Question

Find the image of the point (1,6,3) in the line x/1 = (y-1)/2 = (z - 2)/3.

Answer 1

Let P(1, 6, 3) be the given point and let L be the foot of perpendicular from P to the given line.

The coordinates of a general point on the given line are

\(\frac {x -0}1 = \frac{y - 1}2 = \frac{z - 2}3 = \lambda\)

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,

\(\frac{x_1 + 1}2 = 1\)

\(\frac{y_1 + 6}2 = 3\)

\(\frac{z_1 + 3}2 = 5\)

⇒ x₁ = 1, y₁ = 0, z₁ = 7

Hence, the image of (1, 6, 3) in the given line is (1, 0, 7).

Answer 2

Any point ‘O’ on line AB is given by (k, 2k + 1, 3k + 2)

So, direction ratio of the line OP are k – 1 , 2k – 5, 3k – 1

. ^.. (k – 1) × 1 + (2k – 5) × 2 + (3k – 1) × 3 = 0

=> 14k – 14 = 0

=> k = 1

Hence, co-ordinate of O are (1,3,5)

Now, Let image of P(1,6,3) in the given line be Q (α, β , γ)

So, ‘O’ is the mid point of PQ

So, The image of P is R(1,0,7)