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the probability that exactly any two of A, B and C will hit the target?}
\displaystyle \text{(1) }\frac1{30}\qquad\text{(2) }\frac{11}{30}\qquad\text{(3) }\frac{17}{30}\qquad\text{(4) }\frac{13}{30}
\displaystyle \text{Answer:}
\displaystyle P(\text{exactly two hit})=P(ABC')+P(AB'C)+P(A'BC).
\displaystyle P(ABC')=\frac45\times\frac34\times\frac13=\frac15.
\displaystyle P(AB'C)=\frac45\times\frac14\times\frac23=\frac2{15}.
\displaystyle P(A'BC)=\frac15\times\frac34\times\frac23=\frac1{10}.
\displaystyle \therefore P(\text{exactly two hit})=\frac15+\frac2{15}+\frac1{10}.
\displaystyle =\frac6{30}+\frac4{30}+\frac3{30}=\frac{13}{30}.
\displaystyle \therefore \text{Option (4) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What is the probability that none of them will hit the target?}
\displaystyle \text{(a) }\frac1{30}\qquad\text{(b) }\frac1{60}\qquad\text{(c) }\frac1{15}\qquad\text{(d) }\frac2{15}
\displaystyle \text{Answer:}
\displaystyle P(A')=\frac15,\qquad P(B')=\frac14,\qquad P(C')=\frac13.
\displaystyle P(A'\cap B'\cap C')=\frac15\times\frac14\times\frac13.
\displaystyle =\frac1{60}.
\displaystyle \therefore \text{The probability that none of them hits the target is }\frac1{60}.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the probability that at least one of A, B or C will hit the target?}
\displaystyle \text{(a) }\frac{59}{60}\qquad\text{(b) }\frac25\qquad\text{(c) }\frac35\qquad\text{(d) }\frac1{60}
\displaystyle \text{Answer:}
\displaystyle P(\text{at least one hits})=1-P(\text{none hits}).
\displaystyle =1-\frac1{60}.
\displaystyle =\frac{59}{60}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 2}

\displaystyle \text{The reliability of a COVID PCR test is specified as follows. Of people having COVID, }90\%
\displaystyle \text{test positive, while }10\%\text{ go undetected. Of people free of COVID, }99\%\text{ test negative,}
\displaystyle \text{while }1\%\text{ are incorrectly diagnosed as COVID positive. In a large population, only }0.1\%
\displaystyle \text{have COVID. One person is selected at random and is given the COVID PCR test.} \displaystyle \text{Based on the above information, answer the following questions.}
\displaystyle \\

\displaystyle \textbf{Question 1: }\text{What is the probability that a person tests COVID positive, given that the person}
\displaystyle \text{actually has COVID?}
\displaystyle \text{(a) }0.001\qquad\text{(b) }0.1\qquad\text{(c) }0.8\qquad\text{(d) }0.9
\displaystyle \text{Answer:}
\displaystyle \text{Let }C\text{ denote the event that the person has COVID and }P\text{ denote a positive test result.}
\displaystyle \text{It is given that }90\%\text{ of people having COVID test positive.}
\displaystyle \therefore P(P\mid C)=0.90.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the probability that a person tests COVID positive, given that the person}
\displaystyle \text{does not actually have COVID?}
\displaystyle \text{(a) }0.01\qquad\text{(b) }0.99\qquad\text{(c) }0.1\qquad\text{(d) }0.001
\displaystyle \text{Answer:}
\displaystyle \text{It is given that }1\%\text{ of people without COVID are diagnosed as COVID positive.}
\displaystyle \therefore P(P\mid C')=0.01.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{What is the probability that the person actually does not have COVID?}
\displaystyle \text{(a) }0.998\qquad\text{(b) }0.999\qquad\text{(c) }0.001\qquad\text{(d) }0.111
\displaystyle \text{Answer:}
\displaystyle P(C)=0.1\%=0.001.
\displaystyle \therefore P(C')=1-P(C).
\displaystyle =1-0.001=0.999.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{What is the probability that a person actually has COVID, given that the person tests}
\displaystyle \text{COVID positive?}
\displaystyle \text{(a) }0.83\qquad\text{(b) }0.0803\qquad\text{(c) }0.083\qquad\text{(d) }0.089
\displaystyle \text{Answer:}
\displaystyle P(C)=0.001,\qquad P(C')=0.999.
\displaystyle P(P\mid C)=0.90,\qquad P(P\mid C')=0.01.
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(C\mid P)=\frac{P(C)P(P\mid C)} {P(C)P(P\mid C)+P(C')P(P\mid C')}.
\displaystyle =\frac{(0.001)(0.90)} {(0.001)(0.90)+(0.999)(0.01)}.
\displaystyle =\frac{0.0009}{0.0009+0.00999}.
\displaystyle =\frac{0.0009}{0.01089}\approx0.08264.
\displaystyle \therefore P(C\mid P)\approx0.083.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the probability that a randomly selected person will be diagnosed as COVID}
\displaystyle \text{positive?}
\displaystyle \text{(a) }0.1089\qquad\text{(b) }0.01089\qquad\text{(c) }0.0189\qquad\text{(d) }0.189
\displaystyle \text{Answer:}
\displaystyle P(P)=P(C)P(P\mid C)+P(C')P(P\mid C').
\displaystyle =(0.001)(0.90)+(0.999)(0.01).
\displaystyle =0.0009+0.00999.
\displaystyle =0.01089.
\displaystyle \therefore \text{The probability of being diagnosed COVID positive is }0.01089.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Case Study - 3}

