CBSE Class 12 Sample Paper for 2022 Boards (For Term 2)

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

## Find the vector and the Cartesian equations of the plane containing the point ๐ ฬ+2๐ ฬ−๐ ฬ and parallel to the lines ๐ โ = (๐ ฬ+2๐ ฬ+2๐ ฬ ) + s (2๐ ฬ−3๐ ฬ+2๐ ฬ ) = 0 and ๐ โ = (3๐ ฬ+๐ ฬ−2๐ ฬ ) + t (๐ ฬ−3๐ ฬ+๐ ฬ ) = 0

This question is similar to Misc 13 Chapter 11 Class 12 - Three Dimensional Geometry

### Transcript

Question 10 (Choice 2) Find the vector and the Cartesian equations of the plane containing the point ๐ ฬ+2๐ ฬโ๐ ฬ and parallel to the lines ๐ โ = (๐ ฬ+2๐ ฬ+2๐ ฬ ) + s (2๐ ฬโ3๐ ฬ+2๐ ฬ ) = 0 and ๐ โ = (3๐ ฬ+๐ ฬโ2๐ ฬ ) + t (๐ ฬโ3๐ ฬ+๐ ฬ ) = 0 Given that Plane is parallel to lines ๐ โ = (๐ ฬ+2๐ ฬ+2๐ ฬ ) + s (2๐ ฬโ3๐ ฬ+2๐ ฬ ) and ๐ โ = (3๐ ฬ+๐ ฬโ2๐ ฬ ) + t (๐ ฬโ3๐ ฬ+๐ ฬ ) Thus, Normal will be perpendicular to both their parallel vectors โด Normal of plane = (๐๐ ฬโ๐๐ ฬ+๐๐ ฬ ) ร (๐ ฬโ๐๐ ฬ+๐ ฬ ) = |โ 8(๐ ฬ&๐ ฬ&๐ ฬ@2&โ3&2@1&โ3&1)| = ๐ ฬ (โ3(1) โ (โ3)(2)) โ ๐ ฬ (2(1) โ 1(2)) + ๐ ฬ(2(โ3) โ 1(โ3)) = ๐ ฬ (โ3 + 6) โ ๐ ฬ (2 โ 2) + ๐ ฬ(โ6 + 3) = 3๐ ฬ โ 3๐ ฬ Now, Equation of plane passing through point A whose position vector is ๐ โ & perpendicular to ๐ โ is (๐ โ โ ๐ โ) . ๐ โ = 0 Thus, Equation of plane passing through ๐ โ =๐ ฬ+2๐ ฬโ๐ ฬ & perpendicular to ๐ โ = 3๐ ฬ โ 3๐ ฬ is [๐ โโ(๐ ฬ+๐๐ ฬโ๐ ฬ)]. (๐๐ ฬโ๐๐ ฬ) = 0 Finding Cartesian form Putting ๐ โ = x๐ ฬ + y๐ ฬ + z๐ ฬ [(๐ฅ๐ ฬ+๐ฆ๐ ฬ+๐ง๐ ฬ )โ(๐ ฬ+2๐ ฬโ๐ ฬ)]. (3๐ ฬ โ 3๐ ฬ) = 0 [(๐ฅโ1) ๐ ฬ+(๐ฆโ2) ๐ ฬ+ (๐งโ(โ 1))๐ ฬ ]. (3๐ ฬ โ 3๐ ฬ) = 0 3(x โ 1) + 0 (y โ 2) + (โ3)(z + 1) = 0 3x โ 3 โ 3z โ 3 = 0 3x โ 3z โ 6 = 0 3(x โ z โ 2) = 0 x โ z โ 2 = 0