and 
are collinear then the value of 
is 
Solution: Option(C) To find: Area of $A B C$ Given: $A(3,-2), B(k, 2)$ and $C(8,8)$ The formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} &...
are 
and 
The area of a is 
units 
unitsSolution: Option(B) To find: Area of $A B C$ Given: $A(-2,4), B(2,-6)$ and $C(5,4)$ Formula used: $\Delta=\frac{1}{2}\left|\begin{array}{lll}x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\...
is 
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}3 x-8 & 3 & 3 \\ 3 & 3 x-8 & 3 \\ 3 & 3 & 3 x-8\end{array}\right|=0$ Applying $\mathrm{R}_{1}...
is 
Solution: Option(B) To find: Value of $x$ We have, $\left|\begin{array}{ccc}a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x\end{array}\right|=0$ Applying...
is 
Solution: Option(A) To find: Value of $x$ We have, $\left|\begin{array}{ccc}x-2 & 2 x-3 & 3 x-4 \\ x-4 & 2 x-9 & 3 x-16 \\ x-8 & 2 x-27 & 3 x-64\end{array}\right|=0$ Applying...
is 
Solution: Option(C) To find: Value of $x$ We have, $\left|\begin{array}{lll}\mathrm{x} & 3 & 7 \\ 2 & \mathrm{X} & 2 \\ 7 & 6 & \mathrm{x}\end{array}\right|=0$ Applying...
then 
Solution: Option(C) To find: Value of $\mathrm{x}$ We have, $\left|\begin{array}{ccc}5 & 3 & -1 \\ -7 & x & 2 \\ 9 & 6 & -2\end{array}\right|=0$ Applying $R_{1} \rightarrow 2...
Solution: Option(A) To find: Value of $\left|\begin{array}{ccc}\mathrm{x}+1 & \mathrm{x}+2 & \mathrm{x}+4 \\ \mathrm{x}+3 & \mathrm{x}+5 & \mathrm{x}+8 \\ \mathrm{x}+7 &...
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}a & 1 & b+c \\ b & 1 & c+a \\ c & 1 & a+b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & 1 &...
Solution: Option(A) To find: Value of $\left|\begin{array}{ccc}b+c & a & a \\ b & c+a & b \\ c & c & a+b\end{array}\right|$ We have, $\left|\begin{array}{ccc}b+c & a...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\mathrm{bc} & \mathrm{b}+\mathrm{c} & 1 \\ \mathrm{ca} & \mathrm{a}+\mathrm{c} & 1 \\ \mathrm{ab} &...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1 & 1...
Solution: Option(C) To find: Value of $\left|\begin{array}{lll}b+c & a & b \\ c+a & c & a \\ a+b & b & c\end{array}\right|$ We have,...
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}a & a+2 b & a+2 b+3 c \\ 3 a & 4 a+6 b & 5 a+7 b+9 c \\ 6 a & 9 a+12 b & 11 a+15 b+18 c\end{array}\right|$ We...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}a^{2}+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|$ We have,...
Solution: Option( B) To find: Value of $\left|\begin{array}{ccc}x+y & x & x \\ 5 x+4 y & 4 x & 2 x \\ 10 x+8 y & 8 x & 3 x\end{array}\right|$ We have,...
be distinct positive real numbers then the value of 
is Solution: Option(B) To find: Nature of $\left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|$ We have, $\left|\begin{array}{lll}a & b & c \\...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}\operatorname{sind} & \cos \alpha & \sin (\alpha+\bar{\delta}) \\ \sin \beta & \cos \beta & \sin (\beta+\bar{\delta})...
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$ We have, $\left|\begin{array}{ccc}1 & 1...
Solution: Option(B) To find: Value of $\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 1+3 p+2 q \\ 3 & 6+3 p & 1+6 p+3 q\end{array}\right|$ We have,...
Solution: Option(D) To find: Value of $\left|\begin{array}{lll}a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c\end{array}\right|$ We have, $\left|\begin{array}{lll}a-b...
Solution: Option(24) To find: Value of $\left|\begin{array}{lll}1 ! & 2 ! & 3 ! \\ 2 ! & 3 ! & 4 ! \\ 3 ! & 4 ! & 5 !\end{array}\right|$ We have,...
Solution: Option(A) To find: Value of $\left|\begin{array}{lll}1^{2} & 2^{2} & 3^{2} \\ 2^{2} & 3^{2} & 4^{2} \\ 3^{2} & 4^{2} & 5^{2}\end{array}\right|$ We have,...
is a complex cube root of unity then the value of 
is Solution: Option(0) To find: Value of $\left|\begin{array}{ccc}1 & \omega & 1+\omega \\ 1+\omega & 1 & \omega \\ \omega & 1+\omega & 1\end{array}\right|$ Formula used: (i)...
is a complex root of unity then 
Solution: Option(C) To find: Value of $\left|\begin{array}{ccc}1 & \omega & \omega^{2} \\ \omega & \omega^{2} & 1 \\ \omega^{2} & 1 & \omega\end{array}\right|$ We have,...
Solution: Option(C) To find: Value of $\left|\begin{array}{cc}a+i b & c+i d \\ -c+i d & a-i b\end{array}\right|$ Formula used: $\mathrm{i}^{2}=-1$ We have, $\left|\begin{array}{cc}a+i b...
