Question

Find the image of the point (5,7,3) in the line$\dfrac{{x - 5}}{3} = \dfrac{{y - 29}}{8} = \dfrac{{5 - z}}{5}$.

$(a)$ (9, 13, 15)
$(b)$ (13, 19, 27)
$(c)$ (18, 3, 0)
$(d)$ (0, -11, 30)

Answer 1

Hint: In the above given question, we are asked to calculate the image of the point in the line and we know that the image is formed perpendicular to the mirror. So, by using the given point and the line, the image on the point can be calculated.

Complete step-by-step answer:
We have the given point as P (5, 7, 3) and line$\dfrac{{x - 5}}{3} = \dfrac{{y - 29}}{8} = \dfrac{{5 - z}}{5}$.

Now, let us assume

$\dfrac{{x - 5}}{3} = \dfrac{{y - 29}}{8} = \dfrac{{5 - z}}{5} = k$ … (1)

\[(3k + 15,8k + 29, - 5k + 5)\]

So, any point on this line is given as $Q(3k + 15,8k + 29, - 5k + 5)$.

Therefore, we get the direction ratios of $PQ$ as

\[ = (3k + 15 - 5,8k + 29 - 7, - 5k + 5 - 3)\]

\[ = (3k + 10,8k + 20, - 5k + 2)\]

We are given in the question that $ PQ$ is perpendicular to$\dfrac{{x - 5}}{3} = \dfrac{{y -

29}}{8} = \dfrac{{5 - z}}{5}$.

Therefore, we have

\[ \Rightarrow 3(3k + 10) + 8(8k + 22) - 5( - 5k + 2) = 0\;\;\]

\[ \Rightarrow 9k + 30 + 64k + 176 + 25k - 10 = 0\;\;\]

\[ \Rightarrow 98k + 196 = 0\]
\[ \Rightarrow 98k = - 196\]

\[ \Rightarrow k = \dfrac{{ - 196}}{{98}}\]

\[ \Rightarrow k = - 2\]

By substituting this value in the equation (1), we get

\[ = Q(3( - 2) + 15,8( - 2) + 29, - 5( - 2) + 5)\]

\[ = Q( - 6 + 15, - 16 + 29,10 + 5)\]

\[ = Q(9,13,15)\]

Hence, we have the required solution as option $(a)$ (9, 13, 15).

Note: When we face such a type of problem, the key point is to have a good understanding of the geometry involving the equations of lines, direction ratios, etc. By calculating the direction ratio and using the conditions given in the question, the required solution can be obtained.