Find the coordinates of the image of the point (1 , 6, 3) with respect to the line r ⃗=(j ˆ+2k ˆ)+λ(i ˆ+2j ˆ+3k ˆ); where ' λ ' is a scalar. Also, find the distance of the image from the y-axis.

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Let Point P be (1, 6, 3) Let Q (a, b, c) be the image of point P (1, 6, 3) in the line 𝒓 ⃗ Since line is a mirror Point P & Q are at equal distance from line AB, i.e. PR = QR, i.e. R is the mid point of PQ Image is formed perpendicular to mirror i.e. line PQ is perpendicular to line 𝒓 ⃗ Given line is 𝑟 ⃗=(𝑗 ˆ+2𝑘 ˆ)+𝜆(𝑖 ˆ+2𝑗 ˆ+3𝑘 ˆ) In cartesian form (𝒙 − 𝟎)/𝟏 = (𝒚 − 𝟏)/𝟐 = (𝒛 − 𝟐)/𝟑 Since PQ ⊥ Line 1 (𝑙_1) ∴ PR ⊥ Line 1 (𝒍_𝟏) Coordinates of R Since R lies of line 𝑙_1 ∴ (𝑥 − 0)/1 = (𝑦 − 1)/2 = (𝑧 − 2)/3 = 𝜆 ∴ x = 𝝀 , y = 2𝝀 + 1 and z = 3𝝀 + 2Direction ratios of Line 𝒍_𝟏 Since equation of lines is (𝒙 − 𝟎)/𝟏 = (𝒚 − 𝟏)/𝟐 = (𝒛 − 𝟐)/𝟑 Direction ratios are 1, 2, 3 Direction ratios of Line PR Coordinates of P (1, 6, 3) Coordinates of R R (𝝀, 2𝝀 + 1, 3𝝀 + 2) Direction ratios are 𝜆 – 1, 2𝜆 + 1 – 6 & 3𝜆 + 2 – 3 i.e. 𝝀 – 1, 2𝝀 – 5 & 3𝝀 – 1 14𝜆 – 14 = 0 14𝜆 = 14 𝜆 = 14/14 𝝀 = 1 Now, Coordinates of R = (𝜆, 2𝜆 + 1, 3𝜆 + 2) = (1, 2(1) + 1, 3(1) + 2) = (𝟏, 3, 5) Since R is the midpoint of PQ Coordinates of R = ((𝟏 + 𝒂)/𝟐 " , " (𝟔 + 𝒃)/𝟐 " , " (𝟑 + 𝒄)/𝟐) (1, 3, 5)= ((𝟏 + 𝒂)/𝟐 " , " (𝟔 + 𝒃)/𝟐 " , " (𝟑 + 𝒄)/𝟐) 1 = (1+𝑎)/2 , 3 = (6+𝑏)/2 , 5 = (3+𝑐)/2 2 = 1 + a, 6 = 6 + b , 10 = 3 + c ∴ a = 1, b = 0 and c = 7 Hence, Q(1,0,7) is the required image of P Finding the distance of the image from the 𝒚-axis. Distance of Q(1, 0, 7) from the 𝑦-axis = Distance of parallel point Y and point Q = Distance of point Y (0, 0, 0) and point Q (1, 0, 7) =√(〖(0−1)〗^2 + 〖(0−0)〗^2 + 〖(0−7)〗^2 ) =√(1+49) =√𝟓𝟎 units asdf

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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 14 years. He provides courses for Maths, Science and Computer Science at Teachoo