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\sin 23^{\circ} & -\sin 7^{\circ} \\ \cos 23^{\circ} & \cos 7^{\circ}\end{array}\right|$ Formula used: (i) $\sin (A+B)=\sin A...
Solution: Option(C) To find: Value of $\left|\begin{array}{ll}\cos 15^{\circ} & \sin 15^{\circ} \\ \sin 15^{\circ} & \cos 15^{\circ}\end{array}\right|$ Formula used: (i) $\cos (A+B)=\cos A...
Solution: Option(B) To find: Value of $\left|\begin{array}{cc}\cos 70^{\circ} & \sin 20^{\circ} \\ \sin 70^{\circ} & \cos 20^{\circ}\end{array}\right|$ Formula used: (i) $\cos \theta=\sin...
and 
are collinear, prove that 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
be three points such that area of a is 4 sq units, find the value of .Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the area of a ABC having vertices 
and 
is 35 sq units.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points 
and 
are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
for which the points and are collinear.Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
and 
Solution: Area of a triangle $=\frac{1}{2}\left|\begin{array}{lll}\mathrm{x}_{1} & \mathrm{y}_{1} & 1 \\ \mathrm{x}_{2} & \mathrm{y}_{2} & 1 \\ \mathrm{x}_{3} & \mathrm{y}_{3}...
Solution: $\sin ^{-1}\left(\frac{-1}{2}\right)=-\sin ^{-1}\left(\frac{1}{2}\right)\left[\right.$ Formula: $\left.\sin ^{-1}(-x)=\sin ^{-1}(x)\right]$ $=-\frac{\pi}{6}$
Solution: When two different dice are thrown, the total number of outcomes = 36. Let E1 be the event of getting the sum of the numbers on the two dice is 10. These numbers are (4 ,6), (5,5)...
appears on the third toss if it is given that 
and 
appear respectively on first two tosses.As per the given question, $n\;(A)\;=\;36$ And, let $B$ be the event of $6\;and\;5$ appearing respectively on first two tosses, if the die is tossed three times
a) Current is given as I = V/R from the Ohm’s law Therefore, I = 1A But, I = ne Avd vd = I/neA When the values for the above parameters are substituted, vd = 1/1.6 × 10-4 m/s The KE = (KE of one...
Equivalent resistance of the potentiometer = 50 Ohm + R’ Equivalent voltage across the potentiometer = 10 V Current through the main circuit I = 10/(50 Ohms +R’) Potential difference across wire of...
Power consumption in a day = 10 units Power consumption per hour = 2 units Power consumption = 2 units = 2 kW = 2000 J/s Power consumption in resistors, P = VI Which gives I = 9A We know that...
Kirchhoff’s law is applied at c, I1 = I + I2 Kirchhoff’s law is applied at efbae, 10 – IR – 10I2 = 0 10 = IR + 10I1 Kirchhoff’s law is applied at cbadc, -2-IR+5I2 = 0 2 = 5I2- RI I2 = I1 – I...
Reff is the equivalent internal resistance of the battery Veff is the equivalent voltage of the battery Using Ohm’s law, I = Veff/(Reff + R) When the resistance and effective voltage are increased...
Resistance of wire R = ρ l/A Where A is the cross-sectional area of the conductor L is the length of the conductor ρ is the specific resistance RA = ρl/π(10-3 × 0.5)2 RB = ρl/ π[10-3)2 × (0.5 ×...
Effective emf of two cells = E + E = 2E Effective resistance = R + r1 + r2 Electric current is given as I = 2E/R+r1+r2 Potential difference is given as V1 – E – Ir1 = 0 Which f=gives R = r1 –...
Applying Ohm’s law, equivalent emf of the two cells = 6 – 4 = 2V Equivalent resistance = 2 + 8 = 10 Ω Electric current, I = 6-4/2+8 = 0.2A When the loop is considered in the anti-clockwise...
The refractive index of the cylinder is -1 and is placed in the air of μ = 1 AB is incident at B to the cylinder such that θr will be negative θ1= θi = θr Total deviation of the outcoming ray is...
n(r) = 1 + 2GM/rc2
Let the height of the long vertical column with transparent liquid be h and dx be the thickness The angle at which the ray AB enters is θ Let y be the new height of the liquid (θ + d θ) is the...
i) Power lenses are required to see distant objects 1/f = 1/v – 1/u 1/f = 1/10 f = -10 cm = -0.1 m P = 1/f P = 1/(-0.1) P = -10 diopter ii) When no corrective lens used Pn = Pf + Pa u = -10 cm =...
u = -x1 v = +(D – x1) 1/D – x1 – 1/(-x1) = 1/f u = -x2 v = +(D – x2) 1/D – x2 – 1/(-x2) = 1/f D = x1 + x2 d = x2 – x1 x1 = D – d/2 D – x1 = D + d/2 u = D/2 + d/2 v = D/2 – d/2 m1 = D – d/D + d m2/m1...
1/v = 1/u + 1/f = 1/-50 + 1/25 = 1/50 v = 50 cm Magnification is m = v/u = -50/50 = -1 Therefore, the coordinates of the image are (50 cm, -1 cm) ...
