JEE Main & Advanced

(1) Equation of plane passing through a given point : Equation of plane passing through the point \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] is \[A(x-{{x}_{1}})+B(y-{{y}_{1}})+C(z-{{z}_{1}})=0\], where \[A,\,\,B\] and \[C\] are d.r.’s of normal to the plane.     (2) Equation of plane through three points : The equation of plane passing through three non-collinear points \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\], \[({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}})\] and \[({{x}_{3}},\,{{y}_{3}},\,{{z}_{3}})\] is \[\left| \,\begin{matrix} x & y & z & 1  \\ {{x}_{1}} & {{y}_{1}} & {{z}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & {{z}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & {{z}_{3}} & 1  \\ \end{matrix}\, \right|=0\] or \[\left| \,\begin{matrix} x-{{x}_{1}} & y-{{y}_{1}} & z-{{z}_{1}}  \\ {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}}  \\ {{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}}  \\ \end{matrix}\, \right|=0\].

(1) Area of a triangle : The area of a triangle ABC with vertices \[A({{x}_{1}},{{y}_{1}}),\,\,B\text{ }({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\]. The area of triangle ABC is denoted by \['\Delta '\]and is given as       \[\Delta =\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\    {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|\]\[=\frac{1}{2}\left| \text{ }({{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})\text{ } \right|\]     In equilateral triangle     (i) Having sides a, area is \[\frac{\sqrt{3}}{4}{{a}^{2}}\].     (ii) Having length of perpendicular as 'p' area is \[\frac{({{p}^{2}})}{\sqrt{3}}\] .     (2) Collinear points : Three points \[A({{x}_{1}},{{y}_{1}}),\,\,B({{x}_{2}},{{y}_{2}}),\,C({{x}_{3}},{{y}_{3}})\] are collinear.  If area of triangle is zero, then     (i)  \[\Delta =0\]  \[\Rightarrow \]  \[\frac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1  \\ {{x}_{2}} & {{y}_{2}} & 1  \\ {{x}_{3}} & {{y}_{3}} & 1  \\ \end{matrix} \right|=0\] \[\Rightarrow more...

If point \[P(x,y,z)\] moves according to certain rule, then it may lie in a 3-D region on a surface or on a line or it may simply be a point. Whatever we get, as the region of P after applying the rule, is called locus of P. Let us discuss about the plane or curved surface. If Q be any other point on it’s locus and all points of the straight line PQ lie on it, it is a plane. In other words if the straight line PQ, however small and in whatever direction it may be, lies completely on the locus, it is a plane, otherwise any curved surface.   (1) General equation of plane : Every equation of first degree of the form \[Ax+By+Cz+D=0\] represents the equation of a plane. The coefficients of \[x,\,\,y\] and z i.e., \[A,\,\,B,\,\,C\] are the direction ratios of the normal to the plane. more...

  (1) Skew lines : Two straight lines in space which are neither parallel nor intersecting are called skew lines.     Thus, the skew lines are those lines which do not lie in the same plane.         (2) Line of shortest distance : If \[{{l}_{1}}\] and \[{{l}_{2}}\] are two skew lines, then the straight line which is perpendicular to each of these two non-intersecting lines is called the “Line of shortest distance.”     There is one and only one line perpendicular to each of lines \[{{l}_{1}}\] and \[{{l}_{2}}\].     (3) Shortest distance between two skew lines     Let two skew lines be, \[\frac{x-{{x}_{1}}}{{{l}_{1}}}=\frac{y-{{y}_{1}}}{{{m}_{1}}}=\frac{z-{{z}_{1}}}{{{n}_{1}}}\] and \[\frac{x-{{x}_{2}}}{{{l}_{2}}}=\frac{y-{{y}_{2}}}{{{m}_{2}}}=\frac{z-{{z}_{2}}}{{{n}_{2}}}\]     Therefore, the shortest distance between the lines is given by  \[d=\frac{\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}} & more...

Foot of perpendicular from a point \[A(\alpha ,\,\,\beta ,\,\,\gamma )\]to the line  \[\frac{x-{{x}_{1}}}{l}=\frac{y-{{y}_{1}}}{m}=\frac{z-{{z}_{1}}}{n}\] : If P be the foot of perpendicular, then P is \[(lr+{{x}_{1}},\,mr+{{y}_{1}},\,nr+{{z}_{1}})\]. Find the direction ratios of AP and apply the condition of perpendicularity of AP and the given line. This will give the value of r and hence the point P, which is foot of perpendicular.     Length and equation of perpendicular : The length of the perpendicular is the distance AP and its equation is the line joining two known points A and P.     The length of the perpendicular is the perpendicular distance of given point from that line.     Reflection or image of a point in a straight line : If the perpendicular PL from point P on the given line be produced to more...

