Misc 1 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Misc 1 Find the angle between the lines whose direction ratios are a, b, c and b − c, c − a, a − b. Angle between the lines with direction ratios a1, b1, c1 and a2, b2, c2 is given by cos θ = |(𝒂_𝟏 𝒂_𝟐 + 𝒃_𝟏 𝒃_𝟐 + 𝒄_𝟏 𝒄_𝟐)/(√(𝒂_𝟏^𝟐 + 𝒃_𝟏^𝟐 + 𝒄_𝟏^𝟐 ) √(𝒂_𝟏^𝟐 + 𝒃_𝟏^𝟐 + 𝒄_𝟏^𝟐 ))| Given, 𝑎1 = 𝑎, 𝑏1 = 𝑏, c1 = c and 𝑎2 = 𝑏 − 𝑐, 𝑏2 = 𝑐 − 𝑎, c2 = a – b So, cos θ = |(𝑎(𝑏 − 𝑐) + 𝑏(𝑐 − 𝑎) + 𝑐(𝑎 − 𝑏))/(√(𝑎^2 + 𝑏^2 + 𝑐^2 ) √((𝑏 − 𝑐)2 + (𝐶 − 𝑎)2 + (𝑎 − 𝑏)2))| = |(𝒂𝒃 − 𝒂𝒄 + 𝒃𝒄 − 𝒂𝒃 + 𝒄𝒂 − 𝒃𝒄)/(√(𝑎^2 + 𝑏^2 + 𝑐^2 ) √(𝑏^2 + 𝑐2 − 2𝑏𝑐 + 𝑐^2 + 𝑎^2 − 2𝑐𝑎 + 𝑎^2 + 𝑏^2 − 2𝑎𝑏 ))| = |𝟎/(√(𝑎^2 + 𝑏^2 + 𝑐^2 ) √(𝑏^2 + 𝑐2 − 2𝑏𝑐 + 𝑐^2 + 𝑎^2 − 2𝑐𝑎 + 𝑎^2 + 𝑏^2 − 2𝑎𝑏 ))| = 0 Since cos θ = 0 So, θ = 90° Therefore, angle between the given pair of lines is 90°