Distance at which the bowl should be placed in the disc is given as: d = μ(a2 – b2)/√(a + r)2 – μ(a – r)2
Answer: (a) The speed of the car in the rear is 65 km h–1. Negative refractive index materials react to Snell's law in the exact opposite direction. When a...
Answer: c) the beam of blue light would undergo total internal reflection
Answer: a) 7.5o The distance between the refracting surfaces is negligible with thin prisms, thus the prism angle (A) is very small. Because A = r1 + r2, if A is tiny, both r1 and r2 will be little...
Effective emf of two cells = E + E = 2E Effective resistance = R + r1 + r2 Electric current is given as I = 2E/R+r1+r2 Potential difference is given as V1 – E – Ir1 = 0 Which f=gives R = r1 –...
The current is represented as I = E/R+nR when the resistors are connected in series. Current is expressed as 10I = E/(R+R/n) when the resistors are connected in parallel....
The current is represented as I = E/R+nR when the resistors are connected in series. Current is expressed as 10I = E/(R+R/n) when the resistors are connected in parallel. We get n = 10 by solving...
The graphic depiction is as follows: The resistance r is connected across the external resistance R, and E is the cell's emf. V = ER/R+r is the connection between voltage and R....
The positive terminal of cell E1 is linked to E, and E is connected to X. Furthermore, E1 > E ii) cell E1's negative terminal is linked to X.
P = i2Rc is the power utilised by transmission lines. The resistance of connecting wires is denoted by Rc. P = VI is the formula for calculating power. Power...
Power consumed by the transmission lines is given as P = i2Rc. Where Rc is the resistance of connecting cables. Power is given as P = VI. The transmission of power takes places either during low...
Alloys are used in the making of the standard resistance coils because they have less temperature coefficient of resistance and the temperature sensitivity is also less.
The main considerations in the selection of the wires is the conductivity of the metal, cost of metal, and their availability.
The advantage of using thick metallic strips is that the resistance of these strips is negligible.
The advantage of a null-point in the Wheatstone bridge is that the resistance of the galvanometer is not affected by the balance point. The R unknown is calculated by using Kirchhoff’s rule.
Relaxation time is the time interval between two successive collisions of the electrons.It is defined asτ = mean free path/rms velocity of electrons usually, the drift velocity of the electrons is...
The momentum is not conserved when the charge crosses a junction in an electric circuit. This is because the drift velocity is proportional to the electric field.
The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T
The correct answer is a) the meter bridge can have no other neutral point for this set of resistances c) when the jockey contacts a point on a meter wire to the right of D, current flows from B to...
solution:The correct answer is a) number of charge carriers can change with temperature T b) time interval between two successive collisions can depend on T
solution: The correct answer is a) potential drop across AB is nearly constant as R’ is varied d) I ≥V/r+R always
solution: The correct answer is b) conservation of charge d) the fact that there is no accumulation of charges at a junction
solution: The correct answer is a) drift velocity alone
solution:The correct solution is a) maximum when the battery is connected across 1 cm × 1/2 cm faces
Solution: The correct solution is b) the potentiometer's battery can be set to 15V and R adjusted so that the potential drop across the wire is a little higher than 10V.
solution:The correct answer is c) he should change S to 3 Ω and repeat the experiment
solution: The correct answer is a) the equivalent emf εeq of the two cells is between ε1 and ε2 that is ε1 < εeq < ε2
solution: The correct answer is b) electric field produced by charges accumulated on the surface of wire
Given $x+y \geq 4$ $\Rightarrow y \geq 4-x$ Consider the equation $y=4-x$. Finding points on the coordinate axes: If $x=0$, the $y$ value is 4 i.e, $y=4$ $\Rightarrow$ the point on the $Y$ axis is...
is continues at 
Solution: L.H.L.: $\lim _{x \rightarrow 2-} f(x)=\lim _{x \rightarrow 2-} x^{2}$ $=4$ R.H.L.: $\lim _{\mathrm{x} \rightarrow 2^{*}} \mathrm{f}(\mathrm{x})=\lim _{\mathrm{x} \rightarrow 2^{*}}...
Solution: (i) We can write the given question as, $\sin ^{-1} \frac{1}{2}-2 \sin ^{-1} \frac{1}{\sqrt{2}}=\sin ^{-1} \frac{1}{2}-\sin ^{-1}\left(2 \times \frac{1}{\sqrt{2}}...
Solution: (i) Suppose $\sin ^{-1}\left(\cos \frac{3 \pi}{4}\right)=\mathrm{y}$ Therefore we can write the above equation as $\sin \mathrm{y}=\cos \frac{3 \pi}{4}=-\sin \left(\pi-\frac{3...
Solution: (i) It is given that functions can be written as $\sin ^{-1}\left(\frac{\sqrt{3}-1}{2 \sqrt{2}}\right)=\sin ^{-1}\left(\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}\right) $Taking $1 /...
