Current Affairs JEE Main & Advanced

Every matrix satisfies its characteristic equation e.g. let A be a square matrix then \[|A-xI|=0\]is the characteristics equation of A. If \[{{x}^{3}}-4{{x}^{2}}-5x-7=0\] is the characteristic equation for A, then \[{{A}^{3}}-4{{A}^{2}}+5A-7I=0\].   Roots of characteristic equation for A are called Eigen values of A or characteristic roots of A or latent roots of A.   If \[\lambda \] is characteristic root of A, then \[{{\lambda }^{-1}}\]is characteristic root of \[{{A}^{-1}}\].

(1) Reflexion in the x-axis: If \[P'\,\,(x',y')\]is the reflexion of the point \[P(x,y)\]on the x-axis, then the matrix \[\left[ \begin{matrix} 1 & 0  \\ 0 & -1  \\\end{matrix} \right]\] describes the reflexion of a point \[P(x,y)\]in the x-axis.   (2) Reflexion in the y-axis    Here the matrix is \[\left[ \begin{matrix} -1 & 0  \\ 0 & 1  \\\end{matrix} \right]\]   (3) Reflexion through the origin   Here the matrix is \[\left[ \begin{matrix} -1 & 0  \\ 0 & -1  \\ \end{matrix} \right]\]   (4) Reflexion in the line  \[\mathbf{y=x}\]   Here the matrix is \[\left[ \begin{matrix} 0 & 1  \\ 1 & 0  \\ \end{matrix} \right]\]   (5) Reflexion in the line \[\mathbf{y=}-\mathbf{x}\]   Here the matrix is \[\left[ \begin{matrix} \,\,0 & -1  \\ -1 & \,\,0  \\ \end{matrix} \right]\]   (6) Reflexion in \[y=x\,\mathbf{tan\theta }\]   Here matrix is \[\left[ \begin{matrix} \cos 2\theta  & \sin 2\theta   \\ \sin 2\theta  & -\cos 2\theta   \\ \end{matrix} \right]\]   (7) Rotation through an angle \[\mathbf{\theta }\]   Here matrix is \[\left[ \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right]\]

 We know that if \[x\] and \[y\] axis are rotated through an angle \[\theta \] about the origin the new coordinates are given by   \[x=X\,\cos \theta -Y\sin \theta \] and \[y=X\sin \theta +Y\cos \theta \]   \[\Rightarrow \left[ \begin{matrix} x  \\ y  \\ \end{matrix} \right]=\left[ \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right]\,\left[ \begin{matrix} X  \\ Y  \\ \end{matrix} \right]\Rightarrow \left[ \begin{matrix} \cos \theta  & -\sin \theta   \\ \sin \theta  & \cos \theta   \\ \end{matrix} \right]\]   is the matrix of rotation through an angle \[\theta \].  

If the angle of intersection of two spheres is a right angle, the spheres are said to be orthogonal.     Condition for orthogonality of two spheres :     Let the equation of the two spheres be     \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\]               .....(i)     and         \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2{u}'x+2{v}'y+2{w}'z+{d}'=0\]            .....(ii)     If the sphere (i) and (ii) cut orthogonally, then \[2u{u}'+2v{v}'+2w{w}'=d+{d}',\] which is the required condition.    

• If the spheres \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{a}^{2}}\] and \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}\] \[+2ux+2vy+2wz+d=0\] cut orthogonally, then \[d={{a}^{2}}\].

   

• Two spheres of radii \[{{r}_{1}}\] and \[{{r}_{2}}\] cut orthogonally, then the radius of the common circle is \[\frac{{{r}_{1}}{{r}_{2}}}{\sqrt{r_{1}^{2}+r_{2}^{2}}}\].

