question_answer

Find the foot of perpendicular drawn from the point A (1, 8, 4) to the line joining the points \[B(0,\,\,-\,1,\,\,3)\] and \[C(2,\,\,-3,\,\,1).\]

Answer:

Cartesian equation of a line passing through the points \[B\,(0,\,\,-1,\,\,3)\] and \[C\,(2,\,\,-3,\,\,1)\] is \[\frac{x-{{x}_{1}}}{{{x}_{2}}-{{x}_{1}}}=\frac{y-{{y}_{1}}}{{{y}^{2}}-{{y}_{1}}}=\frac{z-{{z}_{1}}}{{{z}^{2}}-{{z}_{1}}}\] \[\Rightarrow \]   \[\frac{x-0}{2-0}=\frac{y+1}{-\,3+1}=\frac{z-3}{1-3}\] \[\Rightarrow \]   \[\frac{x}{2}=\frac{y+1}{-\,2}=\frac{z-3}{-\,2}\] Now, let ?L? be the foot of the perpendicular from the point A (1, 8, 4) to the given line. The coordinates of the point L on the line BC is given by \[\frac{x}{2}=\frac{y+1}{-\,2}=\frac{z-3}{-\,2}=\lambda \]             \[\Rightarrow \]   \[x=2\lambda ,\]\[y=-\,2\lambda -1,\]\[z=-\,2\lambda +3\]             \[\therefore \]DR?s of AL is             \[(2\lambda -1),\]\[(-\,2\lambda -1-8),\]\[(-\,2\lambda -3-4)\]             i.e.        \[(2\lambda -1),\]\[(-\,2\lambda -9)\,\,(-2\lambda -1)\] \[\therefore \]DR?s of given lines are proportional to \[(2,\,\,-\,2,\,\,-\,2)\] Since, AL is perpendicular to the given line BC. \[\therefore \]                  \[{{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0\] i.e \[2\,\,(2\lambda -1)+(-\,2)\,\,(-\,2\lambda -9)+(-\,2)\,\,(-\,2\lambda -1)=0\] \[\Rightarrow \]               \[4\lambda -2+4\lambda +18+4\lambda +2=0\] \[\Rightarrow \]                           \[12\lambda +18=0\] \[\Rightarrow \]   \[\lambda =\frac{-\,18}{12}=\frac{-\,3}{2}\] \[\Rightarrow \]   \[2\lambda -1=2\left( \frac{-\,3}{2} \right)-1=-\,3-1=-\,4\] \[\Rightarrow \]   \[-\,2\lambda -9=-\,2\left( \frac{-\,3}{2} \right)-9=3-9=-\,6\] \[\Rightarrow \]   \[-\,2\lambda -1=-\,2\left( -\frac{3}{2} \right)-1=3-1=2\] Hence, the coordinates are\[(-\,\,4,\,\,-\,\,6,\,\,2)\].