The equations of the three circles are $(x - x_1)^2 + (y - y_1)^2 = r_1^2 \hspace{30pt}(1)$ $(x - x_2)^2 + (y - y_2)^2 = r_2^2 \hspace{30pt}(2)$ $(x - x_3)^2 + (y - y_3)^2 = r_3^2 \hspace{30pt}(3)$...

Since the two ellipsoid share the same center, then we can take this center to be the origin of the coordinate system, and then the equations of the two ellipsoids will be $r^T Q_1 r = 1$ and $r^T ... View answer 5 votes Let$x = 2 \cos \theta$Then$x + \sqrt{4 - x^2} = 2 \cos \theta + 2 \sin \theta = 2 (\cos \theta + \sin \theta)$Now, the trigonometric function$ a \cos \theta + b \sin \theta $satisfies:$$-\sqrt{... View answer Accepted answer 5 votes$ \theta \cdot x = | \theta | | x | \cos \psi = \dfrac{1}{y}$where$\psi$is the angle between the vector$x$and the vector$\theta$. Since$x$and$y$are fixed, this implies the minimum$| \theta |...

The equation of a circular cone is determined by two things: It's axis represented by the unit vector $\mathbf{a}$, and the semi-vertical angle which is the angle $\theta$ ​between the axis and the ...

From Cauchy-Schwartz inequality, $a^2 + b^2 = (a^2 + b^2) ( \dfrac{1}{2} + \dfrac{1}{2} ) \ge (\dfrac{1}{\sqrt{2}} a + \dfrac{1}{\sqrt{2}} b )^2 = \dfrac{1}{2} (a + b)^2$ Hence $(a + b)^2 \le 2 (a^2 + ... View answer 3 votes$ 3 \sec x - 2 \cot x \gt 0 $Implies$ \dfrac{3}{\cos x} \gt 2 \dfrac{\cos x }{\sin x} $Multiply through by$\sin^2 x \cos^2 x  3 \cos x \sin^2 x \gt 2 \sin x \cos^3 x $Hence,$ \cos x \sin x ( ...

Since $0 \gt a \gt b$ both $a$ and $b$ are negative, so their product is positive. Divide the inequality by $ab$, the inequality sign remains the same. So $\dfrac{0}{ab} \gt \dfrac{a}{ab} \gt \... View answer Accepted answer 3 votes If you rotate both$a$and$b$by the matrix$R$then it is evident (without proof) that$ (R a) \times (R b) = R (a \times b ) $The left hand side$[ R a ]_\times (R b) $and the right hand side is$...

Let $n$ be the number of points. If $n\le 4$ and the points are not collinear, then you can always find an ellipse on which the points lie. If $n \ge 5$, then take any $5$ of the points, and construct ...

The equation of the line is $P(t) = P_0 + t v$ $OP(t) \times v = ((P_0 - O) + t v ) \times v = (P_0 - O) \times v + t (v \times v) = (P_0 - O) \times v$ Thus the cross product is independent of $t$....

Since point $D$ will be on $AC$, then there exists a scalar $t \in \mathbb{R}$ such that $D = A + t (C - A) \hspace{20pt}(1)$ The vector $(C-A)$ is the direction vector of the ray $AC$. Now we want ...

First find the intersection of the line $y = x$ with the circle. Plug in $y = x$ you get $(x - 3)^2 + ( x - 4)^2 = 4$ This simplifies to, $2 x^2 - 14 x + 21 = 0$ By factoring, this becomes, $2 (... View answer 2 votes Any point$P$on the perimeter or the interior of$\triangle ABC$can be written as a linear combination of the coordinates of$A,B,C$(Barycentric coordinates)$P = t A + s B + (1 - t - s) C $where ... View answer Accepted answer 2 votes If you're rotating by$90^\circ$counter clockwise about the origin, then the image of the point$(x, y)$on the line becomes:$(x', y') = (-y, x) $Thus$x = y'$and$y = -x'$. Plug these into the ... View answer 2 votes The point$P = a A + b B + c C + d D$where$a + b + c + d=1$You want to find the intersection of the plane$PCD$with the segment$AB$. Points in plane$PCD$are spanned by two vectors:$CD$and$CP$... View answer 2 votes The actual condition is$ (\mathbf{a_2 } - \mathbf{a_1} ) \cdot (\mathbf{b_2} \times \mathbf{b_1} ) = 0 $But since for any three vectors$\mathbf{a,b,c}$,$ \mathbf{a}\cdot (\mathbf{b} \times \mathbf{...

