MHT CET · Maths · Three Dimensional Geometry
The equation of the line passing through the point \((2,3,-4)\) and perpendicular to
\(\mathrm{XOZ}\) plane is
- A \(x=-2 ; \quad y=3+\lambda ; \quad z=4\)
- B \(\frac{x-2}{1}=\frac{z+4}{1} ; y=3\)
- C \(x=-2 ; \quad y=-3+\lambda ; \quad z=4\)
- D \(x=2 ; \quad y=3+\lambda ; \quad z=-4\)
Answer & Solution
Correct Answer
(D) \(x=2 ; \quad y=3+\lambda ; \quad z=-4\)
Step-by-step Solution
Detailed explanation
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 \()\)
\(\therefore x=2 ; y=3+\lambda ; z=-4\)
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