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. 

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Question 28 Find the distance of point βˆ’2𝑖 ̂ + 3𝑗 ̂ βˆ’ 4π‘˜Β Μ‚ from the line π‘ŸΒ Μ‚ = 𝑖 ̂ + 2𝑗 ̂ βˆ’ π‘˜Β Μ‚ + πœ† (𝑖 ̂ + 3𝑗 ̂ βˆ’ 9π‘˜Β Μ‚) measured parallel to the plane: x – y + 2z – 3 = 0. Let Point A = βˆ’2𝑖 ̂ + 3𝑗 ̂ βˆ’ 4π‘˜Β 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‘ŸΒ Μ‚ = 𝑖 ̂ + 2𝑗 ̂ βˆ’ π‘˜Β Μ‚ + πœ† (𝑖 ̂ + 3𝑗 ̂ βˆ’ 9π‘˜Β 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 ̂ – 𝑗 ̂ + 2π‘˜Β 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 ̂ – 𝑗 ̂ + 2π‘˜Β Μ‚ Direction cosines of AB = (πœ†+3, 3πœ†βˆ’ 1, βˆ’9πœ†+3) Direction cosines of 𝑖 ̂ – 𝑗 ̂ + 2π‘˜Β 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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