Misc 12 - Chapter 11 Class 12 Three Dimensional Geometry (Term 2)

Transcript

Misc 12 Find the coordinates of the point where the line through (3, –4, –5) and (2, –3, 1) crosses the plane 2x + y + z = 7. The equation of a line passing through two points A(𝑥_1, 𝑦_1, 𝑧_1) and B(𝑥_2, 𝑦_2, 𝑧_2) is (𝒙 − 𝒙_𝟏)/(𝒙_𝟐 − 𝒙_𝟏 ) = (𝒚 − 𝒚_𝟏)/(𝒚_𝟐 − 𝒚_𝟏 ) = (𝒛 − 𝒛_𝟏)/(𝒛_𝟐 − 𝒛_𝟏 ) Given the line passes through the points A (3, −4, −5) ∴𝑥_1 = 3, 𝑦_1= −4, 𝑧_1= −5 B (2, −3, 1) ∴𝑥_2 = 2, 𝑦_2= −3, 𝑧_2= 1 So, the equation of line is (𝑥 − 3)/(2 − 3) = (𝑦 − (−4))/(−3 − (−4)) = (𝑧 − (−5))/(1 − (−5)) (𝒙 − 𝟑)/(−𝟏) = (𝒚 + 𝟒)/𝟏 = (𝒛 + 𝟓)/𝟔 = k So, Let (x, y, z) be the coordinates of the point where the line crosses the plane 2x + y + z = 7 Putting value of x, y, z, from (1) in the equation of plane, 2x + y + z = 7 x = −k + 3 2(−k + 3) + (k − 4) + (6k − 5) = 7 −2k + 6 + k − 4 + 6k − 5 = 7 5k − 3 = 7 5k = 7 + 3 5k = 10 ∴ k = 𝟏𝟎/𝟓 = 2 Putting value of k in x, y, z, x = −k + 3 = −2 + 3 = 1 y = k − 4 = 2 − 4 = −2 z = 6k − 5 = 6 × 2 − 5 = 12 − 5 = 7 Therefore, the coordinate of the required point are (1, −2, 7).