\displaystyle \text{In answering a multiple-choice question in a Class XII test, a student either knows the answer or}
\displaystyle \text{guesses. The probability that the student knows the answer is }\frac35\text{ and the probability that the}
\displaystyle \text{student guesses is }\frac25.\text{ A student who guesses answers correctly with probability }\frac13.
\displaystyle \text{Let }E_1,E_2\text{ and }E\text{ denote the events that the student knows the answer, guesses the answer and}
\displaystyle \text{answers correctly, respectively. Based on the above information, answer the following questions.}

\displaystyle \textbf{Question 1: }\text{What is the value of }P(E_1)\text{?}
\displaystyle \text{(a) }\frac25\qquad\text{(b) }\frac13\qquad\text{(c) }1\qquad\text{(d) }\frac35
\displaystyle \text{Answer:}
\displaystyle E_1\text{ is the event that the student knows the answer.}
\displaystyle \text{It is given that }P(E_1)=\frac35.
\displaystyle \therefore \text{Option (d) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{What is the value of }P(E\mid E_1)\text{?}
\displaystyle \text{(a) }\frac13\qquad\text{(b) }1\qquad\text{(c) }\frac23\qquad\text{(d) }\frac4{15}
\displaystyle \text{Answer:}
\displaystyle P(E\mid E_1)=\text{Probability of answering correctly, given that the student knows the answer.}
\displaystyle \text{Since the student knows the answer, }P(E\mid E_1)=1.
\displaystyle \therefore \text{Option (b) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find }\sum_{k=1}^{2}P(E\mid E_k)P(E_k).
\displaystyle \text{(a) }\frac{11}{15}\qquad\text{(b) }\frac4{15}\qquad\text{(c) }\frac15\qquad\text{(d) }1
\displaystyle \text{Answer:}
\displaystyle P(E_1)=\frac35,\qquad P(E_2)=\frac25.
\displaystyle P(E\mid E_1)=1,\qquad P(E\mid E_2)=\frac13.
\displaystyle \sum_{k=1}^{2}P(E\mid E_k)P(E_k)
\displaystyle =P(E\mid E_1)P(E_1)+P(E\mid E_2)P(E_2).
\displaystyle =1\times\frac35+\frac13\times\frac25.
\displaystyle =\frac35+\frac2{15}=\frac9{15}+\frac2{15}=\frac{11}{15}.
\displaystyle \therefore \text{Option (a) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find }\sum_{k=1}^{2}P(E_k).
\displaystyle \text{(a) }\frac13\qquad\text{(b) }\frac15\qquad\text{(c) }1\qquad\text{(d) }\frac35
\displaystyle \text{Answer:}
\displaystyle \sum_{k=1}^{2}P(E_k)=P(E_1)+P(E_2).
\displaystyle =\frac35+\frac25=1.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{What is the probability that the student knows the answer, given that the student}
\displaystyle \text{answered it correctly?}
\displaystyle \text{(a) }\frac2{11}\qquad\text{(b) }\frac53\qquad\text{(c) }\frac9{11}\qquad\text{(d) }\frac{13}{3}
\displaystyle \text{Answer:}
\displaystyle \text{By Bayes' theorem,}
\displaystyle P(E_1\mid E)=\frac{P(E_1)P(E\mid E_1)} {P(E_1)P(E\mid E_1)+P(E_2)P(E\mid E_2)}.
\displaystyle =\frac{\frac35\times1} {\frac35\times1+\frac25\times\frac13}.
\displaystyle =\frac{\frac35}{\frac35+\frac2{15}}.
\displaystyle =\frac{\frac35}{\frac{11}{15}}=\frac35\times\frac{15}{11}.
\displaystyle =\frac9{11}.
\displaystyle \therefore \text{The required probability is }\frac9{11}.
\displaystyle \therefore \text{Option (c) is correct.}
\displaystyle \\

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