Determine whether two lines intersect or not. In case they intersect, the following algorithm is used to find their point of intersection.     Algorithm:     Let the two lines be  \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\]             …..(i)     and  \[\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}}\]                        …..(ii)     Step I : Write the co-ordinates of general points on (i) and (ii). The co-ordinates of general points on (i) and (ii) are given by \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}=\lambda \] and \[\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}}=\mu \] respectively.   i.e., \[({{a}_{1}}\lambda +{{x}_{1}},\,{{b}_{1}}\lambda +{{y}_{1}}+{{c}_{1}}\lambda +{{z}_{1}})\]and \[({{a}_{2}}\mu +{{x}_{2}},\,{{b}_{2}}\mu +{{y}_{2}},\,{{c}_{2}}\mu +{{z}_{2}})\].     Step II : If the lines (i) and (ii) intersect, then they have a common point. \[{{a}_{1}}\lambda +{{x}_{1}}={{a}_{2}}\mu +{{x}_{2}},\,{{b}_{1}}\lambda +{{y}_{1}}={{b}_{2}}\mu +{{y}_{2}}\] and \[{{c}_{1}}\lambda +{{z}_{1}}={{c}_{2}}\mu +{{z}_{2}}\].     Step III : Solve any two of the equations in \[\lambda \] and \[\mu \] obtained in step II. If the values of l and m satisfy the third equation, then the lines (i) and (ii) intersect, otherwise they more...

The unsymmetrical form of a line \[ax+by+cz+d=0,\] \[\,{a}'x+{b}'y+{c}'z+{d}'=0\] can be changed to symmetrical form as follows : \[\frac{x-\frac{b{d}'-{b}'d}{a{b}'-{a}'b}}{b{c}'-{b}'c}=\frac{y-\frac{d{a}'-{d}'a}{a{b}'-{a}'b}}{c{a}'-{c}'a}=\frac{z}{a{b}'-{a}'b}\].

(1) Centroid of a triangle : The centroid of a triangle is the point of intersection of its medians. The centroid divides the medians in the ratio 2 : 1   (vertex : base)         If \[A({{x}_{1}},{{y}_{1}})\], \[B({{x}_{2}},{{y}_{2}})\] and \[C({{x}_{3}},{{y}_{3}})\] are the vertices of a triangle. If G be the centroid upon one of the median (say) AD, then AG : GD = 2 : 1     \[\Rightarrow \] Co-ordinate of G are \[\left( \frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\frac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)\]     (2) Circumcentre : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle. It is the centre of the circle which passes through the vertices of the triangle and so its distance from the vertices of the triangle is the same and this distance is more...

Every equation of the first degree represents a plane. Two equations of the first degree are satisfied by the co-ordinates of every point on the line of intersection of the planes represented by them.     Therefore, the two equations of that line \[ax+by+cz+d=0\] and \[{a}'x+{b}'y+{c}'z+{d}'=0\] together represent a straight line.     (1) Equation of a line passing through a given point     Cartesian equation of a straight line passing through a fixed point \[({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}})\] and having direction ratios \[a,b,c\] is \[\frac{x-{{x}_{1}}}{a}=\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\].     (2) Equation of line passing through two given points     If \[A({{x}_{1}},\,{{y}_{1}},\,{{z}_{1}}),\,B({{x}_{2}},\,{{y}_{2}},\,{{z}_{2}})\] be two given points, the equations to the line AB are \[\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}_{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}_{2}}-{{z}_{1}}}\].

Let \[\theta \]  be the angle between two straight lines AB and AC whose direction  cosines are \[{{l}_{1}},\,{{m}_{1}},\,{{n}_{1}}\] and \[{{l}_{2}},\,{{m}_{2}},\,{{n}_{2}}\] respectively, is given by\[\cos \theta ={{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}\].     If direction ratios of two lines \[{{a}_{1}},\,{{b}_{1}},\,{{c}_{1}}\] and \[{{a}_{2}},\,{{b}_{2}},\,{{c}_{2}}\] are given, then angle between two lines is given by     \[\cos \theta =\frac{{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}.\sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\].     Particular results: We have, \[{{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]     \[=(l_{1}^{2}+m_{1}^{2}+n_{1}^{2})(l_{2}^{2}+m_{2}^{2}+n_{2}^{2})-{{({{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}})}^{2}}\]     \[={{({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})}^{2}}+{{({{m}_{1}}{{n}_{2}}-{{m}_{2}}{{n}_{1}})}^{2}}+{{({{n}_{1}}{{l}_{2}}-{{n}_{2}}{{l}_{1}})}^{2}}\]     \[\Rightarrow \] \[\sin \theta =\pm \sqrt{\sum {{({{l}_{1}}{{m}_{2}}-{{l}_{2}}{{m}_{1}})}^{2}}}\], which is known as Lagrange’s identity.     The value of \[\sin \,\theta \] can easily be obtained by,    \[\sin \theta =\sqrt{{{\left| \begin{matrix} {{l}_{1}} & {{m}_{1}}  \\ {{l}_{2}} & {{m}_{2}}  \\ \end{matrix} \right|}^{2}}+{{\left| \begin{matrix} {{m}_{1}} & {{n}_{1}}  \\ {{n}_{2}} & {{n}_{2}}  \\ \end{matrix} \right|}^{2}}+{{\left| \begin{matrix} {{n}_{1}} & {{l}_{1}}  \\ {{n}_{2}} & {{l}_{2}}  \\ \end{matrix} \right|}^{2}}}\]   If \[{{a}_{1}},\,{{b}_{1}},\,{{c}_{1}}\] and \[{{a}_{2}},\,{{b}_{2}},\,{{c}_{2}}\] are d.r.’s of two given lines, then angle \[\theta \] between them is more...