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Question
Find the coordinates of the image of the point (1, 6, 3) with respect to the line `vecr = (hatj + 2hatk) + λ(hati + 2hatj + 3hatk)`; where 'λ' is a scalar. Also, find the distance of the image from the y – axis.
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Solution
Let P(1, 6, 3) be the given point, and let 'L' be the foot of the perpendicular from 'P' to the given line AB (as shown in the figure below). The coordinates of a general point on the given line are given by
`(x - 0)/1 = (y - 1)/2 = (z - 2)/3` = λ; λ is a scalar, i.e., x = λ, y = 2λ + 1 and z = 3λ + 2
Let the coordinates of L be (λ, 2λ + 1, 3λ + 2).
So, direction ratios of PL are λ – 1, 2λ + 1 – 6 and 3λ + 2 – 3, i.e., λ – 1, 2λ – 5 and 3λ – 1.
Direction ratios of the given line are 1, 2 and 3, which is perpendicular to PL.
Therefore, (λ – 1)1 + (2λ – 5)2 + (3λ – 1)3 = 0
`\implies` 14λ – 14 = 0
`\implies` λ = 1
So, coordinates of L are (1, 3, 5).
Let Q(x1, y1, z1) be the image of P(1, 6, 3) in the given line.
Then, L is the mid-point of PQ.
Therefore, `((x_1 + 1))/2` = 1, `((y_1 + 6))/2` = 3 and `((z_1 + 3))/2` = 5
`\implies` x1 = 1, y1 = 0 and z1 = 7
Hence, the image of P(1, 6, 3) in the given line is (1, 0, 7).
Now, the distance of the point (1, 0, 7) from the y-axis is `sqrt(1^2 + 7^2) = sqrt(50)` units.
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