Question:

To find the square of the distance of the image of the point $(6,1,5)$ in the line $\dfrac{x-1}{3}=\dfrac{y}{2}=\dfrac{z-2}{4}$, from the origin, we proceed as follows: Step 1: First, let's find the equation of the line in vector form. The line can be represented by a point on the line, say $P(1,0,2)$, and a direction vector $\vec{d} = (3,2,4)$. Thus, the line can be written as $\vec{r} = \vec{P} + t\vec{d}$, where $t$ is a parameter. Step 2: The point $M$ on the line that is closest to the point $(6,1,5)$ will be such that the vector joining $M$ and $(6,1,5)$ is perpendicular to the direction vector of the line. To find $M$, we first express the general point on the line as $(1+3t, 2t, 2+4t)$. Step 3: The vector $\vec{MP}$ joining $M$ and $P(6,1,5)$ is given by $\vec{MP} = (6 - (1+3t), 1 - 2t, 5 - (2+4t)) = (5-3t, 1-2t, 3-4t)$. For $\vec{MP}$ to be perpendicular to the direction vector $\vec{d}$, the dot product $\vec{MP} \cdot \vec{d}$ must be zero. Thus, we have: $$ \begin{aligned} (5-3t, 1-2t, 3-4t) \cdot (3,2,4) &= 0 \\ 15 - 9t + 2 - 4t + 12 - 16t &= 0 \\ 29 - 29t &= 0 \\ t &= 1 \end{aligned} $$ Step 4: Substituting $t=1$ into the equation of the line, we find the point $M$: $$ \begin{aligned} x &= 1 + 3t = 1 + 3(1) = 4 \\ y &= 2t = 2(1) = 2 \\ z &= 2 + 4t = 2 + 4(1) = 6 \end{aligned} $$ So, $M = (4,2,6)$. Step 5: The image $I$ of the point $(6,1,5)$ in the line is such that $M$ is the midpoint of the line segment joining $(6,1,5)$ and $I$. If $I = (x_I, y_I, z_I)$, then: $$ \begin{aligned} \frac{x_I + 6}{2} &= 4 \\ \frac{y_I + 1}{2} &= 2 \\ \frac{z_I + 5}{2} &= 6 \end{aligned} $$ Solving these equations gives: $$ \begin{aligned} x_I + 6 &= 8 \\ y_I + 1 &= 4 \\ z_I + 5 &= 12 \end{aligned} $$ Thus, $x_I = 2$, $y_I = 3$, and $z_I = 7$. So, $I = (2,3,7)$. Step 6: The distance $d$ of the point $I$ from the origin is given by: $$ \begin{aligned} d^2 &= (2-0)^2 + (3-0)^2 + (7-0)^2 \\ d^2 &= 4 + 9 + 49 \\ d^2 &= 62 \end{aligned} $$ Therefore: $62$ <div class="key-concept"><strong>Key Concept:</strong> Foot $M$ of perpendicular from $A(6,1,5)$ to line: $M=(3\lambda+1,2\lambda,4\lambda+2)$. $\overrightarrow{AM}\cdot(3,2,4)=0$: $(3\lambda-5)(3)+(2\lambda-1)(2)+(4\lambda-3)(4)=0\Rightarrow29\lambda=29\Rightarrow\lambda=1$.</div> <div class="trap-box"><strong>Trap:</strong> $M=(4,2,6)$. Image $I=2M-A=(2,3,7)$. Distance² from origin $=4+9+49=62$.</div>