The point A(3, -2, 4) is shifted parallel to the line x 3 = y-1 2 = z+1 3 by a distance 1…
Mathematics · JEE Main · NTA Exams — Three Dimensional Geometry
The point A(3, -2, 4) is shifted parallel to the line \(\frac{x}{\sqrt{3}}=\frac{y-1}{2}=\frac{z+1}{3}\) by a distance 1 to the point P. The coordinates of P in the new position are :
- \(\left(3 \pm \frac{\sqrt{3}}{4},-2 \pm \frac{1}{2}, 4 \pm \frac{3}{4}\right)\)
- \((3+\sqrt{3}, 3,2)\)
- \((3-\sqrt{3},-1,-4)\)
- None of these
Answer
(A) (3 3 4 ,-2 1 2 , 4 3 4 )
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