3D Geometry
Line — Point at Given Distance from Another Point
nta_pyq_2024_apr
Grade 12
Question:
Let the point, on the line passing through the points $P(1,-2,3)$ and $Q(5,-4,7)$, farther from the origin and at distance of 9 units from the point $P$, be $(\alpha,\beta,\gamma)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to:
Step-by-Step Solution
Key Concept: Direction of $PQ$: $(4,-2,4)$, unit vector $=(2,-1,2)/3$. Point at distance 9 from $P$ in direction away from origin: $(1,-2,3)+9\cdot(2/3,-1/3,2/3)=(1+6,-2-3,3+6)=(7,-5,9)$.
$(\alpha,\beta,\gamma)=(7,-5,9)$. $\alpha^2+\beta^2+\gamma^2=155$.
Correct Answer: 3