# Current Affairs JEE Main & Advanced

#### Angle Between Line and Plane

The angle $\theta$ between the line $\frac{x-\alpha }{l}=\frac{y-\beta }{m}=\frac{z-\gamma }{n}$, and the plane $ax+by+cz+d=0$, is given by $\sin \theta =\frac{al+bm+cn}{\sqrt{({{a}^{2}}+{{b}^{2}}+{{c}^{2}})}\sqrt{({{l}^{2}}+{{m}^{2}}+{{n}^{2}})}}$.     (i) The line is perpendicular to the plane if and only if $\frac{a}{l}=\frac{b}{m}=\frac{c}{n}$.     (ii) The line is parallel to the plane if and only if $al+bm+cn=0$.     (iii) The line lies in the plane if and only if $al+bm+cn=0$ and $a\alpha +b\beta +c\gamma +d=0$.

#### Intersection Point of a Line and Plane

Algorithm for finding the point of intersection of a line and a plane     Step I : Write the co-ordinates of any point on the line in terms of some parameters $r$ (say).     Step II : Substitute these co-ordinates in the equation of the plane to obtain the value of $r$.     StepIII : Put the value of $r$ in the co-ordinates of the point in step I.

#### Equation of Plane Through a Given Line

(1) If equation of the line is given in symmetrical form as $\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}$, then equation of plane is $a(x-{{x}_{1}})+b(y-{{y}_{1}})+c(z-{{z}_{1}})=0$                             .....(i)     where $a,\,\,b,\,\,c$ are given by $al+bm+cn=0$           .....(ii)     (2) If equations of line is given in general form as ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0={{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}$, then the equation of plane passing through these line is $({{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}})$ $+\lambda ({{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}})=0$.     (3) Equation of plane through a given line parallel to another line : Let the d.c.’s of the other line be ${{l}_{2}},\,{{m}_{2}},\,{{n}_{2}}$. Then, since the plane is parallel to the given line, normal is perpendicular.     \ $a{{l}_{2}}+b{{m}_{2}}+c{{n}_{2}}=0$                                                      ……(iii)     Hence, the plane from (i), (ii) and (iii) is $\left| \,\begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix}\, \right|=0$.

#### Coplanar Lines

Lines are said to be coplanar if they lie in the same plane or a plane can be made to pass through them.     Condition for the lines to be coplanar:     If the lines $\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}$ and $\frac{x-{{x}_{2}}}{{{l}_{2}}}=$ $\frac{y-{{y}_{2}}}{{{m}_{2}}}=$ $\frac{z-{{z}_{2}}}{{{n}_{2}}}$ are coplanar, then $\left| \,\begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix}\, \right|=0$.     The equation of the plane containing them is  $\left| \,\begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix}\, \right|=0$ or $\left| \,\begin{matrix} x-{{x}_{2}} & y-{{y}_{2}} & z- {{z}_{2}} \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}} \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}} \\ \end{matrix}\, \right|=0$.

#### Image of a Point in a Plane

Let P and Q be two points and let $\pi$ be a plane such that     (i) Line PQ is perpendicular to the plane $\pi ,$ and     (ii) Mid-point of PQ lies on the plane $\pi$.     Then either of the point is the image of the other in the plane$\pi$.     To find the image of a point in a given plane, we proceed as follows     (i) Write the equations of the line passing through P and normal to the given plane as  $\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}$.       (ii) Write the co-ordinates of image Q as $({{x}_{1}}+ar,\,{{y}_{1}},\,+br,\,{{z}_{1}}+cr)$.     (iii) Find the co-ordinates of the mid-point R of PQ.     (iv) Obtain the value of r by putting the co-ordinates of R in the equation of the plane.     (v) Put the value of r in the co-ordinates of Q.

#### Equation of Planes Bisecting Angle Between Two Given Planes

Equations of planes bisecting angles between the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+d=0$ are $\frac{{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}}{\sqrt{(a_{1}^{2}+b_{1}^{2}+c_{1}^{2})}}=$ $\pm \frac{{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}}{\sqrt{(a_{2}^{2}+b_{2}^{2}+c_{2}^{2})}}$.     (i) If angle between bisector plane and one of the plane is less than ${{45}^{o}}$, then it is acute angle bisector, otherwise it is obtuse angle bisector.     (ii) If ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}$ is negative, then origin lies in the acute angle between the given planes provided ${{d}_{1}}$ and ${{d}_{2}}$ are of same sign and if ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}$ is positive, then origin lies in the obtuse angle between the given planes.

#### Locus

The curve described by a point which moves under given condition or conditions is called its locus.     Equation to the locus of a point : The equation to the locus of a point is the relation, which is satisfied by the coordinates of every point on the locus of the point.     Algorithm to find the locus of a point     Step I : Assume the coordinates of the point say $(h,\,\,k)$ whose locus is to be found.     Step II : Write the given condition in mathematical form involving $h,\,\,k$.     Step III : Eliminate the variable (s), if any.     Step  IV : Replace $h$ by $x$ and $k$ by $y$ in the result obtained in step III. The equation so obtained is the locus of the point which moves under some stated condition (s).

