3D Geometry
Foot of perpendicular; reflection; sum of coordinates
MJMT_Full_Test_05
Grade 12
Question:
If the length of the perpendicular from point $P(a, 4, 2)$, $a>0$, to the line $\dfrac{x+1}{2}=\dfrac{y-3}{3}=\dfrac{z-1}{-1}$ is $2\sqrt{6}$ units, and $Q(\alpha_1,\alpha_2,\alpha_3)$ is the image of $P$ on this line, then $a+\displaystyle\sum_{i=1}^3 \alpha_i$ equals
Step-by-Step Solution
Key Concept: Find the foot of perpendicular from $P$ to the line using the parameter $\lambda$. Use distance $=2\sqrt{6}$ to find $a$. Then reflect $P$ through the foot to get $Q$.
$a=5$, $M=(1,6,0)$, $Q=(-3,8,-2)$. Sum $=5-3+8-2=8$.
Correct Answer: 8