The equation of the line passing through the point $(2,3,-4)$ and perpendicular to $\mathrm{XOZ}$ plane is

The equation of the line passing through the point $(2,3,-4)$ and perpendicular to $\mathrm{XOZ}$ plane is
  1. $x=-2 ; \quad y=3+\lambda ; \quad z=4$
  2. $\frac{x-2}{1}=\frac{z+4}{1} ; y=3$
  3. $x=-2 ; \quad y=-3+\lambda ; \quad z=4$
  4. $x=2 ; \quad y=3+\lambda ; \quad z=-4$

Solution

Direction cosines of normal are $\cos \alpha, \cos \beta, \cos \gamma$ are $\cos 90^{\circ}, \cos 0^{\circ}, \cos 90^{\circ}$ i.e. $0,1,0$ The line is parallel to normal of the plane. $\therefore$ Required line is $\frac{x-2}{0}=\frac{y-3}{1}=\frac{z+4}{0}=\lambda$ $\cdots($ say $)$ $\therefore x=2 ; y=3+\lambda ; z=-4$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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