Miscellaneous

Chapter 11 Class 12 Three Dimensional Geometry
Serial order wise

### Transcript

Question 10 If the points (1, 1 , p) and (ā3 , 0, 1) be equidistant from the plane š ā. (3š Ģ + 4š Ģ ā 12š Ģ) + 13 = 0, then find the value of p. The distance of a point with position vector š ā from the plane š ā.š ā = d is |(š ā.š ā ā š)/|š ā | | Given, the points are The equation of plane is š ā. (3š Ģ + 4š Ģ ā 12š Ģ) + 13 = 0 š ā.(3š Ģ + 4š Ģ ā 12š Ģ) = ā13 (1, 1, p) So, (š_1 ) ā = 1š Ģ + 1š Ģ + pš Ģ (ā3, 0, 1) So, (š_2 ) ā = ā3š Ģ + 0š Ģ + 1š Ģ āš ā.(3š Ģ + 4š Ģ ā 12š Ģ) = 13 š ā.(ā3š Ģ ā 4š Ģ + 12š Ģ) = 13 Comparing with š ā.š ā = d, š ā = ā3š Ģ ā 4š Ģ + 12š Ģ d = 13 Magnitude of š ā = ā((ā3)^2+(ā4)^2+ć12ć^2 ) |š ā | = ā(9+16+144) = ā169 = 13 Distance of point (šš) ā from plane |((š1) ā"." š ā" " ā š)/|š ā | | = |((1š Ģ + 1š Ģ + šš Ģ ).(ā3š Ģā4š Ģ+12š Ģ )ā13)/13| = |((1Ćā3)+(1Ćā4) +(šĆ12)ā13)/13| = |(ā3ā4+12šā13)/13| = |(12š ā 20)/13| Distance of point (šš) ā from plane |((š2) ā"." š ā ā š)/|š ā | | = |((ā3š Ģ +0š Ģ +1š Ģ ).(ā3š Ģā4š Ģ+12š Ģ )ā13)/13| = |((ā3Ćā3)+(0Ćā4) +(1Ć12)ā13)/13| = |(9 + 0 +12ā13)/13| = |8/13| = 8/13 Since the plane is equidistance from both the points, |(ššš ā šš)/šš| = š/šš |12šā20| = 8 (12p ā 20) = Ā± 8 12p ā 20 = 8 12p = 8 + 20 12p = 28 p = 28/12 p = 7/3 12p ā 20 = ā8 12p = ā8 + 20 12p = 12 p = 12/12 p = 1 Answer does not match at end. If mistake, please comment