question_answer

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

\[=\sqrt{{{\left( \frac{170}{49}-2 \right)}^{2}}+{{\left( \frac{78}{49}-3 \right)}^{2}}+{{\left( \frac{10}{49}-4 \right)}^{2}}}\]

\[=\sqrt{{{\left( \frac{72}{49} \right)}^{2}}+{{\left( -\frac{69}{49} \right)}^{2}}+{{\left( \frac{-\,186}{49} \right)}^{2}}}\]

\[=\sqrt{\frac{5184}{49\times 49}+\frac{4761}{49\times 49}+\frac{34596}{49\times 49}}=\sqrt{\frac{44541}{49\times 49}}\] \[=\sqrt{\frac{909}{49}}=\frac{3\sqrt{101}}{7}\,\,units\]