Let A is set of all possible planes passing through four vertices of given cube. Find number of ways of selecting four planes from set A, which are linearly dependent and one common point. (If planes P 1 = 0, P 2 = 0, P 3 = 0 and P 4 = 0 can be writen as aP 1 + bP 2 + cP 3 + dP 4 = 0, where all a, b, c, d are not equal to zero, then we say planes P 1 , P 2 , P 3 , P 4 are linearly dependent planes).
Text Solution
Verified by Experts135
(135)
Sol.
Number of plane passing through four vertices of cube equals to 12 (In which 6 are faces and 6 are perpendicular bisector of line segment joining end points of face diagonals)
Here four plane are linearly dependent because of two cases.
Four planes have common point of intersection.
Now there are 9 point through which six planes of set A are passing in which 8 are vertices and 1 is body centre. Except these 9 points no point exist through which at least 4 planes passing of set A.
⇒ Total number of ways of selecting 4 linearly dependent planes = 6 C 4 × 9 = 135
──────────────────────────────────────────────────────────────────────────────────────────
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems