in the cube shown below, which lines are skew?
options:
\overleftrightarrow{nr} and \overleftrightarrow{pq}
\overleftrightarrow{qr} and \overleftrightarrow{mq}
\overleftrightarrow{pq} and \overleftrightarrow{op}
\overleftrightarrow{ko} and \overleftrightarrow{mq}

Question

Accepted Answer

# Explanation:
## Step1: Recall skew lines definition
Skew lines are non - parallel, non - intersecting, and lie in different planes.
## Step2: Analyze $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$
$\overleftrightarrow{NR}$ is vertical (along the edge from $N$ to $R$) and $\overleftrightarrow{PQ}$ is horizontal (top - face edge). But $\overleftrightarrow{NR}\parallel\overleftrightarrow{MQ}$ and $\overleftrightarrow{PQ}$ is parallel to some other edges, and actually $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$ are not skew (they are parallel? Wait, no, in a cube, $\overleftrightarrow{NR}$ is parallel to $\overleftrightarrow{KO}$ and $\overleftrightarrow{LM}$ and $\overleftrightarrow{PQ}$ is parallel to $\overleftrightarrow{OP}$, $\overleftrightarrow{LM}$, $\overleftrightarrow{KN}$. Wait, maybe I made a mistake. Let's re - analyze each pair:
- For $\overleftrightarrow{QR}$ and $\overleftrightarrow{MQ}$: $\overleftrightarrow{QR}$ and $\overleftrightarrow{MQ}$ intersect at $Q$, so they are not skew.
- For $\overleftrightarrow{PQ}$ and $\overleftrightarrow{OP}$: $\overleftrightarrow{PQ}$ and $\overleftrightarrow{OP}$ intersect at $P$, so they are not skew.
- For $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$: $\overleftrightarrow{KO}$ is a vertical edge (from $K$ to $O$) and $\overleftrightarrow{MQ}$ is a vertical edge? Wait, no. Wait, $KO$ is from $K$ to $O$ (left - front vertical), $MQ$ is from $M$ to $Q$ (right - back vertical). They are non - parallel, non - intersecting, and in different planes.
- For $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$: $\overleftrightarrow{NR}$ is from $N$ to $R$ (left - back vertical) and $\overleftrightarrow{PQ}$ is from $P$ to $Q$ (top - right horizontal). Wait, actually $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$: Let's check if they are parallel. The direction of $\overleftrightarrow{NR}$ is vertical, direction of $\overleftrightarrow{PQ}$ is horizontal (along the top face). Wait, no, in a cube, the top face edges are horizontal, and the vertical edges are vertical. Wait, maybe I messed up the edges. Let's label the cube properly: Let's assume the bottom face is $K - L - M - N$ (with $K$ bottom - left, $L$ bottom - front, $M$ bottom - right, $N$ bottom - back), and top face is $O - P - Q - R$ (with $O$ top - left, $P$ top - front, $Q$ top - right, $R$ top - back). Then:
    - $KO$: from $K$ (bottom - left) to $O$ (top - left) - vertical.
    - $MQ$: from $M$ (bottom - right) to $Q$ (top - right) - vertical. Wait, no, $MQ$ is from $M$ (bottom - right) to $Q$ (top - right), so $KO$ and $MQ$ are parallel? Wait, that can't be. Wait, maybe the labels are different. Wait, the original diagram: $K$ is bottom - left, $L$ is bottom - front, $M$ is bottom - right, $N$ is bottom - back; $O$ is top - left, $P$ is top - front, $Q$ is top - right, $R$ is top - back. Then:
        - $\overleftrightarrow{KO}$: $K$ to $O$ (left - side vertical).
        - $\overleftrightarrow{MQ}$: $M$ to $Q$ (right - side vertical). Wait, they are parallel? No, that's not right. Wait, maybe the pair $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$: Wait, no, let's check the other pairs again.
        - $\overleftrightarrow{QR}$ and $\overleftrightarrow{MQ}$: Intersect at $Q$, so not skew.
        - $\overleftrightarrow{PQ}$ and $\overleftrightarrow{OP}$: Intersect at $P$, so not skew.
        - $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$: $\overleftrightarrow{NR}$ is from $N$ (bottom - back) to $R$ (top - back), $\overleftrightarrow{PQ}$ is from $P$ (top - front) to $Q$ (top - right). Wait, $\overleftrightarrow{NR}$ is vertical (along the back - edge), $\overleftrightarrow{PQ}$ is horizontal (top - face, front - to - right). They are non - parallel, non - intersecting, and in different planes? Wait, no, maybe I made a mistake. Wait, the correct pair: Let's recall that skew lines are in different planes, not parallel, not intersecting.
        - $\overleftrightarrow{KO}$ (from $K$ to $O$) and $\overleftrightarrow{MQ}$ (from $M$ to $Q$): $KO$ is on the left - front - vertical, $MQ$ is on the right - back - vertical. They don't intersect, and they are not parallel (since their directions are the same? Wait, no, in a cube, all vertical edges are parallel. Wait, that's a mistake. All vertical edges in a cube are parallel. So $\overleftrightarrow{KO}\parallel\overleftrightarrow{NR}\parallel\overleftrightarrow{LM}\parallel\overleftrightarrow{MQ}$? No, $MQ$ is a vertical edge? Wait, $M$ is bottom - right, $Q$ is top - right, so $MQ$ is vertical. $KO$ is bottom - left to top - left, vertical. So they are parallel. So that's not skew.
