Let two vertices of a triangle ABC be (2, 4, 6) and (0, -2, -5), and its centroid be (2, 1…

Question

Let two vertices of a triangle ABC be (2, 4, 6) and (0, -2, -5), and its centroid be (2, 1, -1). If the image of the third vertex in the plane $x+2y+4z=11$ is $$(\alpha,\beta,\gamma)$$

Accepted Answer

Given that two vertex of a $\triangle A B C$ be $A(2,4,6)$ and

$B(0,-2,-5)$ and $G=$ centroid $=(2,1,-1)$
[Given]

Let, the other vertex is $C(x, y, z)$

According to the question,

$$
\begin{aligned}
& \frac{2+0+x}{3}=2 \\
&\Rightarrow x  =4
\end{aligned}
$$

$$
\begin{aligned}
&\frac{4-2+y}{3}=1 \\
&\Rightarrow y=1
\end{aligned}
$$

$$
\begin{aligned}
& \frac{6-5+z}{3}=-1 \\
&\Rightarrow z=1
\end{aligned}
$$

Hence, third vertex is $C(4,1,-4)$

Now, if image of $C(4,1,-4)$ in the plane $x+2 y+4 z=11$ is $D(\alpha, \beta, \gamma)$.

So, $\frac{\alpha-4}{1}=\frac{\beta-1}{2}=\frac{\gamma+4}{4}=k$
(say)

$$
\Rightarrow \alpha=k+4, \beta=2 k+1, \gamma=4 k-4
$$

Then, co-ordinates of $D$ is $(k+4,2 k+1,4 k-4)$

Let, $E$ be the mid-point of $C D$, which lies on the plane $x+2 y+4 z=11$.

Then, co-ordinates of $E$ is

$$
\left(\frac{k+8}{2}, \frac{2 k+2}{2}, \frac{4 k-8}{2}\right)=\left(\frac{k+8}{2}, k+1,2 k-4\right)
$$

Since, $E$ lies on the plane, $x+2 y+4 z-11=0$

$$
\begin{aligned}
& \text { So, } \frac{k+8}{2}+2 k+2+8 k-16-11=0 \\
& \Rightarrow \frac{k+8+4 k+4+16 k-54}{2}=0
\end{aligned}
$$

$$
\[\begin{gathered} \Rightarrow  21 k  =42 \ \Rightarrow  k  =2 \end{gathered}$$
\]

Hence, $D \equiv(6,5,4) \equiv(\alpha, \beta, \gamma)$

So, $\alpha=6, \beta=5, \gamma=4$

Now, $\alpha \beta+\beta \gamma+\gamma \alpha=6 \times 5+5 \times 4+4 \times 6$

$=30+20+24=74$

Solutions · 1