CBSE Class 12 Sample Paper for 2020 Boards

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## (d) rΒ  = (2iΒ  + 3jΒ  ) + Ξ» (βiΒ  + 3jΒ  + 5kΒ  )

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Question 10 The equation of the line in vector form passing through the point (β1, 3, 5) and parallel to line (π₯ β 3)/2 = (π¦ β 4)/3, z = 2. is (a) π β = (βπ Μ + 3π Μ + 5π Μ) + π (2π Μ + 3π Μ + π Μ) (b) π β = (βπ Μ + 3π Μ + 5π Μ) + π (2π Μ + 3π Μ) (c) π β = (2π Μ + 3π Μ β 2π Μ) + π (βπ Μ + 3π Μ + 5π Μ) (d) π β = (2π Μ + 3π Μ) + π (βπ Μ + 3π Μ + 5π Μ) Equation of a line passing through a point with position vector π β and parallel to vector π β is π β = π β + ππ β Here, Point is (β1, 3, 5) So, π β = βπ Μ + 3π Μ + 5π Μ And it is parallel to line (π₯ β 3)/2 = (π¦ β 4)/3, z = 2 This means that z-component of equation is 0 So, line is (π₯ β 3)/2 = (π¦ β 4)/3 = (π§ β 2)/0 So, π β = 2π Μ + 3π Μ + 0π Μ = 2π Μ + 3π Μ Thus, Equation of line is π β = (βπ Μ + 3π Μ + 5π Μ) + π (2π Μ + 3π Μ) So, (b) is the correct answer