        - Wait, let's take $\overleftrightarrow{KO}$ and $\overleftrightarrow{PQ}$: No, $\overleftrightarrow{KO}$ is vertical, $\overleftrightarrow{PQ}$ is horizontal. Wait, maybe the correct pair is $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ is wrong. Wait, let's re - examine the first pair: $\overleftrightarrow{NR}$ (from $N$ to $R$) and $\overleftrightarrow{PQ}$ (from $P$ to $Q$). $NR$ is back - vertical, $PQ$ is top - front - to - top - right. They are not parallel, not intersecting, and in different planes. Wait, no, maybe the answer is $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ is wrong. Wait, the correct pair: Let's check each option again:
            - Option 1: $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$: $\overleftrightarrow{NR}$ is parallel to $\overleftrightarrow{KO}$, $\overleftrightarrow{PQ}$ is parallel to $\overleftrightarrow{OP}$, $\overleftrightarrow{LM}$, $\overleftrightarrow{KN}$. So $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$ are parallel? No, $\overleftrightarrow{NR}$ is vertical, $\overleftrightarrow{PQ}$ is horizontal. Wait, no, in 3D, two lines with the same direction vector are parallel. The direction vector of $\overleftrightarrow{NR}$ is $(0,0,1)$ (assuming $z$ - axis vertical), and the direction vector of $\overleftrightarrow{PQ}$ is $(0,1,0)$ (assuming $y$ - axis horizontal on the top face). So they are not parallel. And they don't intersect. So they are skew? Wait, maybe I was wrong earlier.
            - Option 4: $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$: Direction vector of $\overleftrightarrow{KO}$ is $(0,0,1)$, direction vector of $\overleftrightarrow{MQ}$ is $(0,0,1)$, so they are parallel. So not skew.
            - Option 2: $\overleftrightarrow{QR}$ and $\overleftrightarrow{MQ}$: Intersect at $Q$, so not skew.
            - Option 3: $\overleftrightarrow{PQ}$ and $\overleftrightarrow{OP}$: Intersect at $P$, so not skew.
            - Wait, so the correct pair should be $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ is wrong. Wait, maybe the first pair $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$ is correct? Wait, no, in a cube, $\overleftrightarrow{NR}$ is parallel to $\overleftrightarrow{KO}$, $\overleftrightarrow{LM}$, $\overleftrightarrow{MQ}$ (vertical edges), and $\overleftrightarrow{PQ}$ is parallel to $\overleftrightarrow{OP}$, $\overleftrightarrow{LM}$, $\overleftrightarrow{KN}$ (horizontal top - face and bottom - face edges). Wait, $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$: $\overleftrightarrow{NR}$ is along the back - vertical, $\overleftrightarrow{PQ}$ is along the top - front - right. They are in different planes, not parallel, not intersecting. So they are skew? But earlier I thought $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ were parallel. Wait, $\overleftrightarrow{KO}$ is from $K$ (bottom - left) to $O$ (top - left), $\overleftrightarrow{MQ}$ is from $M$ (bottom - right) to $Q$ (top - right). So their direction vectors are $(0,0,1)$ (assuming $z$ - axis up), so they are parallel. So $\overleftrightarrow{KO}\parallel\overleftrightarrow{MQ}$, so they are not skew.
            - So the correct pair is $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ is wrong. Wait, the only pair that is non - intersecting, non - parallel is $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ is wrong. Wait, maybe the answer is $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$? No, they are parallel. Wait, I think I made a mistake in the direction vectors. Let's take coordinates: Let $K=(0,0,0)$, $L=(1,0,0)$, $M=(1,1,0)$, $N=(0,1,0)$, $O=(0,0,1)$, $P=(1,0,1)$, $Q=(1,1,1)$, $R=(0,1,1)$. Then:
                - $\overleftrightarrow{NR}$: from $N=(0,1,0)$ to $R=(0,1,1)$, direction vector $\vec{v}_{NR}=(0,0,1)$
                - $\overleftrightarrow{PQ}$: from $P=(1,0,1)$ to $Q=(1,1,1)$, direction vector $\vec{v}_{PQ}=(0,1,0)$
                - $\overleftrightarrow{QR}$: from $Q=(1,1,1)$ to $R=(0,1,1)$, direction vector $\vec{v}_{QR}=(- 1,0,0)$
                - $\overleftrightarrow{MQ}$: from $M=(1,1,0)$ to $Q=(1,1,1)$, direction vector $\vec{v}_{MQ}=(0,0,1)$
                - $\overleftrightarrow{PQ}$: from $P=(1,0,1)$ to $Q=(1,1,1)$, direction vector $\vec{v}_{PQ}=(0,1,0)$
                - $\overleftrightarrow{OP}$: from $O=(0,0,1)$ to $P=(1,0,1)$, direction vector $\vec{v}_{OP}=(1,0,0)$
                - $\overleftrightarrow{KO}$: from $K=(0,0,0)$ to $O=(0,0,1)$, direction vector $\vec{v}_{KO}=(0,0,1)$

