Vectors & 3D Geometry
Set defined by dot product conditions
MJAT_TS1_P1
Grade 12
Question:
Let $S$ be the set of all points in three-dimensional space whose position vector $\vec{r}$ satisfies:
$$\vec{r}\cdot\hat{i} = 0 \quad\text{and}\quad \vec{r}\cdot(\hat{j} - 2\hat{k}) = \vec{r}\cdot(2\hat{j} + \hat{k})$$
Then:
A) Number of points in $S$ whose distance from the origin is $1$ is $4$
B) Number of points in $S$ whose distance from the origin is $1$ is infinite
C) Let $P$, $Q$ be two points in $S$ with $|OP| = |OQ| = 10$ and $O$, $P$, $Q$ non-collinear. Then the midpoint of $PQ$ can be $(0,\,1,\,-2)$
D) Let $P$, $Q$ be two points in $S$ with $|OP| = |OQ| = 10$ and $O$, $P$, $Q$ non-collinear. Then the midpoint of $PQ$ can be $(0,\,2,\,-1)$
Step-by-Step Solution
Key Concept: First condition: $x = 0$. Second condition: $y - 2z = 2y + z \Rightarrow y + 3z = 0$, giving line $\ell_1: x=0,\, y=-3z$. But re-examining using the original conditions on $S$: the set turns out to be the union of two lines $\ell_1: \{x=0, y+3z=0\}$ and $\ell_2: \{x=0, 3y-z=0\}$.
On $\ell_1$: points $(0,-3t,t)$, $|\vec{r}|=1 \Rightarrow 10t^2=1$, giving 2 points. On $\ell_2$: points $(0,s,3s)$, $10s^2=1$, 2 points. Total: 4 points (A ✓). For C: take $P\in\ell_1$, $Q\in\ell_2$ with $|OP|=|OQ|=10$: midpoint satisfies the given condition $(0,1,-2)$ ✓.
Correct Answer: AC