Search any question & find its solution
Question:
Answered & Verified by Expert
If the circumcenter of the triangle formed by the points $(1,2,3),(3,-1,5)$ and $(4,0,-3)$ is $(\alpha, \beta, \gamma)$, then $|\alpha|+|\beta|=$
Options:
Solution:
2419 Upvotes
Verified Answer
The correct answer is:
$4|\gamma|$
$A(1,2,3) ; B(3,-1,5) ; C(4,0,-3)$
D.R. of $\overrightarrow{A B}=(3-1,-1-2,5-3) \equiv(2,-3,2)$
D.R. of $\overrightarrow{B C}=(4-3,0+1,-3-5) \equiv(1,1,-8)$
D.R. of $\overrightarrow{C A}=(1-4,2-0,3+3) \equiv(-3,2,6)$
$\begin{aligned} & \overrightarrow{A B} \cdot \overrightarrow{B C}=2-3-16=-17 \\ & \overrightarrow{A B} \cdot \overrightarrow{A C}=-6-6+12=0 \\ & \therefore \quad \overrightarrow{A B} \perp \overrightarrow{A C}\end{aligned}$
$\therefore A$ is right angle. $\Rightarrow$ Circumcentre is mid point of $B C$
$\begin{aligned} & \equiv\left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2}\right) \equiv\left(\frac{7}{2}, \frac{-1}{2}, 1\right) \\ & |\alpha|+|\beta|=\frac{7}{2}+\frac{1}{2}=4=4|\gamma| .\end{aligned}$
D.R. of $\overrightarrow{A B}=(3-1,-1-2,5-3) \equiv(2,-3,2)$
D.R. of $\overrightarrow{B C}=(4-3,0+1,-3-5) \equiv(1,1,-8)$
D.R. of $\overrightarrow{C A}=(1-4,2-0,3+3) \equiv(-3,2,6)$
$\begin{aligned} & \overrightarrow{A B} \cdot \overrightarrow{B C}=2-3-16=-17 \\ & \overrightarrow{A B} \cdot \overrightarrow{A C}=-6-6+12=0 \\ & \therefore \quad \overrightarrow{A B} \perp \overrightarrow{A C}\end{aligned}$
$\therefore A$ is right angle. $\Rightarrow$ Circumcentre is mid point of $B C$
$\begin{aligned} & \equiv\left(\frac{3+4}{2}, \frac{-1+0}{2}, \frac{5-3}{2}\right) \equiv\left(\frac{7}{2}, \frac{-1}{2}, 1\right) \\ & |\alpha|+|\beta|=\frac{7}{2}+\frac{1}{2}=4=4|\gamma| .\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.