3D Geometry
Line intersecting plane and yz-plane; distance
Grade Class 12
Question:
The line $(x,y,z)=(2,-3,4)+\lambda(1,2,-3)$ intersects $2x+3y-z=13$ at $P$ and $yz$-plane at $Q$. If $PQ=a\sqrt{b}$, $a,b\in\mathbb{N}$, $a>3$, then $\dfrac{a+b}{3}$ equals
Step-by-Step Solution
Key Concept: At $P$: $2(2+\lambda)+3(-3+2\lambda)-(4-3\lambda)=13\Rightarrow 2+2\lambda-9+6\lambda-4+3\lambda=13\Rightarrow 11\lambda=24\Rightarrow\lambda=2$. $P=(4,1,-2)$. At $Q$: $x=0\Rightarrow\lambda=-2$. $Q=(0,-7,10)$. $PQ=\sqrt{16+64+144}=\sqrt{224}=4\sqrt{14}$. $a=4,b=14$. $(a+b)/3=18/3=6$.
$\dfrac{a+b}{3}=6$.
Correct Answer: 6