Question

For any two points \( P, Q \in \mathbb{R}^3 \), let \( \vec{P, Q} \) denote the vector from the point \( P \) to the point \( Q \) and let \( d(P, Q) \) denote the length of \( \vec{P, Q} \). Let \( F: \mathbb{R}^3 \to \mathbb{R}^3 \) be a linear transformation such that

• A. \( F \) is an injective map.
• B. \( F \) is a surjective map.
• C. For any four points \( P, Q, R, S \in \mathbb{R}^3 \), we have \[ \vec{(F(P), F(Q))} \cdot \vec{(F(R), F(S))} = \vec{(P, Q)} \cdot \vec{(R, S)}, \] where \( \cdot \) denotes the usual dot product in \( \mathbb{R}^3 \).
• D. For any four points \( P, Q, R, S \in \mathbb{R}^3 \), we have \[ \vec{(F(P), F(Q))} \times \vec{(F(R), F(S))} = \vec{(P, Q)} \times \vec{(R, S)}, \] where \( \times \) denotes the usual cross product in \( \mathbb{R}^3 \).

Hint:

Isometries in \( \mathbb{R}^3 \) preserve lengths, angles, and the cross product, but they are not always surjective, particularly when they involve transformations like reflections or rotations that do not cover the entire space.

Answer 1

The Correct Option is B

Step 1: Use the distance condition.
We are told $F$ is linear and keeps all lengths the same, so $|F(v)| = |v|$ for every vector $v$. A linear map that preserves length is an orthogonal map.

Step 2: Injective is forced.
If $F(v)=0$ then $|v|=|F(v)|=0$, so $v=0$. The kernel is just zero, hence $F$ is one to one. So statement A holds.

Step 3: Surjective is also forced.
On a finite space like $\mathbb{R}^3$, a one to one linear map is automatically onto. So $F$ being surjective is actually true as well, not a property that can fail.

Step 4: Dot product is kept.
An orthogonal map preserves both lengths and angles, so $F(u)\cdot F(w) = u\cdot w$. Thus the dot product statement is true.

Step 5: Cross product can flip.
For the cross product, an orthogonal map keeps it only when the map preserves orientation. A reflection reverses orientation and then $F(u)\times F(w) = -\,u\times w$. So this is the property that need not hold.
\[ \boxed{\text{The cross product relation is the one that can fail}} \]