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reasoning & proof review
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name two planes that intersect in \\(\overleftrightarrow{qr}\\) in the figure to the right.
choose the correct answer below.
a. plane qts and plane rvu
b. plane tsr and plane uqt
c. plane qts and plane rsw
d. plane rsw and plane rvu
Brief Explanations
To determine which two planes intersect in \(\overleftrightarrow{QR}\), we analyze each option:
- Option A: Plane QTS and plane RVU. Plane QTS contains \(QR\) (since \(Q\) and \(R\) are vertices of this plane), and plane RVU: Let's check if \(QR\) is in both. Wait, actually, plane TSR (which is a base plane) and plane UQT (the front - left vertical plane) both contain \(\overleftrightarrow{QR}\)? Wait, no, let's re - examine. Plane TSR: points \(T\), \(S\), \(R\), \(Q\) (since it's a rectangular face, \(Q\), \(R\), \(S\), \(T\) form a face). Plane UQT: points \(U\), \(Q\), \(T\), \(X\) (a vertical face). The line \(\overleftrightarrow{QR}\) is part of the base edge. Plane TSR (the bottom face, with vertices \(Q\), \(R\), \(S\), \(T\)) and plane UQT (the front - left face, with vertices \(U\), \(Q\), \(T\), \(X\)): Do they share \(\overleftrightarrow{QR}\)? Wait, \(Q\) and \(R\) are in plane TSR (as \(Q\) and \(R\) are on the bottom edge) and in plane UQT? No, \(R\) is not in plane UQT. Wait, maybe I made a mistake. Let's check each option:
- Option A: Plane QTS (vertices \(Q\), \(T\), \(S\)) and plane RVU (vertices \(R\), \(V\), \(U\)). Do they share \(\overleftrightarrow{QR}\)? \(Q\) is in QTS, \(R\) is in RVU, but does the line \(\overleftrightarrow{QR}\) lie in both? No.
- Option B: Plane TSR (vertices \(T\), \(S\), \(R\), \(Q\)) and plane UQT (vertices \(U\), \(Q\), \(T\), \(X\)). The line \(\overleftrightarrow{QR}\) has points \(Q\) and \(R\). \(Q\) is in both plane TSR (since \(Q\) is a vertex of TSR) and plane UQT (since \(Q\) is a vertex of UQT), and \(R\) is in plane TSR. Wait, \(R\) is not in plane UQT. So maybe my initial analysis is wrong. Wait, the line \(\overleftrightarrow{QR}\) is along the edge from \(Q\) to \(R\). Plane TSR is the bottom face ( \(Q\), \(R\), \(S\), \(T\)) and plane UQT is the front - left face ( \(U\), \(Q\), \(T\), \(X\)). The intersection of two planes is a line. The line \(\overleftrightarrow{QT}\) is in both plane TSR and plane UQT, but we need \(\overleftrightarrow{QR}\). Wait, maybe I messed up the plane labels. Let's look at the cube: \(Q\), \(R\), \(S\), \(W\), \(V\), \(U\), \(T\), \(X\) (assuming it's a rectangular prism). The bottom face is \(Q\), \(R\), \(S\), \(T\) (plane TSR or QRS T), the front face (left - front) is \(Q\), \(T\), \(X\), \(U\) (plane UQT X), the right - front face is \(R\), \(S\), \(W\), \(V\) (plane RSW V), the top face is \(X\), \(W\), \(V\), \(U\) (plane XWV U), the left - back face is \(U\), \(X\), \(T\), \(Q\) (plane UQT X) and the right - back face is \(V\), \(W\), \(S\), \(R\) (plane RSW V). Now, the line \(\overleftrightarrow{QR}\) is the edge from \(Q\) to \(R\) on the bottom. Plane TSR (bottom face: \(Q\), \(R\), \(S\), \(T\)) and plane UQT (front - left face: \(U\), \(Q\), \(T\), \(X\)): The line \(\overleftrightarrow{QT}\) is common, but \(\overleftrightarrow{QR}\): \(Q\) is in both, \(R\) is in plane TSR. Wait, maybe the correct option is B. Let's check other options:
- Option C: Plane QTS ( \(Q\), \(T\), \(S\)) and plane RSW ( \(R\), \(S\), \(W\)). Do they share \(\overleftrightarrow{QR}\)? \(Q\) is in QTS, \(R\) is in RSW, but \(\overleftrightarrow{QR}\) is not in both.
- Option D: Plane RSW ( \(R\), \(S\), \(W\)) and plane RVU ( \(R\), \(V\), \(U\)). Do they share \(\overleftrightarrow{QR}\)? \(R\) is in both, but \(Q\) is not in either of these planes (except RSW? No, \(Q\) is not in RSW). So the correct option is B as plane TSR (which contains \(Q\) and \(R\) as part of its edge) and plane UQT (which contains \(Q\) and the line extendi…
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B. plane TSR and plane UQT