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} &...
Solution: 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 \\...
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}...
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...
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...
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...
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,...
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,...
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)...
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...
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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: 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: $\left.\sin ^{-1}(-x)=\sin ^{-1}(x)\right]$ $=-\frac{\pi}{6}$
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
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...
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...
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 $y$...
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) $10 x+20 y \leq 120 x+2 y \leq 12 \ldots$...
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...
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...
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 $\mathrm{O}(0,0), \mathrm{A}(7,0), \mathrm{B}(3,4)$...
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,...
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...
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...
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 \\...
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,...
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} \\...
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:...
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...
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{...
Therefore, the correct option is option(c).
SOLUTION: Therefore, yje correct option is option(c).
Given question is Therefore, the correct option is OPTION(C)
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...
Therefore, its degree is three.
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
$x=\frac{{{y}^{3}}}{3}+c$ $x=\frac{{{y}^{2}}}{3}+\frac{c}{y}$
It is given that, Hence the correct answer is C.
Here the correct answer is option d
It is given that Hence proved.
let us assume,
Let us consider
Assume,
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...
we know that,
we know that,
Let us consider
Let us consider
let us consider,
we know ,
It is given that,
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:
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...
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
(C) x = v y
The required solution of the differential equation.
Let us consider,
We know that,
x2 + y2 = Cx
Firstly let us consider,
It is given that,
It is given that
It is given that,
Firstly consider the LHS, We have,
Firstly,
The correct option is option(A) ex + e-y = C
It is given that:
It is given that: