Question 4 - Case Based Questions (MCQ) - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Last updated at July 29, 2021 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)

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.