1. Chapter 11 Class 12 Three Dimensional Geometry
2. Serial order wise

Transcript

Misc 22 (Method 1) Distance between the two planes : 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D) 2﷮ ﷮29﷯﷯ units Distance between two parallel planes Ax + By + Cz = 𝑑﷮1﷯ and Ax + By + Cz = 𝑑﷮2﷯ is 𝒅﷮𝟏﷯ − 𝒅﷮𝟐﷯﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯ ﷯﷯ So, Distance between the two planes = 4 − 6﷮ ﷮ 2﷮2﷯ + 3﷮2﷯ + 4﷮2﷯﷯﷯﷯ = −2﷮ ﷮4 + 9 + 16﷯﷯﷯ = 𝟐﷮ ﷮𝟐𝟗﷯﷯ Hence, (D) is the correct option Misc 22 (Method 2) Distance between the two planes : 2x + 3y + 4z = 4 and 4x + 6y + 8z = 12 is (A) 2 units (B) 4 units (C) 8 units (D) 2﷮ ﷮29﷯﷯ units Distance of a point ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) from the plane Ax + By + Cz = D is 𝑨 𝒙﷮𝟏﷯ + 𝑩 𝒚﷮𝟏﷯ + 𝑪 𝒛﷮𝟏﷯− 𝑫﷮ ﷮ 𝑨﷮𝟐﷯ + 𝑩﷮𝟐﷯ + 𝑪﷮𝟐﷯﷯﷯﷯ Let us take a point P ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) on the plane 2x + 3y + 4z = 4 2 𝑥﷮1﷯ + 3 𝑦﷮1﷯ + 4 𝑧﷮1﷯ = 4 Now, to find the distance of point P form plane 4x + 6y + 8z = 12, Comparing with Ax + By + Cz = D, A = 4, B = 6, C = 8, D = 12 Distance of P ( 𝑥﷮1﷯, 𝑦﷮1﷯, 𝑧﷮1﷯) from the plane 4x + 6y + 8z = 12 = 4 𝑥﷮1﷯+ 6𝑦﷮1﷯ + 8 𝑧﷮1﷯− 12﷮ ﷮ 4﷮2﷯ + 6﷮2﷯ + 8﷮2﷯﷯﷯﷯ = 2 𝟐 𝒙﷮𝟏﷯ + 𝟑𝒚﷮𝟏﷯ + 𝟒 𝒛﷮𝟏﷯﷯− 12﷮ ﷮16 + 36 + 64﷯﷯﷯ = 2 × 𝟒 − 12﷮ ﷮116﷯﷯﷯ = 8 − 12﷮ ﷮4 × 29﷯﷯﷯ = −4﷮2 ﷮29﷯﷯﷯ = 𝟐﷮ ﷮𝟐𝟗﷯﷯ Hence, (D) is the correct option