Question

Find the image of the point 1, 6,3 in the line
1 2
1 2 3
x y  z 
  . Also find the equation of the line
joining the given point and its image.

Answer 1

Answer on Question #44599 – Math - Analytic Geometry

Problem.

Find the image of the point $(1, 6, 3)$ in the line

1 2

1 2 3

$x y - z -$

$\equiv =$. Also find the equation of the line

joining the given point and its image.

Remark.

The statement isn't correctly formatted. I suppose that the correct statement is

"Find the image of the point $(1, 6, 3)$ in the line

$$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}.$$

Also find the equation of the line joining the given point and its image."

Solution.

Let $A(1,6,3)$ and $l\colon \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$. Suppose that the perpendicular from $A$ to the line $l$ intersects the line $l$ at $B$ and image of point $A$ is $C$. The equation of the line $l$ can be rewritten, as

$$\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t$$

or $x = t, y = 2t - 1, z = 3t - 2$ where $t \in \mathbb{R}$.

Therefore point $B$ has coordinates $(k, 2k + 1, 3k + 2)$, where $k$ is unknown parameter.

The vector $\overrightarrow{AB}$ has coordinates $(k - 1, 2k + 1 - 6, 3k + 2 - 3) = (k - 1, 2k - 5, 3k - 1)$.

The direction vector of the line $l$ has coordinates $(1,2,3)$.

The line $l$ and the line $\overrightarrow{AB}$ are perpendicular, so the inner product of their direction vectors equals 0. Hence

$$(k - 1) \cdot 1 + (2k - 5) \cdot 2 + (3k - 1) \cdot 3 = 0$$

or $k = 1$. Therefore $B(1,3,5)$.

The point $C$ is the midpoint of the segment $AB$. If $C$ has coordinates $(x, y, z)$, then $\frac{x + 1}{2} = 1$, $\frac{y + 6}{2} = 3$, $\frac{z + 3}{2} = 5$ or $x = 1, y = 0, z = 7$. $C(1, 0, 7)$.

The vector $\overrightarrow{AC}$ has coordinates $(1 - 1, 0 - 6, 7 - 3) = (0, -6, 4)$.

The equation of line $AC$ is $x = 1, y = -6t + 6, z = 4t + 3$, where $t \in \mathbb{R}$.

Answer: $(1,0,7), x = 1, y = -6t + 6, z = 4t + 3$, where $t \in \mathbb{R}$.

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