Q.(a) Find the image $A'$ of the point $A(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$. Also, find the equation of the line joining $A$ and $A'$.

Part (a)

Line $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=\lambda$, general point $M(\lambda,2\lambda+1,3\lambda+2)$. $\vec{AM}=(\lambda-1,2\lambda-5,3\lambda-1)\perp(1,2,3)$:

$M$ is the midpoint of $AA'$, so $A'=2M-A=(1,0,7)$. Direction $\vec{AA'}=(0,-6,4)\parallel(0,-3,2)$: …

Part (a)

Write the line as $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=\lambda$, so a general point is $M(\lambda,2\lambda+1,3\lambda+2)$ and the direction is $(1,2,3)$.

$\vec{AM}=(\lambda-1,\,2\lambda-5,\,3\lambda-1)$ must be perpendicular to $(1,2,3)$:

So the foot of the perpendicular is $M(1,3,5)$. As $M$ is the midpoint of $AA'$,

Direction $\vec{AA'}=(0,-6,4)\parallel(0,-3,2)$, so line $AA'$ is …

Method: Parametrise the line, then impose the geometric condition

Problems asking for a point related to a line — the foot of a perpendicular, the image (reflection) of a point in the line, or a point on the line at a fixed distance from another point — all reduce to one move: write a general point of the line using a single parameter, then turn the geometric requirement into one equation in that parameter.

Steps

Step 1: Write the general point of the line.

Set every ratio of the symmetric equation equal to a parameter $\lambda$. For $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}=\lambda$ the general point is

and the direction vector is $\vec d=(a,b,c)$. Every point of the line is one value of $\lambda$.

Step 2: Convert the geometric condition into an equation in $\lambda$.

• Foot of perpendicular / image of a point $A$: the vector $\vec{AM}$ must be perpendicular to the line, so $\vec{AM}\cdot\vec d=0$ — a linear equation in $\lambda$. …

Common Mistakes

Mistake 1: Reporting the foot of the perpendicular $M$ as the image $A'$.

Why it's wrong: $M$ is only the midpoint between the point and its image, not the image itself. Correct approach: reflect through $M$ with $A'=2M-A$ — in part (a), $M(1,3,5)$ gives $A'(1,0,7)$, not $(1,3,5)$.

Mistake 2: Forgetting the perpendicularity condition when locating $M$.

Why it's wrong: any point on the line is some point, but only the one with $\vec{AM}\cdot\vec d=0$ is the foot. Correct approach: impose $\vec{AM}\perp\vec d$ to fix $\lambda$, don't guess $M$. …