Misc 11 - Class 12

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﷯﷯