Question 35 - CBSE Class 12 Sample Paper for 2026 Boards - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

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Question 35 The two lines (𝑥−1)/3=−𝑦,𝑧+1=0 and (−𝑥)/2=(𝑦+1)/2=𝑧+2 intersect at a point whose 𝑦-coordinate is 1 . Find the co-ordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.Given line (𝑥−1)/3=−𝑦,𝑧+1=0 Writing in normal form (𝒙 − 𝟏)/𝟑=𝒚/(−𝟏)=(𝒛 + 𝟏)/𝟎 And second line 𝒙/(−𝟐)=(𝒚 + 𝟏)/𝟐=(𝐳 + 𝟐)/𝟏 To find point of intersection, we find general point of both lines and equate them If point is in Line (1) (𝒙 − 𝟏)/𝟑=𝒚/(−𝟏)=(𝒛 + 𝟏)/𝟎 General point is (𝒙 − 𝟏)/𝟑=𝒚/(−𝟏)=(𝒛 + 𝟏)/𝟎 = p So, x = 3p + 1 y = –p z = –1 If point is in Line (2) 𝒙/(−𝟐)=(𝒚 + 𝟏)/𝟐=(𝐳 + 𝟐)/𝟏 General point is 𝒙/(−𝟐)=(𝒚 + 𝟏)/𝟐=(𝐳 + 𝟐)/𝟏 = q So, x = –2q y = 2q – 1 z = q – 2 Equating z-coordinates –1 = q – 2 –1 + 2 = q 1 = q q = 1 Putting q = 1, in x, y, z x = –2q = –2 × 1 = –2 y = 2q – 1 = 2 × 1 – 1 = 2 – 1 = 1 z = q – 2 = 1 – 2 = –1 Thus, point of intersection is (–2, 1, –1) We need to find vector equation of a line perpendicular to both the given lines and passing through this point of intersection. The vector equation of a line passing through a point with position vector 𝑎 ⃗ and parallel to a vector 𝑏 ⃗ is 𝒓 ⃗ = 𝒂 ⃗ + 𝜆𝒃 ⃗ The line passes through (–2, 1, −1) So, 𝒂 ⃗ = –2𝒊 ̂ + 𝒋 ̂ − 𝒌 ̂ Given, line is perpendicular to both lines ∴ 𝑏 ⃗ is perpendicular to both lines We know that 𝑥 ⃗ × 𝑦 ⃗ is perpendicular to both 𝑥 ⃗ & 𝑦 ⃗ So, 𝒃 ⃗ is cross product of both lines (𝑥 − 1)/3=𝑦/(−1)=(𝑧 + 1)/0 and 𝑥/(−2)=(𝑦 + 1)/2=(z + 2)/1 Required normal = |■8(𝑖 ̂&𝑗 ̂&𝑘 ̂@3&−1&0@−2&2&1)| = 𝑖 ̂ (–1(1) – 2(0)) – 𝑗 ̂ (3(1) – (–2)(0)) + 𝑘 ̂(3(2) – (–2) (–1)) = 𝑖 ̂ (–1) – 𝑗 ̂ (3) + 𝑘 ̂(6 – 2) = –𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂ Thus, 𝒃 ⃗ = –𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂ Now, Putting value of 𝑎 ⃗ & 𝑏 ⃗ in formula 𝑟 ⃗ = 𝑎 ⃗ + 𝜆𝑏 ⃗ ∴ 𝑟 ⃗ = (–2𝒊 ̂ + 𝒋 ̂ − 𝒌 ̂) + 𝜆 (–𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂) Therefore, the equation of the line is (–2𝒊 ̂ + 𝒋 ̂ − 𝒌 ̂) + 𝜆 (–𝒊 ̂ – 3𝒋 ̂ + 4𝒌 ̂)