Question 35 (B) - CBSE Class 12 Sample Paper for 2025 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Question 35 (B) Find the image of the point (𝟏,𝟐,𝟏) with respect to the line (𝑥−3)/1=(𝑦+1)/2=(𝑧−1)/3. Also find the equation of the line joining the given point and its image.Let Point P be (1, 2, 1) Let Q (a, b, c) be the image of point P (1, 2, 1) 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 (𝒙 − 𝟑)/𝟏=(𝒚 + 𝟏)/𝟐=(𝒛 − 𝟏)/𝟑 Since PQ ⊥ Line 1 (𝑙_1) ∴ PR ⊥ Line 1 (𝒍_𝟏) Coordinates of R Since R lies of line 𝑙_1 ∴ (𝑥 − 3)/1=(𝑦 + 1)/2=(𝑧 − 1)/3 = 𝜆 ∴ x = 𝝀+𝟑 , y = 2𝝀 – 1 and z = 3𝝀 + 1 So, coordinates of R = (𝝀 + 3, 2𝝀 – 1, 3𝝀 + 1) Now, Since PR ⊥ AB 1(𝝀 + 2) + 2(2𝝀 – 3) + 3 (3𝝀) = 0 𝜆 + 2 + 4𝜆 – 6 + 9𝜆 = 0 Direction ratios of Line 𝒍_𝟏 Since equation of lines is (𝒙 − 𝟑)/𝟏=(𝒚 + 𝟏)/𝟐=(𝒛 − 𝟏)/𝟑 Direction ratios are 1, 2, 3 Direction ratios of Line PR Coordinates of P (1, 2, 1) Coordinates of R (𝜆 + 3, 2𝜆 – 1, 3𝜆 + 1) Direction ratios are 𝜆 + 3 – 1, 2𝜆 – 1 – 2 & 3𝜆 + 1 – 1 i.e. 𝝀 + 2, 2𝝀 – 3 & 3𝝀 14𝜆 – 4 = 0 14𝜆 = 4 𝜆 = 4/14 𝝀 = 2/7 Now, Coordinates of R = (𝜆 + 3, 2𝜆 – 1, 3𝜆 + 1) = (2/7+3, 2(2/7)−1, 3(2/7)+1) = ((2 + 21)/7, (4 − 7)/7, (6 + 7)/7) = (𝟐𝟑/𝟕, (−𝟑)/𝟕, 𝟏𝟑/𝟕) Since R is the midpoint of PQ Coordinates of R = ((𝟏 + 𝒂)/𝟐 " , " (𝟐 + 𝒃)/𝟐 " , " (𝟏 + 𝒄)/𝟐) (𝟐𝟑/𝟕, (−𝟑)/𝟕, 𝟏𝟑/𝟕) = ((𝟏 + 𝒂)/𝟐 " , " (𝟐 + 𝒃)/𝟐 " , " (𝟏 + 𝒄)/𝟐) 𝟐𝟑/𝟕 = (𝟏 + 𝒂)/𝟐 23 × 2 = 7 + 7a 46 = 7 + 7a 46 – 7 = 7a 39 = 7a 7a = 39 a = 𝟑𝟗/𝟕 (−𝟑)/𝟕 = (𝟐 + 𝒃)/𝟐 -3 × 2 = 14 + 7b –6 = 14 + 7b –6 – 14 = 7b –20 = 7b 7b = –20 b = (−𝟐𝟎)/𝟕 𝟏𝟑/𝟕 = (𝟏 + 𝒄)/𝟐 13 × 2 = 7 + 7c 26 = 7 + 7c 26 – 7 = 7c 19 = 7c 7c = 19 c = 𝟏𝟗/𝟕 Hence, Coordinates of Q = (a, b, c) = (𝟑𝟗/𝟕, (−𝟐𝟎)/𝟕, 𝟏𝟗/𝟕) ∴ Q(𝟑𝟗/𝟕, (−𝟐𝟎)/𝟕, 𝟏𝟗/𝟕) is the required image of P Finding the equation of the line joining the given point and its image. Cartesian equation of a line passing through two points P(x1, y1, z1) and Q (x2, y2, z2) is (𝑥 − 𝑥1)/(𝑥2 − 𝑥_1 ) = (𝑦 − 𝑦1)/(𝑦2 − 𝑦1) = (𝑧 − 𝑧1)/(𝑧2 − 𝑧1) Since the line passes through P (1, 2, 1) x1 = 1, y1 = 2, z1 = 1 And also passes through Q (𝟑𝟗/𝟕, (−𝟐𝟎)/𝟕, 𝟏𝟗/𝟕) x2 = 𝟑𝟗/𝟕, y2 = (−𝟐𝟎)/𝟕, z2 = 𝟏𝟗/𝟕 Equation of line is (𝒙 −𝟏)/(𝟑𝟗/𝟕 −𝟏) = (𝒚 −𝟐)/( (−𝟐𝟎)/𝟕 −𝟐) = (𝒛 − 𝟏)/(𝟏𝟗/𝟕 −𝟏) (𝑥 −1)/((39 − 7)/7) = (𝑦 −2)/( (−20 − 14)/7) = (𝑧 − 1)/((19 − 7)/7 ) (𝑥 −1)/(32/7) = (𝑦 −2)/( (−34)/7) = (𝑧 − 1)/(12/7 ) 7 × (𝑥 −1)/32 = 7 × (𝑦 −2)/( −34) = 7 × (𝑧 − 1)/(12 ) We can remove constant 7 (𝑥 −1)/32 = (𝑦 −2)/( −34) = (𝑧 − 1)/(12 ) (𝑥 −1)/(2 × 16) = (𝑦 −2)/(2 × −17) = (𝑧 − 1)/(2 × 6 ) 1/2 × (𝑥 −1)/16 = 1/2 × (𝑦 −2)/( −17) = 1/2 × (𝑧 − 1)/(6 ) We can remove constant 1/2 (𝒙 −𝟏)/𝟏𝟔 = (𝒚 −𝟐)/( −𝟏𝟕) = (𝒛 − 𝟏)/(𝟔 )