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Find the distance of point − 2i + 3j − 4k from the line

 r  = i  + 2j  − k  + λ (i  + 3j − 9k )

measured parallel to the plane: x – y + 2z – 3 = 0. 

This is a question of CBSE Sample Paper - Class 12 - 2017/18.

You can download the question paper here  https://www.teachoo.com/cbse/sample-papers/


Find distance of point -2i + 3j - 4k from line r = i+2j-k + (i+3j-9k)

Question 28 - CBSE Class 12 Sample Paper for 2018 Boards - Part 2
Question 28 - CBSE Class 12 Sample Paper for 2018 Boards - Part 3 Question 28 - CBSE Class 12 Sample Paper for 2018 Boards - Part 4 Question 28 - CBSE Class 12 Sample Paper for 2018 Boards - Part 5

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Question 28 Find the distance of point βˆ’2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ from the line π‘Ÿ Μ‚ = 𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ π‘˜ Μ‚ + πœ† (𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 9π‘˜ Μ‚) measured parallel to the plane: x – y + 2z – 3 = 0. Let Point A = βˆ’2𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 4π‘˜ Μ‚ ∴ A = (–2, 3, –4) Let point B the point on line such that AB is parallel to the plane Equation of line is π‘Ÿ Μ‚ = 𝑖 Μ‚ + 2𝑗 Μ‚ βˆ’ π‘˜ Μ‚ + πœ† (𝑖 Μ‚ + 3𝑗 Μ‚ βˆ’ 9π‘˜ Μ‚) In Cartesian form, (π‘₯ βˆ’ 1)/1=(𝑦 βˆ’ 2)/3=(𝑧 + 1)/(βˆ’9) Since point B lies on line It’s coordinates will be (π‘₯ βˆ’ 1)/1=(𝑦 βˆ’ 2)/3=(𝑧 + 1)/(βˆ’9)=πœ† So, x = πœ†+1 y = 3πœ†+2 z = βˆ’9πœ†βˆ’1 So, B = (πœ†+1, 3πœ†+2, βˆ’9πœ†βˆ’1) Now, Line AB is parallel to the plane So, it will be perpendicular to plane’s normal ∴ Line AB is perpendicular to 𝑖 Μ‚ – 𝑗 Μ‚ + 2π‘˜ Μ‚ Finding equation of line AB Equation of a line passing through two points (x1, y1, z1) and (x1, y1, z1) (π‘₯ βˆ’ π‘₯1)/(π‘₯2 βˆ’ π‘₯1) = (𝑦 βˆ’ 𝑦1)/(𝑦2 βˆ’ 𝑦1) = (𝑧 βˆ’ 𝑧1)/(𝑧2 βˆ’ 𝑧1) We use (x1, y1, z1) = Point A = (–2, 3, –4) & (x2, y2, z2) = B = (πœ†+1, 3πœ†+2, βˆ’9πœ†βˆ’1) So, equation of AB is (π‘₯ βˆ’ (βˆ’2))/((πœ†+1) βˆ’ (βˆ’2)) = (𝑦 βˆ’ 3)/((3πœ†+2) βˆ’ (3)) = (𝑧 βˆ’ (βˆ’4))/((βˆ’9πœ†βˆ’1) βˆ’ (βˆ’4)) (π‘₯ + 4)/(πœ† + 3) = (𝑦 βˆ’ 3)/(3πœ† βˆ’ 1) = (𝑧 + 4)/(βˆ’9πœ† + 3) From (1) Line AB is perpendicular to 𝑖 Μ‚ – 𝑗 Μ‚ + 2π‘˜ Μ‚ Direction cosines of AB = (πœ†+3, 3πœ†βˆ’ 1, βˆ’9πœ†+3) Direction cosines of 𝑖 Μ‚ – 𝑗 Μ‚ + 2π‘˜ Μ‚ = (1, –1, 2) Thus, (πœ†+ 3) Γ— 1 + (3πœ†βˆ’ 1) Γ— –1 + (βˆ’9πœ†+3) Γ— 2 = 0 πœ†+ 3 – 3πœ† + 1 βˆ’18πœ†+ 6 = 0 3 + 1 + 6 = 18πœ†+ 3πœ† – πœ† 10 = 20πœ† 10/20 = πœ† πœ† = 1/2 So, point B is B = (πœ†+1, 3πœ†+2, βˆ’9πœ†βˆ’1) = (1/2 + 1, 3(1/2)+2, –9(1/2)βˆ’1) = (3/2, 7/2,(βˆ’11)/2) Thus, Distance between A(–2, 3, –4) and B (3/2, 7/2,(βˆ’11)/2) AB = √((3/2βˆ’(βˆ’2))^2+(7/2βˆ’3)^2+((βˆ’11)/2βˆ’(βˆ’4))^2 ) = √((3/2+2)^2+(7/2βˆ’3)^2+((βˆ’11)/2+4)^2 ) = βˆšπŸ“πŸ—/𝟐 unit

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.