Question 4 - Case Based Questions (MCQ) - Chapter 11 Class 12 Three Dimensional Geometry
Last updated at April 16, 2024 by Teachoo
Suppose the floor of a hotel is made up of mirror polished Salvatore stone. There is a large crystal chandelier attached to the ceiling of the hotel room. Consider the floor of the hotel room as a plane having the equation x – y + z = 4 and the crystal chandelier is suspended at the point (1, 0, 1).
Based on the above answer the following:
Question 1
Find the direction ratios of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4
(a) (–1, –1, 1)
(b) (1, –1, –1)
(c) (–1, –1, –1)
(d) (1, –1, 1)
Question 2
Find the length of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4.
(a) 2/√3 units
(b) 4/√3 units
(c) 6/√3 units
(d) 8/√3 units
Question 3
The equation of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4 is
(a) (x - 1)/2 = (y + 3)/(-1) = (z + 5)/3
(b) (x - 1)/(-2) = (y + 3)/(-1) = (z - 5)/2
(c) (x - 1)/1 = y/(-1) = (z - 1)/1
(d) (x - 1)/2 = y/(-2) = (z - 1)/1
Question 4
The equation of the plane parallel to the plane x – y + z = 4, which is at a unit distance from the point (1, 0, 1) is
(a) x - y + z + (2-√3)
(b) x - y + z - (2+√3)
(c) x - y + z + (2+√3)
(d) Both (a) and (c)
Question 5
The direction cosine of the normal to the plane x – y + z = 4 is
Question Suppose the floor of a hotel is made up of mirror polished Salvatore stone. There is a large crystal chandelier attached to the ceiling of the hotel room. Consider the floor of the hotel room as a plane having the equation x – y + z = 4 and the crystal chandelier is suspended at the point (1, 0, 1). Based on the above answer the following:Question 1 Find the direction ratios of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4 (a) (–1, –1, 1) (b) (1, –1, –1) (c) (–1, –1, –1) (d) (1, –1, 1) (d) (1, –1, 1)
Question 2 Find the length of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4. (a) 2/√3 units (b) 4/√3 units (c) 6/√3 units (d) 8/√3 units(a) 2/√3 units
Question 3 The equation of the perpendicular from the point (1, 0, 1) to the plane x – y + z = 4 is (a) (𝑥 − 1)/2=(𝑦 + 3)/(−1)=(𝑧 + 5)/3 (b) (𝑥 − 1)/(−2)=(𝑦 + 3)/(−1)=(𝑧 − 5)/2 (c) (𝑥 − 1)/1=𝑦/(−1)=(𝑧 − 1)/1 (d) (𝑥 − 1)/2=𝑦/(−2)=(𝑧 − 1)/1 (c) (𝑥 − 1)/1=𝑦/(−1)=(𝑧 − 1)/1
Question 4 The equation of the plane parallel to the plane x – y + z = 4, which is at a unit distance from the point (1, 0, 1) is (a) 𝑥−𝑦+𝑧+(2−√3) (b) 𝑥−𝑦+𝑧−(2+√3) (c) 𝑥−𝑦+𝑧+(2+√3) (d) Both (a) and (c) (d) Both (a) and (c)
Question 5 The direction cosine of the normal to the plane x – y + z = 4 is (a) (1/√3,(−1)/√3,(−1)/√3) (b) (1/√3,(−1)/√3,1/√3) (c) ((−1)/√3,(−1)/√3,1/√3) (d) ((−1)/√3,(−1)/√3,(−1)/√3)(b) (1/√3,(−1)/√3,1/√3)
Made by
Davneet Singh
Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo
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