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JEE Main 2022 · Mathematics · Three Dimensional Geometry · medium

The distance of the point (1,2,3) from the plane x-2y+2z=19, measured along the line with direction ratios 2:3:6, is

Single correct option

The distance of the point (1,2,3)(1,2,3) from the plane x−2y+2z=19x-2y+2z=19, measured along the line with direction ratios 2:3:62:3:6, is
  1. A. 163\dfrac{16}{3}
  2. B. 77
  3. C. 1414
  4. D. 2121
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Answer: (C) 1414

Points on the line through (1,2,3)(1,2,3) with direction (2,3,6)(2,3,6) are (1+2t, 2+3t, 3+6t)(1+2t,\,2+3t,\,3+6t), and the distance travelled is ∣t∣22+32+62=7∣t∣|t|\sqrt{2^2+3^2+6^2}=7|t|. This point lies on the plane when (1+2t)−2(2+3t)+2(3+6t)=19(1+2t)-2(2+3t)+2(3+6t)=19, i.e. 3+8t=193+8t=19, so t=2t=2. The required distance is 7×2=147\times2=14. Note that the perpendicular distance, ∣1−4+6−19∣3=163\frac{|1-4+6-19|}{3}=\frac{16}{3}, is a different quantity and appears as a distractor.
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