Solution: $(i)$ Let $\sin ^{-1}\left(\frac{-\sqrt{3}}{2}\right)=y$ Therefore $\sin y=\left(\frac{-\sqrt{3}}{2}\right)$ $\begin{array}{l} =-\sin \left(\frac{\pi}{3}\right) \\ =\sin...
and Model 
. Model takes 6 man-hours to make per unit, while Model Y takes 10 man-hours per unit. There is a total of 450 man-hour available per week. Handling and Marketing costs are Rs 2000 and Rs 1000 per unit for Models 
and 
respectively. The total funds available for these purposes are Rs 80,000 per week. Profits per unit for Models 
and 
are Rs 1000 and Rs 500 , respectively. How many bikes of each model should the manufacturer produce so as to yield a maximum profit? Find the maximum profit.Solution: Let us take $\mathrm{x}$ an $\mathrm{y}$ to be the no. of models of bike produced by the manufacturer. From the question we have, Model $x$ takes 6 man-hours to make per unit Model $y$...
subject to 
.Solution: It is given that: $Z=x+y$ subject to constraints, $x+4 y \leq 8$ $2 x+3 y \leq 12,3 x+y \leq 9, x \geq 0, y \geq 0$ Now construct a constrain table for the above, we have Here, it can be...
Solution: According to the solution of exercise 15, we have Maximize $Z=x+y$, subject to the constraints $2 x+3 y \leq 120 \ldots$ (i) $8 x+5 y \leq 400 \ldots$ (ii) $x \geq 0, y \geq 0$ Let's...
Solution: According to the solution of exercise 14, we have Maximize $Z=200 x+120 y$ subject to constrains $\begin{array}{l} 3 x+y \leq 600 \ldots \text { (i) } \\ x+y \leq 300 \ldots \text { (ii) }...
Solution: According to the solution of exercise 13, we have The objective function for maximum profit $\mathrm{Z}=100 \mathrm{x}+170 \mathrm{y}$ Subject to constraints, $x+4 y \leq 1800 \ldots(i)$...
Solution: According to the solution of exercise 12, we have The objective function for minimum cost is $\mathrm{Z}=400 \mathrm{x}+200 \mathrm{y}$ Subject to the constrains; $5 x+2 y \geq 30 \ldots...
Solution: According to the solution of exercise 11, we have Maximize $Z=50 x+60 y$ subject to the constraints $20 x+10 y \leq 2002 x+y \leq 20 \ldots$ (i) $10 x+20 y \leq 120 x+2 y \leq 12 \ldots$...
hour. He has to spend Rs 2 per km on petrol. If he rides it at a faster speed of 
hour, the petrol cost increases to Rs 3 per km. He has at most Rs 120 to spend on petrol and one hour’s time. He wishes to find the maximum distance that he can travel. Express this problem as a linear programming problem.Solution: Suppose the man covers $\mathrm{x} \mathrm{km}$ on his motorcycle at the speed of $50 \mathrm{~km} / \mathrm{hr}$ and covers $\mathrm{y} \mathrm{km}$ at the speed of $50 \mathrm{~km} /...
Solution: Suppose $\mathrm{x}$ and $\mathrm{y}$ to be the number of sweaters of type $\mathrm{A}$ and type $\mathrm{B}$ respectively. The following constraints are: $360 x+120 y \leq 72000...
screws requires 8 minutes of threading on the threading machine and 2 minutes on the slotting machine. In a week, each machine is available for 60 hours. On selling these screws, the company gets a profit of Rs 100 per box on type A screws and Rs 170 per box on type B screws. Formulate this problem as a LPP given that the objective is to maximize profit.Solution: Suppose that the company manufactures $\mathrm{x}$ boxes of type A screws and $y$ boxes of type B screws. The below table is constructed from the information provided:...
Solution: Suppose $\mathrm{x}$ and $\mathrm{y}$ to be the number of large and small vans respectively. The below constrains table is constructed from the information provided:...
Solution: Suppose $\mathrm{x}$ units of type A and $y$ units of type $\mathrm{B}$ electric circuits be produced by the manufacturer. The table is constructed from the information provided:...
Solution: It is seen from the given figure, that the corner points are as follows: $\mathrm{R}(7 / 2,3 / 4), \mathrm{Q}(3 / 2,15 / 4), \mathrm{P}(3 / 13,24 / 13)$ and $\mathrm{S}(18 / 7,2 / 7)$ On...
at each of the corner points of this region. Find the minimum value of 
, if it exists.Solution: It is given that: $Z=4 x+y$ In the figure given, $\mathrm{ABC}$ is the feasible region which is open unbounded. Here, we get $x+y=3\dots \dots(i)$ and $\quad x+2 y=4 \quad \ldots$ (ii) On...
.Solution: It is clearly seen that the evaluating table for the value of $Z$, the maximum value of $Z$ is 47 at $(3,2)$
.Solution: It is seen from the given figure, that the feasible region is $\mathrm{ABCA}$. Corner points are $\mathrm{C}(0,3), \mathrm{B}(0,5)$ and for A, we have to solve equations $x+3 y=9$ and...
