A) \[\frac{x}{1}=\frac{y}{-1}=\frac{z}{-2}\]
B) \[\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z}{-2}\]
C) \[\frac{x-1}{1}=\frac{y+1}{-1}=\frac{z}{1}\]
D) \[\frac{x}{-2}=\frac{y}{1}=\frac{z}{2}\]
Correct Answer: B
Solution :
Let equation of the required line be \[\frac{x-{{x}_{1}}}{a}=-\frac{y-{{y}_{1}}}{b}=\frac{z-{{z}_{1}}}{c}\] ...(i) Given two lines \[\frac{x}{1}=-\frac{y}{-1}=\frac{z}{1}\] ?(ii) and \[\frac{x-1}{0}=\frac{y+1}{0}=\frac{z}{1}\] ?(iii) Since the line (i) is perpendicular to both the lines (ii) and (iii), therefore \[ab+c=0\] ...(iv) \[2b+c=0\] ...(v) From (iv) and (v) c = 2b and a + b = 0, which are not satisfy by options (c) and (d). Hence options (c) and (d) are rejected. Thus point \[({{x}_{1}},{{y}_{1}},{{z}_{1}})\]on the required line will be either (0, 0, 0) or (1, ?1, 0). Now foot of the perpendicular from point (0, 0, 0) to the line (iii) = (1, ?2r ? 1, r)
The direction ratios of the line joining the points (0, 0, 0) and (1, ?2r ? 1, r) are 1, ? 2r ? 1, r Since sum of the x and y-coordinate of direction ratio of the required line is 0. \[\therefore \]\[1-2r-1=0,\Rightarrow r=0\] Hence direction ratio are 1, ? 1, 0 But the z-direction ratio of the required line is twice the y-direction ratio of the required line i.e. 0 = 2 (?1), which is not true. Hence the shortest line does not pass through the point (0, 0, 0). Therefore option (a) is also rejected.
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