The equation of an ellipse in the plane is given by $(r - E)^T Q (r - E) = 1 \hspace{20pt} (1)$ where $E$ is the center, and $Q$ is a $2 \times 2$ symmetric matrix. At any point on the ellipse the ...

The parametric equation of the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ is $r = (x, y) = ( a \cos t , b \sin t ) \hspace{20pt} t \in \mathbb{R}$ Now $P_1 = (x_1, y_1) = (a \cos t_1, b \sin ... View answer Accepted answer 2 votes I found an excellent paper that addresses this problem in the$3D$case, but the methods developed there are applicable in a straight forward manner to the$2D$case presented in this question. We ... View answer Accepted answer 2 votes$AT = AP$so$AT = 2$Taking$B$to be the origin of the Cartesian coordinate system, then$A = (0, 3)$The unit direction vector from A down the hypotenuse is$(\cos C, - \sin C) = (4/5, -3/5) T = ...

Let the two lines be $\mathbf{P_1}(t)$ and $\mathbf{P_2}(t)$ then we can write $\mathbf{P_1}(t) = \mathbf{Q_1} + t \mathbf{V_1}$ $\mathbf{P_2} (t) = \mathbf{Q_2} + t \mathbf{V_2}$ It can be ...

Start by parameterizing $\mathbf{k_2}$ in terms of $\mathbf{k_1}$. Let $\mathbf{u_1}, \mathbf{u_2}$ be two mutually orthogonal unit vectors that are also orthogonal to $\mathbf{k_1}$, then any vector ...

$p(t) = (1 - t)^2 P_0 + 2 t (1 - t) P_1 + t^2 P_2$ Taking the derivative with respect to $t$ gives us the tangent vector $p'(t) = - 2 (1 - t) P_0 + 2 (1 - 2 t) P_1 + 2 t P_2$ At maximum $x$, the $x$...

Here is a brute force method to find the new coordinates of $p_3$. Assuming that $\phi$ is the azimuthal angle measured counter clockwise from the positive $x$ axis, and $\theta$ is the polar angle ...

Let the radius of Earth be $R_e$, and let the radius of the orbit of the satellite be $R \gt R_e$. Choose the Cartesian frame to have its origin at the center of Earth, and with $x$ axis pointing ...
Let the three lines be given by parametric equations $R_i(t) = P_{i} + t V_{i} ,\hspace{20pt} i = 1, 2, 3$ where the direction vectors $\{ V_i \}$ are assumed to be unit vectors. Given the distance $... View answer Accepted answer 2 votes$ u \cdot v = \|u\| \|v\| \cos \theta $So$ 3 = (1)(2) \cos \theta $which implies$\cos \theta = \dfrac{3}{2} $which is not possible. View answer Accepted answer 2 votes The figure above summarizes the situation. We have two similar triangles, the big one with height$h$and base$9$, and the small one with height$(h-8)$and base$5$. Therefore, by equal ratios of ... View answer 2 votes Let$v = [ \sqrt{a_1} , \sqrt{a_2} , ...., \sqrt{a_n} ]^T $and$w = [ \dfrac{1}{\sqrt{a_1}}, \dfrac{1}{\sqrt{a_2}}, ..., \dfrac{1}{\sqrt{a_n}}]^T$The by Cauchy-Schwartz inequality,$ v \cdot w \le | ...