.Solution: It is given that: $\mathrm{Z}=5 \mathrm{x}+7 \mathrm{y}$ and feasible region $\mathrm{OABC}$. Corner points of the feasible region are $\mathrm{O}(0,0), \mathrm{A}(7,0), \mathrm{B}(3,4)$...
if the feasible region (shaded) for a LPP is shown in Fig. 12.7.Solution: OAED is the feasible region, as shown in the figure At $A, y=0$ in eq. $2 x+y=104$ we obtain, $\mathrm{x}=52$ This is a corner point $A=(52,0)$ At $D, x=0$ in eq. $x+2 y=76$ we obtain,...
subject to the constraints: 
.Solution: It is given that: $\mathrm{Z}=13 \mathrm{x}-15 \mathrm{y}$ and the constraints $\mathrm{x}+\mathrm{y} \leq 7$, $2 x-3 y+6 \geq 0, x \geq 0, y \geq 0$ Taking $x+y=7$, we have...
, subject to the constraints: 
Solution: It is given that: $\mathrm{Z}=\mathrm{Z}=11 \mathrm{x}+7 \mathrm{y}$ and the constraints $\mathrm{x}$ $\leq 3, y \leq 2, x \geq 0, y \geq 0$ Plotting all the constrain equations it can be...
, subject to the constraints: 
Solution: It is given that: $Z=3 x+4 y$ and the constraints $x+y \leq 1, x \geq 0$ $\mathrm{y} \geq 0$ Taking $x+y=1$, we have $$\begin{tabular}{|l|l|l|} \hline$x$ & 1 & 0 \\ \hline$y$ & 0 & 1 \\...
subject to the constraints: 
Solution: It is given that: $\mathrm{Z}=11 \mathrm{x}+7 \mathrm{y}$ and the constraints $2 \mathrm{x}+\mathrm{y} \leq 6, \mathrm{x} \leq 2, \mathrm{x} \geq 0, \mathrm{y} \geq 0$ Let $2 x+y=6$...
Solution: (a) The length of perpendicular from the point $(2,3,-5)$ on the plane $x+2 y-2 z=9 \Rightarrow x+2 y-2 z-9=0$ is $\frac{\left|a x_{1}+b y_{1}+c z_{1}+d\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}$...
Solution: (a) $4 x+8 y+z-8=0$ and $y+z-4=0$ It is given that The eq. of the given planes are $4 x+8 y+z-8=0 \text { and } y+z-4=0$ It is known to us that, two planes are $\perp$ if the direction...
Solution: (a) $2 x-2 y+4 z+5=0$ and $3 x-3 y+6 z-1=0$ It is given that The eq. of the given planes are $2 x-2 y+4 z+5=0$ and $x-2 y+5=0$ It is known to us that, two planes are $\perp$ if the...
Solution: Let's say that the eq. of the plane that passes through the two-given planes $x+y+z=1$ and $2 x+3 y+4 z=5$ is $\begin{array}{l} (x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \\ (2 \lambda+1) x+(3...
Solution: It is given that Eq. of the plane passes through the intersection of the plane is given by $(3 x-y+2 z-4)+\lambda(x+y+z-2)=0$ and the plane passes through the points $(2,2,1)$ Therefore,...
-axis and parallel to ZOX plane.Solution: It is known to us that the equation of the plane $\mathrm{ZOX}$ is $\mathrm{y}=0$ Therefore, the equation of plane parallel to $\mathrm{ZOX}$ is of the form, $\mathrm{y}=\mathrm{a}$ As the...
Solution: (a) It is given that, The points are $(1,1,-1),(6,4,-5),(-4,-2,3)$. Let, $\begin{array}{l} =\left|\begin{array}{ccc} 1 & 1 & -1 \\ 6 & 4 & -5 \\ -4 & -2 & 3...
Solution: (a) It is given that, The equation of the plane. Let $\overrightarrow{\mathrm{r}}$ be the position vector of $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})$ is given by $\vec{r}=x...
Solution: (a) $2 x+3 y-z=5$ It is given that The eq. of the plane, $2 x+3 y-z=5 \ldots$. (1) The direction ratio of the normal $(2,3,-1)$ Using the formula,...
Solution: Consider the given equations $\begin{array}{l} \Rightarrow \vec{r}=(1-t) \hat{i}+(t-2) \hat{j}+(3-2 t) \hat{k} \\ \vec{r}=\hat{i}-t \hat{i}+t \hat{j}-2 \hat{j}+3 \hat{k}-2 t \hat{k} \\...
and 
Solution: It is known to us that the shortest distance between two lines $\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}$ and $\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$ is given as:...
and 
are perpendicular to each other.Solution: The equations of the given lines are $\frac{\mathrm{x}-5}{7}=\frac{\mathrm{y}+2}{-5}=\frac{\mathrm{z}}{1}$ and $\frac{\mathrm{x}}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}}{3}$ Two lines...
and 
are at right angles.Solution: The standard form of a pair of Cartesian lines is:...
Solution: It is given that Let's calculate the vector form: The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
Given, \[X\text{ }=\text{ }0,\text{ }1,\text{ }2,\text{ }3\] P(X = r) \[={{~}^{n}}{{C}_{r}}~{{p}^{r}}~{{q}^{n-r}}\] Where \[n\text{ }=\text{ }3,\text{ }p\text{ }=\text{ 1/2},\text{ }q\text{ }=\text{...