#### Angle Between Two Planes

Angle between the planes is defined as angle between normals to the planes drawn from any point. Angle between the planes ${{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}z+{{d}_{1}}=0$ and ${{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}z+{{d}_{2}}=0$ is ${{\cos }^{-1}}\left( \frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{(a_{1}^{2}+b_{1}^{2}+c_{1}^{2})(a_{2}^{2}+b_{2}^{2}+c_{2}^{2})}} \right)$     (i) If ${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$, then the planes are perpendicular to each other.     (ii) If $\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}=\frac{{{c}_{1}}}{{{c}_{2}}}$, then the planes are parallel to each other.

#### Foot of Perpendicular from a Point $\mathbf{A(\alpha ,}\,\,\mathbf{\beta ,}\,\,\mathbf{\gamma )}$to a given plane $\mathbf{ax+by+cz+d=0}$.

If AP be the perpendicular from A to the given plane, then it is parallel to the normal, so that its equation is $\frac{x-\alpha }{a}=\frac{y-\beta }{b}=\frac{z-\gamma }{c}=r$ , (Say)     Any point P on it is $(ar+\alpha ,\,br+\beta ,\,cr+\gamma )$. It lies on the given plane and we find the value of $r$ and hence the point P.     (1) Perpendicular distance : The length of the perpendicular from the point $P({{x}_{1}},\,{{y}_{1}},{{z}_{1}})$ to the plane $ax+by+cz+d=0$ is $\left| \,\frac{a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\, \right|$.     Distance between two parallel planes $Ax+By+Cz+{{D}_{1}}=0$ and $Ax+By+Cz+{{D}_{2}}=0$ is $\frac{{{D}_{2}}\tilde{\ }{{D}_{1}}}{\sqrt{{{A}^{2}}+{{B}^{2}}+{{C}^{2}}}}$.     (2) Position of two points w.r.t. a plane : Two points $P({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})$ and $Q({{x}_{2}},\,{{y}_{2}},{{z}_{2}})$ lie on the same or opposite sides of a plane $ax+by+cz+d=0$ according to $a{{x}_{1}}+b{{y}_{1}}+c{{z}_{1}}+d$ and $a{{x}_{2}}+b{{y}_{2}}+c{{z}_{2}}+d$ are of same or opposite signs. The plane divides the line joining the points P and Q externally or internally according to P and Q are lying on same or opposite sides of the plane.

#### Transformation of Axes

(1) Shifting of origin without rotation of axes : Let $P\equiv (x,y)$with respect to axes $OX$ and $OY$.     Let $O'\equiv (\alpha ,\beta )$ with respect to axes $OX$ and $OY$ and let $P\equiv (x',y')$ with respect to axes $O'X'$  and  $O'Y',$  where $OX$ and  $O'X'$  are parallel and $OY$ and  $O'Y'$ are parallel.         Then $x=x'+\alpha ,\text{ }y=y'\,+\beta$     or  $x'=x-\alpha ,\text{ }y'=y-\beta$     Thus if origin is shifted to point $(\alpha ,\beta )$ without rotation of axes, then new equation of curve can be obtained by putting $x+\alpha$ in place of $x$ and $y+\beta$ in place of  $y$.     (2) Rotation of axes without changing the origin : Let  $O$ be the origin. Let $P\equiv (x,y)$ with respect to axes $OX$ and $OY$ and let $P\equiv (x',y')$ with respect to axes $OX'$ and $OY'$ where $\angle X'OX=\angle YOY'=\theta$         then      $x=x'\cos \theta -y'\sin \theta$       $y=x'\sin \theta +y'\cos \theta$                and      $x'=x\cos \theta +y\sin \theta$       $y'=-x\sin \theta +y\cos \theta$     The above relation between $(x,y)$ and $(x',y')$ can be easily obtained with the help of following table
 $x\downarrow$ $y\downarrow$ $x'\to$ $y'\to$ $\cos \theta$ $-\sin \theta$ $\sin \theta$ $\cos \theta$
(3) Change of origin and rotation of axes : If origin is changed to $O'(\alpha ,\beta )$ and axes are rotated about the new origin $O'$ by an angle $\theta$ in the anti-clockwise sense such that the new co-ordinates of $P(x,y)$ become $(x',y')$ then the equations of transformation will be $x=\alpha +x'\cos \theta -y'\sin \theta$ and  $y=\beta +x'\sin \theta +y'\cos \theta$         (4) Reflection (Image of a point) : Let $(x,y)$be any point, then its image with respect to     (i) x-axis $\Rightarrow$ $(x,-y)$ more...

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