JEE Main & Advanced

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 \[a\alpha +b\beta +c\gamma +d=0\].

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.

(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 \[\left| \,\begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}}  \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}}  \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}}  \\ \end{matrix}\, \right|=0\].

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 \[\left| \,\begin{matrix} x-{{x}_{2}} & y-{{y}_{2}} & z- {{z}_{2}}  \\ {{l}_{1}} & {{m}_{1}} & {{n}_{1}}  \\ {{l}_{2}} & {{m}_{2}} & {{n}_{2}}  \\ \end{matrix}\, \right|=0\].

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) more...

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.

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 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.

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 more...

(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 \[OY\] more...