Let the equations of the sphere and the straight line be      \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\]               ……(i)     and         \[\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}=r\],  (say)                    …..(ii)     Any point on the line (ii) is \[(\alpha +lr,\beta +mr,\gamma +nr)\].     If this point lies on the sphere (i) then we have,     \[{{(\alpha +lr)}^{2}}+{{(\beta +mr)}^{2}}+{{(\gamma +nr)}^{2}}+2u(\alpha +lr)+2v(\beta +mr)\]\[+2w(\gamma +nr)+d=0\]     or, \[{{r}^{2}}[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}]+2r[l(u+\alpha )+m(v+\beta )]+n(w+\gamma )]\] \[+\,({{\alpha }^{2}}+{{\beta }^{2}}+{{\gamma }^{2}}+2u\alpha +2v\beta +2w\gamma +d)=0\] ……(iii)     This is a quadratic equation in r and so gives two values of r and therefore the line (ii) meets the sphere (i) in two points which may be real, coincident and imaginary, according as root of (iii) are so.     If l, m, n are the actual d.c.’s of the line, then \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\] and then the equation (iii) can be simplified.

A plane touches a given sphere if the perpendicular distance from the centre of the sphere to the plane is equal to the radius of the sphere. The plane \[lx+my+nz=p\] touches the sphere \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\], If \[{{(ul+vm+wn-p)}^{2}}\]= \[({{l}^{2}}+{{m}^{2}}+{{n}^{2}})({{u}^{2}}+{{v}^{2}}+{{w}^{2}}-d)\].

Consider a sphere intersected by a plane. The set of points common to both sphere and plane is called a plane section of a sphere. The plane section of a sphere is always a circle. The equations of the sphere and the plane taken together represent the plane section.         Let C be the centre of the sphere and M be the foot of the perpendicular from C on the plane. Then M is the centre of the circle and radius of the circle is given by \[PM=\sqrt{C{{P}^{2}}-C{{M}^{2}}}\].     The centre M of the circle is the point of intersection of the plane and line CM which passes through C and is perpendicular to the given plane.     Centre : The foot of the perpendicular from the centre of the sphere to the plane is the centre of the circle.     \[{{(\text{Radius of circle})}^{\text{2}}}\text{= (Radius of sphere}{{)}^{\text{2}}}\text{-- (Perpendicular from centre of spheres on the plane}{{)}^{\text{2}}}\]     Great circle : The section of a sphere by a plane through the centre of the sphere is a great circle. Its centre and radius are the same as those of the given sphere.

(1) Equation of sphere with given centre and radius : The equation of a sphere with centre  \[(a,b,c)\] and radius R is \[{{(x-a)}^{2}}+{{(y-b)}^{2}}+{{(z-c)}^{2}}={{R}^{2}}\]                         ……(i)     If the centre is at the origin, then equation (i) takes the form \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{R}^{2}}\], which is known as the standard form of the equation of the sphere.     (2) Diameter form of the equation of a sphere : If \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] and \[({{x}_{2}},\,{{y}_{2}},{{z}_{2}})\] are the co-ordinates of the extremities of a diameter of a sphere, then its equation is \[(x-{{x}_{1}})(x-{{x}_{2}})+(y-{{y}_{1}})(y-{{y}_{2}})+(z-{{z}_{1}})(z-{{z}_{2}})=0\].

The general equation of a sphere is \[{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2ux+2vy+2wz+d=0\] with centre \[(u,v,w)\] i.e., \[(-(1/2)\] coefficient of \[x,\,\,-(1/2)\] coefficient of \[y,\,\,-(1/2)\] coefficient of z) and, radius \[=\sqrt{{{u}^{2}}+{{v}^{2}}+{{w}^{2}}-d}\].

If P be the point of intersection of given line and plane and Q be the foot of the perpendicular from any point on the line to the plane then PQ is called the projection of given line on the given plane.   Image of line about a plane: Let line is \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\] \[\frac{z-{{z}_{1}}}{{{c}_{1}}}\], plane is \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+d=0\].   Find point of intersection (say P) of line and plane. Find image (say Q) of point \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] about the plane. Line PQ is the reflected line.