Find the coordinates of the point, where the line through \[(3,\,\,-\,4,\,\,-5)\] and \[(2,\,\,-\,3,\,\,1)\] crosses the plane \[2x+y+z=7.\] |
OR |
Find the foot of perpendicular from the point |
(2, 3, 4) to the line \[\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}.\] Also, find the length of the perpendicular segment. |
Answer:
The equation of the line joining\[(3,\,\,-\,4,\,\,-\,5)\]and \[(2,\,\,-\,3,\,\,1)\]is \[\frac{x-3}{-\,1}=\frac{y+4}{1}=\frac{z+5}{6}=\lambda \,\,(say)\] \[\Rightarrow \] \[x=3-\lambda ,\,\,y=\lambda -4,\,\,z=6\lambda -5\] Therefore, any point on the line is of the form \[(3-\lambda ,\,\,\lambda -4,\,\,6\lambda -5)\]. This point lies on the plane\[2x+y+z=7\] Therefore,\[2\,(3-\lambda )+(\lambda -4)+(6\lambda -5)=7\] \[\Rightarrow \] \[5\lambda -3=7\] \[\Rightarrow \] \[\lambda =2\] Hence, the coordinates of the required point are \[(3-2,\,\,2-4,\,\,6\times 2)-5\] i.e. \[(1,\,\,-2,\,\,7)\]. OR Given, \[\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}\] ?(i) Let L be the foot of the perpendicular drawn from the point (2, 3, 4) to the given line. The coordinate of a general point on line (i) are given by \[\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\lambda \] \[x=4-2\lambda ,\]\[y=6\lambda \]and \[z=1-3\lambda \] Let the coordinate of\[L=(4-2\lambda ,\,\,6\lambda ,\,\,1-3\lambda )\] \[\therefore \]Direction ratio of PL are proportional to \[4-2\lambda -2,\]\[6\lambda -\,3,\]\[1-3\lambda -4\] i.e. \[2-2\lambda ,\]\[6\lambda -\,3,\]\[-3-3\lambda \] Direction ratios of the given line are proportional to\[-\,2,\,\,6,\,\,-\,3\]. \[\therefore \]PL is perpendicular to the given line. \[\therefore \]\[(-\,2)\,\,(2-2\pi )+6\,(6\lambda -3)+(-\,3)\,\,(-\,3-3\lambda )=0\] \[\Rightarrow -\,4+4\lambda +36\lambda -18+9+9\lambda =0\Rightarrow 49\lambda -13=0\]\[\Rightarrow \] \[\lambda =\frac{13}{49}\] So, the coordinate of \[L=\left( 4-2\left( \frac{13}{49} \right),\,\,6\left( \frac{13}{49} \right),\,\,1-3\left( \frac{13}{49} \right) \right)\] \[=\left( \frac{170}{49},\,\,\frac{78}{49},\,\,\frac{10}{49} \right)\] \[\therefore \]Length of perpendicular \[=\sqrt{{{\left( \frac{170}{49}-2 \right)}^{2}}+{{\left( \frac{78}{49}-3 \right)}^{2}}+{{\left( \frac{10}{49}-4 \right)}^{2}}}\] \[=\sqrt{{{\left( \frac{72}{49} \right)}^{2}}+{{\left( -\frac{69}{49} \right)}^{2}}+{{\left( \frac{-\,186}{49} \right)}^{2}}}\] \[=\sqrt{\frac{5184}{49\times 49}+\frac{4761}{49\times 49}+\frac{34596}{49\times 49}}=\sqrt{\frac{44541}{49\times 49}}\] \[=\sqrt{\frac{909}{49}}=\frac{3\sqrt{101}}{7}\,\,units\]
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