![\[{{e}^{x}}~dy\text{ }+\text{ }\left( y\text{ }{{e}^{x}}~+\text{ }2x \right)\text{ }dx\text{ }=\text{ }0\text{ }isA.\text{ }x\text{ }ey\text{ }+\text{ }{{x}^{2}}~=\text{ }C\text{ }B.\text{ }x\text{ }ey\text{ }+\text{ }{{y}^{2}}~=\text{ }C\text{ }C.\text{ }y\text{ }ex\text{ }+\text{ }{{x}^{2}}~=\text{ }C\text{ }D.\text{ }y\text{ }ey\text{ }+\text{ }{{x}^{2}}~=\text{ }C\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5083fc3f37724ded34eb973899cb501_l3.png "Rendered by QuickLaTeX.com")
Therefore, the correct option is option(c).
SOLUTION: Therefore, yje correct option is option(c).
![\[\mathbf{A}.\text{ }\mathbf{xy}\text{ }=\text{ }\mathbf{C}\text{ }\mathbf{B}.\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{C}{{\mathbf{y}}^{\mathbf{2}}}~\mathbf{C}.\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{Cx}\text{ }\mathbf{D}.\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{C}{{\mathbf{x}}^{\mathbf{2}}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1d706555279c605d673ea5651ada04da_l3.png "Rendered by QuickLaTeX.com")
Given question is Therefore, the correct option is OPTION(C)
![\[\left( \mathbf{1}\text{ }+\text{ }{{\mathbf{e}}^{\mathbf{2x}}} \right)\text{ }\mathbf{dy}\text{ }+\text{ }\left( \mathbf{1}\text{ }+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)\text{ }{{\mathbf{e}}^{\mathbf{x}}}~\mathbf{dx}\text{ }=\text{ }\mathbf{0}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a741e3dfd2b281497b24b3de70542d10_l3.png "Rendered by QuickLaTeX.com")
On integrating, we get, ⇒ sin-1x + sin-1y = C
(x -a)2 + (y –a)2 = a2 …………1 differentiating eq 1 with respect to x, we get, \[\begin{array}{*{35}{l}} 2\left( x-a \right)\text{ }+\text{ }2\left( y-a \right)\text{ }dy/dx~=\text{ }0 \\ \Rightarrow...
![\[{{\mathbf{x}}^{\mathbf{2}}}~-\text{ }{{\mathbf{y}}^{\mathbf{2}}}~=\text{ }\mathbf{c}\text{ }{{\left( {{\mathbf{x}}^{\mathbf{2}}}~+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)}^{\mathbf{2}}}~\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3aba527b09857b2a939cd844ff559f93_l3.png "Rendered by QuickLaTeX.com")
![\[({{\mathbf{x}}^{\mathbf{3}}}-\mathbf{3x}{{\mathbf{y}}^{\mathbf{2}}})\text{ }\mathbf{dx}\text{ }=\text{ }\left( {{\mathbf{y}}^{\mathbf{3}}}-\mathbf{3}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{y} \right)\text{ }\mathbf{dy},\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f17e91ba3477eee6c20ed07cc8909f13_l3.png "Rendered by QuickLaTeX.com")
(iv) 
(ii) 
(ii) 
Therefore, its degree is three.
is 
Therefore, the correct option is OPTION(D)
Therefore, the correct option is OPTION(C).
Thus, equation (2) becomes: \[\begin{array}{*{35}{l}} 0\text{ }+\text{ }2\text{ }-\text{ }4\text{ }=\text{ }C\text{ }{{e}^{0}} \\ \Rightarrow ~C\text{ }=\text{ }-2 \\ \end{array}\] Substituting C...
Since, the curve passes through origin. Thus, equation 2 becomes 1 = C Substituting C = 1 in equation 2, we get, \[x\text{ }+\text{ }y\text{ }+\text{ }1\text{ }=\text{ }{{e}^{x}}\] Therefore, the...
⇒ x = 3y2 + Cy
$x=\frac{{{y}^{3}}}{3}+c$ $x=\frac{{{y}^{2}}}{3}+\frac{c}{y}$
![\[\theta \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-28c6cbc3df8e19af9d0e93bce5df4065_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{a}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f875b3d145d74fcad5c39bf1d50af3ef_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{b}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-80569b5b72a6d73d223867aa2d36880d_l3.png "Rendered by QuickLaTeX.com")
![\[\left| \overrightarrow{a}.\overrightarrow{b} \right|=\left| \overrightarrow{a}\times \overrightarrow{b} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f4eaf9cf090a16866e02da8ae1036cdb_l3.png "Rendered by QuickLaTeX.com")
![\[0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4f4fbf425b3e7b1e844d7922169d13c_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{\pi }{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-971c5664ffae1358e7ee692b1c572379_l3.png "Rendered by QuickLaTeX.com")
![\[\frac{\pi }{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4fd1d21095f8f8c17211b419ebb4d9b0_l3.png "Rendered by QuickLaTeX.com")
![\[\pi \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-68a8339cbb5250772acfbc5da356942d_l3.png "Rendered by QuickLaTeX.com")
![\[\widehat{i}.(\widehat{j}\times \widehat{k})+\widehat{j}.(\widehat{i}\times \widehat{k})+\widehat{k}(\widehat{i}\times \widehat{j})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fec772ab5ec0808c89d7a1b01f17091d_l3.png "Rendered by QuickLaTeX.com")
![\[-1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-644e6ba3fd2149684207ec19c2f45db3_l3.png "Rendered by QuickLaTeX.com")
![\[1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-64318110f3a374fb786ef713c0b0c575_l3.png "Rendered by QuickLaTeX.com")
![\[3\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-df8bcc09db12acd0f6bda88b4b8135fb_l3.png "Rendered by QuickLaTeX.com")
It is given that, Hence the correct answer is C.
