question_answer

A perpendicular to the pont P(1, 0,3) is drawn to the line joining the points A (4,7,1) and B(3,5,3). The foot of perpendicular to the line is

A)  \[(5,7,1)\]

B)  \[\left( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right)\]

C)  \[\left( \frac{7}{3},\frac{5}{3},\frac{7}{3} \right)\]

D)  \[\left( \frac{5}{2},\frac{2}{3},\frac{7}{3} \right)\]

Correct Answer: B

Solution :

 Let the equation of AB is \[\frac{x-4}{3-4}=\frac{y-7}{5-7}=\frac{z-1}{3-1}=r\] \[\Rightarrow \] \[\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2}=r\] \[\Rightarrow \] \[x=-r+4,y=7-2r\] and         \[z=2r+1\] \[\because \]PC is perpendicular to AB \[\therefore \] \[(3-4)(x-1)+(5-7)(y-0)\] \[+(3+1)(z-3)=0\] \[\Rightarrow \] \[-1(x-1)-2y+2(z-3)=0\] \[\Rightarrow \] \[-(-r+4)+1-2(7-2r)+2(2r+1)-6=0\] \[\Rightarrow \] \[r-4+1-14+4r+4r+2-6=0\] \[\Rightarrow \] \[9r-21=0\] \[\Rightarrow \] \[r=\frac{21}{9}=\frac{7}{3}\] \[\therefore \] \[x=-\frac{7}{3}+4=\frac{5}{3}\] \[y=7-2\times \frac{7}{3}=\frac{7}{3}\] and     \[z=2\times \frac{7}{3}+1=\frac{17}{3}\] \[\therefore \]Required point\[=\left( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right)\]