Question 35 (Choice 1) - CBSE Class 12 Sample Paper for 2024 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Let Point P be (1, 6, 3) Let Q (a, b, c) be the image of point P (1, 6, 3) in the line 𝒓 ⃗ Since line is a mirror Point P & Q are at equal distance from line AB, i.e. PR = QR, i.e. R is the mid point of PQ Image is formed perpendicular to mirror i.e. line PQ is perpendicular to line 𝒓 ⃗ Given line is 𝑟 ⃗=(𝑗 ˆ+2𝑘 ˆ)+𝜆(𝑖 ˆ+2𝑗 ˆ+3𝑘 ˆ) In cartesian form (𝒙 − 𝟎)/𝟏 = (𝒚 − 𝟏)/𝟐 = (𝒛 − 𝟐)/𝟑 Since PQ ⊥ Line 1 (𝑙_1) ∴ PR ⊥ Line 1 (𝒍_𝟏) Coordinates of R Since R lies of line 𝑙_1 ∴ (𝑥 − 0)/1 = (𝑦 − 1)/2 = (𝑧 − 2)/3 = 𝜆 ∴ x = 𝝀 , y = 2𝝀 + 1 and z = 3𝝀 + 2Direction ratios of Line 𝒍_𝟏 Since equation of lines is (𝒙 − 𝟎)/𝟏 = (𝒚 − 𝟏)/𝟐 = (𝒛 − 𝟐)/𝟑 Direction ratios are 1, 2, 3 Direction ratios of Line PR Coordinates of P (1, 6, 3) Coordinates of R R (𝝀, 2𝝀 + 1, 3𝝀 + 2) Direction ratios are 𝜆 – 1, 2𝜆 + 1 – 6 & 3𝜆 + 2 – 3 i.e. 𝝀 – 1, 2𝝀 – 5 & 3𝝀 – 1 14𝜆 – 14 = 0 14𝜆 = 14 𝜆 = 14/14 𝝀 = 1 Now, Coordinates of R = (𝜆, 2𝜆 + 1, 3𝜆 + 2) = (1, 2(1) + 1, 3(1) + 2) = (𝟏, 3, 5) Since R is the midpoint of PQ Coordinates of R = ((𝟏 + 𝒂)/𝟐 " , " (𝟔 + 𝒃)/𝟐 " , " (𝟑 + 𝒄)/𝟐) (1, 3, 5)= ((𝟏 + 𝒂)/𝟐 " , " (𝟔 + 𝒃)/𝟐 " , " (𝟑 + 𝒄)/𝟐) 1 = (1+𝑎)/2 , 3 = (6+𝑏)/2 , 5 = (3+𝑐)/2 2 = 1 + a, 6 = 6 + b , 10 = 3 + c ∴ a = 1, b = 0 and c = 7 Hence, Q(1,0,7) is the required image of P Finding the distance of the image from the 𝒚-axis. Distance of Q(1, 0, 7) from the 𝑦-axis = Distance of parallel point Y and point Q = Distance of point Y (0, 0, 0) and point Q (1, 0, 7) =√(〖(0−1)〗^2 + 〖(0−0)〗^2 + 〖(0−7)〗^2 ) =√(1+49) =√𝟓𝟎 units asdf