![\[\overrightarrow{a}+\overrightarrow{b}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-47f90085f39b958a039f796aceae72d2_l3.png "Rendered by QuickLaTeX.com")
![\[\theta =\frac{\pi }{4}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc4ad60471520b848d28165e99d4ee0e_l3.png "Rendered by QuickLaTeX.com")
![\[\theta =\frac{\pi }{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6e401bf6731e55f12cfbb80436b54987_l3.png "Rendered by QuickLaTeX.com")
![\[\theta =\frac{\pi }{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-024d7e0a3efe01b953a5c076a7a5c638_l3.png "Rendered by QuickLaTeX.com")
![\[\theta =\frac{2\pi }{3}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-110cc2690a75e8489e60c8cab4590ada_l3.png "Rendered by QuickLaTeX.com")
Here the correct answer is option d
![\[\overrightarrow{a}.\overrightarrow{b}\ge 0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b4dfaf184691a3ef98fcc0711022cbab_l3.png "Rendered by QuickLaTeX.com")
![\[0<\theta <\frac{\pi }{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1fe8ec8f0f58a1e2bdd8049591505ed7_l3.png "Rendered by QuickLaTeX.com")
![\[0\le \theta \le \frac{\pi }{2}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6f0a816c563c90f283e9520d68aa9ae0_l3.png "Rendered by QuickLaTeX.com")
![\[0<\theta <\pi \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-db329ef216355048be6e558280aae73a_l3.png "Rendered by QuickLaTeX.com")
![\[0\le \theta \le \pi \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6068637deeff1adbedd1def6ede23295_l3.png "Rendered by QuickLaTeX.com")
![\[(\overrightarrow{a}+\overrightarrow{b}).(\overrightarrow{a}-\overrightarrow{b})={{\left| \overrightarrow{a} \right|}^{2}}+{{\left| \overrightarrow{b} \right|}^{2}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-330d901690d8cf6234407058062682fb_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{a}\ne 0,\overrightarrow{b}\ne 0\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8b1236addf6d9c4c1412a8aa7b5efc47_l3.png "Rendered by QuickLaTeX.com")
It is given that Hence proved.
![\[\overrightarrow{c}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-28e5ac2bb179eb1056bcdac7884094ee_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9e11e6c21752131f88e75db095b19841_l3.png "Rendered by QuickLaTeX.com")
let us assume,
![\[\widehat{i}+\widehat{j}+\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9182cc705b62215c9f98552abb7afa12_l3.png "Rendered by QuickLaTeX.com")
![\[2\widehat{i}+4\widehat{j}-5\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a6681754ddd5c454be051d9879e7154e_l3.png "Rendered by QuickLaTeX.com")
![\[\lambda \widehat{i}+2\widehat{j}+3\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a940bd8c10c3cea99b5fc268c8c30895_l3.png "Rendered by QuickLaTeX.com")
![\[\lambda \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1e5278423fe0cd24b56228ebd2dbfcd2_l3.png "Rendered by QuickLaTeX.com")
Let us consider
![\[\overrightarrow{a}=\widehat{i}+4\widehat{j}+2\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-edd4804415df8cbdae1204c94d7db6c6_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{b}=3\widehat{i}-2\widehat{j}+7\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c06b242e8db2c0dc3306d8d293afa6bc_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{c}=2\widehat{i}-\widehat{j}+4\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1257de77b94154648a39fbd4f389df23_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{d}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-6e24df3bbd1902b7f7ac0f8454c6d0dd_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{c}.\overrightarrow{d}=15\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4672432386a0d5666406085e8fa67cc6_l3.png "Rendered by QuickLaTeX.com")
Assume,
![\[\frac{1}{\sqrt{3}}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fc27ffa7784ce20e0f96b0c1052c51c5_l3.png "Rendered by QuickLaTeX.com")
Firstly, Let’s assume a vector to be equally inclined to axes OX, OY, and OZ at angle \[\alpha \]. Then, the direction cosines of the vector are \[\cos \alpha \],\[\cos \alpha \]and \[\cos \alpha...
