Misc 11 - Chapter 11 Class 12 Three Dimensional Geometry

Transcript

Misc 11 (Method 1) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane. The equation of a line passing through two points with position vectors 𝑎﷯ & 𝑏﷯ is 𝒓﷯ = 𝒂﷯ + 𝜆( 𝒃﷯ − 𝒂﷯) Given, the line passes through ( 𝑏﷯ − 𝑎﷯) = (3 𝑖﷯ + 4 𝑗﷯ + 1 𝑘﷯) − (5 𝑖﷯ + 1 𝑗﷯ + 6 𝑘﷯) = (3 −5) 𝑖﷯ + (4 − 1) 𝑗﷯ + (1 − 6) 𝑘﷯ = −2 𝑖﷯ + 3 𝑗﷯ − 5 𝑘﷯ ∴ 𝒓﷯ = (5 𝒊﷯ + 𝒋﷯ + 6 𝒌﷯) + 𝜆 (−2 𝒊﷯ + 3 𝒋﷯ − 5 𝒌﷯) Let the coordinates of the point where the line crosses the ZX plane be (x, 0, z) So, 𝒓﷯ = x 𝒊﷯ + 0 𝒋﷯ + z 𝒌﷯ Since point lies in line, it will satisfy its equation, Putting (2) in (1) x 𝑖﷯ + 0 𝑗﷯ + z 𝑘﷯ = 5 𝑖﷯ + 𝑗﷯ + 6 𝑘﷯ −2𝜆 𝑖﷯ + 3𝜆 𝑗﷯ − 5𝜆 𝑘﷯ x 𝑖﷯ + 0 𝑗﷯ + z 𝑘﷯ = (5 −2𝜆) 𝑖﷯ + (1 + 3𝜆) 𝑗﷯ + (6 − 5𝜆) 𝑘﷯ Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 3𝜆 3𝜆 = −1 ∴ 𝜆 = −𝟏﷮𝟑﷯ Now, x = 5 − 2𝜆 = 5 − 2 × −1﷮3﷯ = 5 + 2﷮3﷯ = 17﷮13﷯ z = 6 − 5𝜆 = 6 − 5 × −1﷮3﷯ = 6 + 5﷮3﷯ = 23﷮3﷯ Therefore, the coordinate of the required point are 𝟏𝟕﷮𝟑﷯,𝟎, 𝟐𝟑﷮𝟑﷯﷯ Misc 11 (Method 2) Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX-plane. 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 So, the equation of line is 𝑥 − 5﷮3 − 5﷯ = 𝑦 − 1﷮4 − 1﷯ = 𝑧 − 6﷮11 − 6﷯ 𝒙 − 𝟓﷮−𝟐﷯ = 𝒚 − 𝟏﷮𝟑﷯ = 𝒛 − 𝟔﷮−𝟓﷯ = k So, Since the line crosses the ZX plane at (x, 0, z), y = 0 3k + 1 = 0 3k = −1 ∴ k = −𝟏﷮𝟑﷯ So, x = –2k + 5 = −2 × −1﷮3﷯ + 5 = 2﷮3﷯ + 5 = 17﷮3﷯ y = 0 & z = −5k + 6 = −5 × −1﷮3﷯ + 6 = 5﷮3﷯ + 6 = 23﷮3﷯ therefore, the coordinate of the required point are 𝟏𝟕﷮𝟑﷯,𝟎, 𝟐𝟑﷮𝟑﷯﷯