A) \[(1, - 2, 5)\]
B) \[(1, 0, 5)\]
C) \[(0, 3, -5)\]
D) \[(-1, -3, 0)\]
Correct Answer: B
Solution :
Equation of the plane containing the line \[\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}\]is \[a(x-1)+b(y-2)+c(z-3)=0\] ..(i) where \[a.1+b.2+c.3=0\] i.e.,\[a+2b+3c=0\] .... (ii) Since the plane (i) parallel to the line \[\frac{x}{1}=\frac{y}{1}=\frac{z}{4}\] \[\therefore \]\[a.1+b.1+c.4=0\]i.e., \[a+b+4c=0\] ... (iii) From (ii) and (iii), \[\frac{a}{8-3}=\frac{b}{3-4}=\frac{c}{1-2}=k\](let) \[\therefore \]\[a=5k,b=-k,c=-k\] On putting the value of a, b and c in equation (i),\[5(x-1)-(y-2)-(z-3)=0\] \[\Rightarrow \]\[5x-y-z=0\] ... (iv) when x = 1, y = 0 and z = 5; then L.H.S. of equation (iv) \[=5x-y-2\] \[=5\times 1-0-5\] \[=0\] = R.H.S. of equation (iv) Hence coordinates of the point (1, 0, 5) satisfy the equation plane represented by equations (iv), Therefore the plane passes through the point (1,0,5)You need to login to perform this action.
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