![\[2\widehat{i}-4\widehat{j}+5\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-4766f36f7805e520fba218aa3951172e_l3.png "Rendered by QuickLaTeX.com")
![\[\widehat{i}-2\widehat{j}-3\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-9271dbf9cd7a1e0d4b6579035bdd8b97_l3.png "Rendered by QuickLaTeX.com")
we know that,
![\[(2\overrightarrow{a}+\overrightarrow{b})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f89637921312deab284f70a94ca009eb_l3.png "Rendered by QuickLaTeX.com")
![\[(\overrightarrow{a}-3\overrightarrow{b})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-e04b8bab322f492b3c06c77069263be5_l3.png "Rendered by QuickLaTeX.com")
![\[1:2\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-98057744c90982b103f0706671126b04_l3.png "Rendered by QuickLaTeX.com")
we know that,
![\[(1,-2,-8)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-489c52c99a8f3fe514ef8367dd634d09_l3.png "Rendered by QuickLaTeX.com")
![\[(5,0,-2)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-099ce1dc50848cdaa3e15192948c9f3e_l3.png "Rendered by QuickLaTeX.com")
![\[(11,3,7)\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-844c5d13f81314eb0793e67539dc0207_l3.png "Rendered by QuickLaTeX.com")
Let us consider
![\[\overrightarrow{a}=\widehat{i}+\widehat{j}+\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-35d857b3023b451e1d87a34e4b17337b_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{b}=2\widehat{i}-\widehat{j}+3\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-bb09664bac6f93e8d54203ba79800eb9_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{c}=\widehat{i}-2\widehat{j}+\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-c64e04d81b624079d16b7d4525a27c85_l3.png "Rendered by QuickLaTeX.com")
![\[2\overrightarrow{a}-\overrightarrow{b}+3\overrightarrow{c}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-1c9977111157a87f7a4a78223e574283_l3.png "Rendered by QuickLaTeX.com")
Let us consider
![\[\overrightarrow{a}=2\widehat{i}+3\widehat{j}-\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-441cbddb8a4a7e4497122556cfb6a7b7_l3.png "Rendered by QuickLaTeX.com")
![\[\overrightarrow{b}=\widehat{i}-2\widehat{j}+\widehat{k}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-30accd4139b3230fd0fe12fd0c6b06ae_l3.png "Rendered by QuickLaTeX.com")
let us consider,
![\[x(\widehat{i}+\widehat{j}+\widehat{k})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-217e3c386db4107a945319020f578916_l3.png "Rendered by QuickLaTeX.com")
we know ,
![\[\overrightarrow{a}=\overrightarrow{b}+\overrightarrow{c}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-7eb9b063f3664d4111d9371d73353778_l3.png "Rendered by QuickLaTeX.com")
![\[\left| \overrightarrow{a} \right|=\left| \overrightarrow{b} \right|+\left| \overrightarrow{c} \right|\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-867d699a5dde3474d65f59a94fbddd72_l3.png "Rendered by QuickLaTeX.com")
It is given that,
![\[4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3c0fa656ae249b0b1e03c8cd2e7be583_l3.png "Rendered by QuickLaTeX.com")
![\[{{30}^{\circ }}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-34cd53219eaf0763feb7ca785cfc2855_l3.png "Rendered by QuickLaTeX.com")
It is given that, Let O and B be the initial and final positions of the girl respectively. Then, the girl’s position can be shown as:
![\[({{x}_{1}},{{y}_{1}},{{z}_{1}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8e59b5607a276459e63c9744ae8cf498_l3.png "Rendered by QuickLaTeX.com")
![\[({{x}_{2}},{{y}_{2}},{{z}_{2}})\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-34b4e10a50394db0a5a87bc7c7c59790_l3.png "Rendered by QuickLaTeX.com")
let us consider,
let us consider,
Solution: Given: The origin $(0,0,0)$ and the point $(5,-2,3)$ It is known to us that The vector eq. of as line which passes through two points whose position vectors are $\vec{a}$ and $\vec{b}$ is...
and parallel to the line given by 
Solution: It is given that The points $(-2,4,-5)$ It is known that Now, the Cartesian equation of a line through a point $\left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right)$ and having...
Solution: Given that, Line passes through the point $(1,2,3)$ and is parallel to the vector. It is known to us that Vector eq. of a line that passes through a given point whose position vector is...
Solution: The points $(4,7,8),(2,3,4)$ and $(-1,-2,1),(1,2,5)$. Consider $A B$ be the line joining the points, $(4,7,8),(2,3,4)$ and $C D$ be the line through the points $(-1,-2$, 1), $(1,2,5)$. So...
Solution: Given that The points $(1,-1,2),(3,4,-2)$ and $(0,3,2),(3,5,6)$. Let's consider $A B$ be the line joining the points, $(1,-1,2)$ and $(3,4,-2)$, and $C D$ be the line through the points...
D. y2dx + (x2 – x y – y2) dy = 0
can be solved by making the substitution. (A) y = v x (B) v = y x (C) x = v y (D) x = v(C) x = v y
The required solution of the differential equation.
![[x{{\sin }^{2}}x(\frac{y}{x})-y]dx+xdy=0;y=\frac{\pi }{4}whenx=1](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf6ad73b49448015b760c7d33bebe7d1_l3.png "Rendered by QuickLaTeX.com")
![\[{{\mathbf{x}}^{\mathbf{2}}}\mathbf{dy}\text{ }+\text{ }\left( \mathbf{x}\text{ }\mathbf{y}\text{ }+\text{ }{{\mathbf{y}}^{\mathbf{2}}} \right)\mathbf{dx}\text{ }=\text{ }\mathbf{0};\text{ }\mathbf{y}\text{ }=\text{ }\mathbf{1}\text{ }\mathbf{when}\text{ }\mathbf{x}\text{ }=\text{ }\mathbf{1}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-3e161df78c925b8b559ca101f5b1ad69_l3.png "Rendered by QuickLaTeX.com")