Find the image of the point (1, 6, 3) on the line \[\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}.\] |
Also, write the equation of the line joining the given point and its image and find the length of segment joining the given point and its image. |
OR |
Find the foot of the perpendicular from the point (0, 2, 3) on the line \[\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}.\] |
Also, find the length of the perpendicular. |
Answer:
Given, point P = (1, 6, 3) and equation of line AB is \[\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}.\] Let T(x, y, z) be the image of the point P(1, 6, 3) and Q be the foot of perpendicular PQ on line AB. Then, \[\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=\lambda \,\,(say)\] \[\Rightarrow \] \[x=\lambda ,\] \[y=2\lambda +1,\] \[z=3\lambda +2\] Then, coordinates of \[Q=(\lambda ,\,\,2\lambda +1,\,\,3\lambda +2)\] ...(ii) Now, DR's of line \[PQ=(\lambda -1,\,\,2\lambda +1\,-6,\,\,3\lambda +2-3)\] \[=(\lambda -1,\,\,2\lambda -5,\,\,3\lambda -1)\] ?(iii) Since, \[PQ\bot AB.\] \[\therefore \] \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] Where \[{{a}_{1}}=\lambda -1,\] \[{{b}_{1}}=2\lambda -5,\] \[{{c}_{1}}=3\lambda -1\] and \[{{a}_{2}}=1,\]\[{{b}_{2}}=2,\] \[{{c}_{2}}=3\] \[\therefore \] \[1(\lambda -1)+2(2\lambda -5)+3(3\lambda -1)=0\] \[\Rightarrow \] \[\lambda -1+4\lambda -10+9\lambda -3=0\] \[\Rightarrow \] \[14\lambda -14=0\] \[\Rightarrow \] \[\lambda =1\] On putting \[\lambda =1\] in Eq. (ii) we get Coordinates of \[Q=(1,\text{ }2+1,\text{ }3+2)=\] (1, 3, 5) Also, Q is the mid-point of PT. \[\therefore \] \[Q=\left( \frac{x+1}{2},\,\,\frac{y+6}{2},\,\,\frac{z+3}{2} \right)\] \[\left[ \begin{align} & \because \,\,\,\text{mid-point}\,\,\text{of}\,\,({{x}_{1}},\,\,{{y}_{1}}\,\,{{z}_{1}})\,\,and\,\,({{x}_{2}},\,\,{{y}_{2}},\,\,{{z}_{2}}) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\,\frac{{{y}_{1}}+{{y}_{2}}}{2},\,\,\frac{{{z}_{1}}+{{z}_{2}}}{2} \right) \\ \end{align} \right]\] But Q = (1, 3, 5) \[\therefore \] \[\left( \frac{x+1}{2},\,\,\frac{y+6}{2},\,\,\frac{z+3}{2} \right)\] \[=(1,\,\,3,\,\,5)\] \[\Rightarrow \] \[\frac{x+1}{2}=1,\] \[\frac{y+6}{2}=3,\] \[\frac{z+3}{2}=5\] \[\Rightarrow \] \[x=2-1,\] \[y=6-6,\] \[z=10-3\] \[\Rightarrow \] x = 1, y = 0 and z = 1 So, coordinates of T = (x, y, z) = (1, 0, 7) Hence, coordinates of image of the point P(1, 6, 3) is T(1, 0, 7). Length of the segment joining P and T, \[PT=\sqrt{{{(1-1)}^{2}}+{{(0-6)}^{2}}+{{(7-3)}^{2}}}\] \[=\sqrt{0+36+16}\] \[=\sqrt{52}=2\sqrt{13}\] units DR's of line \[PT=(0,\,\,-\,6,\,\,4)\] DR's of line \[PQ=[1-1,\,\,2(1)-5,\,\,3(1)-1]\] [from Eq. (iii)] \[=(0,\,\,-3,\,\,2)\] Hence, the equation of line PT is \[\frac{x-1}{0}=\frac{y-6}{-\,3}=\frac{z-3}{2}.\] OR Let L be the foot of the perpendicular drawn from the point P(0, 2, 3) to the given line. The coordinates of a general point on the line \[\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}\] are given by or, \[x=5\lambda -3,\] \[y=2\lambda +1,\] \[z=3\lambda -4.\] Let the coordinates of L be \[(5\lambda -3,\,\,2\lambda +1,\,\,3\lambda -4).\] Therefore, direction ratios of PL are proportional to \[5\lambda -3-0,\] \[2\lambda +1-2,\] \[3\lambda -4-3\] i.e. \[5\lambda -3,\] \[2\lambda -1,\] \[3\lambda -7.\] \[\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3},\] \[L(5\lambda -3,\,\,2\lambda +1,\,\,3\lambda -4)\] Direction ratios of the given line are proportional to 5, 2, 3. But, PL is perpendicular to the given line. \[\therefore \]\[5(5\lambda -3+2(2\lambda -1)+3(3\lambda -7)=0\] \[\Rightarrow \] \[\lambda =1\] Putting \[\lambda =1\] in \[(5\lambda -3,\,\,2\lambda +1,\,\,3\lambda -4),\] the coordinates of L are \[(2,\,\,3,\,\,-\,1).\] \[\therefore PL=\sqrt{{{(2-0)}^{2}}+{{(3-2)}^{2}}+{{(-1-3)}^{2}}}=\sqrt{21}\,\,\text{unit}.\]Hence, length of the perpendicular from P on the given line is \[PL=\sqrt{27}\] units.
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