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Find the image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.
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Let P(1, 6, 3) be the given point and let L be the foot of a perpendicular from P to the given line.
The coordinates of a general point on the given line are
`(x - 0)/1 = (y - 1)/2 = (z - 2)/3 = lambda`
i.e., x = λ, y = 2λ + 1, z = 3λ + 2.
If the coordinates of L are (λ, 2λ + 1, 3λ + 2), then the direction ratios of PL are λ – 1, 2λ – 5, 3λ – 1.
But the direction ratios of given line which is perpendicular to PL are 1, 2, 3.
Therefore, (λ – 1)1 + (2λ – 5)2 + (3λ – 1)3 = 0, which gives λ = 1.
Hence 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
`(z_1 + 3)/2` = 5
⇒ x1 = 1, y1 = 0, z1 = 7
Hence, the image of (1, 6, 3) in the given line is (1, 0, 7).
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