If the points (1, 1, k) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0, then a possible value of k is: A. 0 B. 1 C. 2 D. 3 Answer: B) 1 Explanation: Distance 1: |3(1) + 4(1) − 12k + 13| / 13 = |20 − 12k| / 13. Distance 2: |3(−3) + 0 − 12(1) + 13| / 13 = |−9 − 12 + 13| / 13 = 8 / 13. So |20 − 12k| = 8. 20 − 12k = 8 → 12k = 12 → k = 1. Or 20 − 12k = −8 → 12k = 28 → k = 7/3. k=1 is an option.

imo class 12 three dimensional geometry

If the points (1, 1, k) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0, then a possible value of k is:

VAVidaara Admin Asked 6d ago 0 views 0 answers

#IMO #Class 12 #Mathematics #Three-Dimensional Geometry #mathematical-reasoning

If the points (1, 1, k) and (−3, 0, 1) are equidistant from the plane 3x + 4y − 12z + 13 = 0, then a possible value of k is:

A. 0
B. 1
C. 2
D. 3

Answer: B) 1

Explanation: Distance 1: |3(1) + 4(1) − 12k + 13| / 13 = |20 − 12k| / 13. Distance 2: |3(−3) + 0 − 12(1) + 13| / 13 = |−9 − 12 + 13| / 13 = 8 / 13. So |20 − 12k| = 8. 20 − 12k = 8 → 12k = 12 → k = 1. Or 20 − 12k = −8 → 12k = 28 → k = 7/3. k=1 is an option.

0 Answers

Discussion (0)

No comments yet — start the discussion.

← Back to all questions