A variable plane which remains at a constant distance 3p from the origin cut the coordinate axes at A, B and C. Show that the locus of the centroid of \[\Delta ABC\] is \[{{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}={{p}^{-2}}.\] |
OR |
If the lines \[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\] and \[\frac{x-3}{2}=\frac{y-k}{2}=\frac{z}{1}\] intersect, find the value of k and hence, find the equation of the plane containing these lines. |
Answer:
Let the equation of plane in intercept form be \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\] \[\Rightarrow \] \[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}-1=0\] ...(i) It cut the axes on the point A(a, 0, 0), B(0, b, 0)and (0, 0, c). Let \[(\alpha ,\,\,\beta ,\,\,\gamma )\] be the centroid of \[\Delta ABC.\] Then, \[\alpha =\frac{a+0+0}{3},\] \[\beta =\frac{0+b+0}{3}\] and \[\gamma =\frac{0+0+c}{3}\] \[\Rightarrow \] \[a=3\alpha ,\] \[b=3\beta \] and \[c=3\gamma \] Now, the perpendicular distance from origin (0, 0, 0) to the plane (i) is \[\Rightarrow \] \[\Rightarrow \] \[\frac{1}{3p}=\sqrt{\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}}\] On squaring both sides, we get \[\frac{1}{9{{p}^{2}}}=\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}+\frac{1}{{{c}^{2}}}\] On putting the values of a, b and c, we get \[\frac{1}{9{{p}^{2}}}=\frac{1}{{{(3\alpha )}^{2}}}+\frac{1}{{{(3\beta )}^{2}}}+\frac{1}{{{(3\gamma )}^{2}}}\] \[\Rightarrow \] \[\frac{1}{9{{p}^{2}}}=\frac{1}{9{{\alpha }^{2}}}+\frac{1}{9{{\beta }^{2}}}+\frac{1}{9{{\gamma }^{2}}}\] \[\Rightarrow \] \[\frac{1}{9{{p}^{2}}}=\frac{1}{9}\left( \frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}+\frac{1}{{{\gamma }^{2}}} \right)\] \[\Rightarrow \] \[\frac{1}{{{p}^{2}}}=\frac{1}{{{\alpha }^{2}}}+\frac{1}{{{\beta }^{2}}}+\frac{1}{{{\gamma }^{2}}}\] Hence, the locus of the centroid \[(\alpha ,\,\,\beta ,\,\,\gamma )\] is \[\frac{1}{{{p}^{2}}}=\frac{1}{{{x}^{2}}}+\frac{1}{{{y}^{2}}}+\frac{1}{{{z}^{2}}}\] \[[\text{replace}\,\,\alpha ,\,\,\beta ,\,\,\gamma \,\,by\,\,x,\,\,y,\,\,z]\] \[\therefore \] \[{{p}^{-2}}={{x}^{-2}}+{{y}^{-2}}+{{z}^{-2}}\] Hence proved. OR Given equation of lines are \[\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}\] ?(i) and \[\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}\] ?(ii) Since, these lines intersect therefore the shortest distance between them will be zero. Now, comparing these lines with \[\frac{x-{{x}_{1}}}{{{a}_{1}}}=\frac{y-{{y}_{1}}}{{{b}_{1}}}=\frac{z-{{z}_{1}}}{{{c}_{1}}}\] and \[\frac{x-{{x}_{2}}}{{{a}_{2}}}=\frac{y-{{y}_{2}}}{{{b}_{2}}}=\frac{z-{{z}_{2}}}{{{c}_{2}}},\] we get \[{{x}_{1}}=1,\] \[{{y}_{1}}=-\,1,\] \[{{z}_{1}}=1\] \[{{x}_{2}}=3,\] \[{{y}_{2}}=k,\] \[{{z}_{2}}=0\] \[{{a}_{1}}=2,\] \[{{b}_{1}}=3,\] \[{{c}_{1}}=4\] \[{{a}_{2}}=1,\] \[{{b}_{2}}=2,\] \[{{c}_{2}}=1\] We know that if two lines intersect, then shortest distance between them = 0 \[\therefore \] \[\therefore \] \[\Rightarrow \] \[\Rightarrow \] \[2(3-8)-(k\,+1)(2-4)-1(4-3)=0\] \[\Rightarrow \] \[2(-\,5)-(k+1)(-\,2)-1(1)=0\] \[\Rightarrow \] \[-\,10+2(k+1)-1=0\] \[\Rightarrow \] \[2(k+1)=11\] \[\Rightarrow \] \[k=\frac{11}{2}-1=\frac{9}{2}\] Now, let the required equation of plane be \[A(x-1)+B(y+1)+C(z-1)=0\] where, A, B and C are DR's ratios of normal and \[(1,\,\,-\,1,\,\,1)\] is the point on the line (i) [\[\therefore \] equation of plane is \[A(x-{{x}_{1}})+B(y-{{y}_{1}})+C(z-{{z}_{1}})=0\]] As we know that, the normal to the plane is also a normal to the lines lying in that plane, therefore \[2A+3B+4C=0\] and \[A+2B+C=0\] \[[\because \,\,{{A}_{1}}{{A}_{2}}+{{B}_{1}}{{B}_{2}}+{{C}_{1}}{{C}_{2}}=0]\] On solving above equations, we get \[\frac{A}{3-8}=\frac{-\,B}{2-4}=\frac{C}{4-3}\] \[\Rightarrow \] \[\frac{A}{-\,5}=\frac{-\,B}{-\,2}=\frac{C}{1}\] \[\Rightarrow \] \[\frac{A}{-\,5}=\frac{B}{2}=\frac{C}{1}\] Thus, the required equation of the plane is \[-\,5(x\,-1)+2(y\,+1)+1(z\,-1)=0\] \[\Rightarrow \] \[-\,5x+5+2y+2+z-1=0\] \[\Rightarrow \] \[5x-5-2y-2-z+1=0\] \[\Rightarrow \] \[5x-2y-z=6\]
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