Now, check each pair:
            - $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$: $\vec{v}_{NR}=(0,0,1)$, $\vec{v}_{PQ}=(0,1,0)$. Not parallel. Do they intersect? Let's see if there are $t_1,t_2$ such that $N + t_1\vec{v}_{NR}=P + t_2\vec{v}_{PQ}$. So $(0,1,0)+t_1(0,0,1)=(1,0,1)+t_2(0,1,0)$. So $0 = 1$ (from $x$ - coordinate), which is impossible. So they are skew? But earlier I thought $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$ were parallel. $\overleftrightarrow{KO}$ has direction $(0,0,1)$, $\overleftrightarrow{MQ}$ has direction $(0,0,1)$, so they are parallel. $\overleftrightarrow{QR}$ and $\overleftrightarrow{MQ}$: $Q$ is a common point, so they intersect. $\overleftrightarrow{PQ}$ and $\overleftrightarrow{OP}$: $P$ is a common point, so they intersect. $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$: parallel. So the only skew pair is $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$? Wait, no, in the coordinates, $\overleftrightarrow{KO}$ is $(0,0,0)+t(0,0,1)$, $\overleftrightarrow{MQ}$ is $(1,1,0)+t(0,0,1)$. They are parallel (same direction vector) and do not intersect (since $x = 0$ for $KO$ and $x = 1$ for $MQ$, $y = 0$ for $KO$ and $y = 1$ for $MQ$). So they are parallel, not skew. $\overleftrightarrow{NR}$ is $(0,1,0)+t(0,0,1)$, $\overleftrightarrow{PQ}$ is $(1,0,1)+t(0,1,0)$. No intersection, not parallel. So they are skew. But wait, the fourth option is $\overleftrightarrow{KO}$ and $\overleftrightarrow{MQ}$, which are parallel. The first option: $\overleftrightarrow{NR}$ and $\overleftrightarrow{PQ}$ are skew? Wait, maybe I made a mistake in the coordinate assignment. Let's re - assign coordinates: Let $N=(0,0,0)$, $R=(0,0,1)$, $P=(1,1,1)$, $Q=(0,1,1)$. Then $\overleftrightarrow{NR}$ is from $(0,0,0)$ to $(0,0,1)$, direction $(0,0,1)$. $\overleftrightarrow{PQ}$ is from

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QUESTION IMAGE

Question

Explanation:

Step1: Recall skew lines definition

Step2: Analyze \(\overleftrightarrow{NR}\) and \(\overleftrightarrow{PQ}\)

Answer:

Step1: Recall skew lines definition

Step2: Analyze \(\overleftrightarrow{NR}\) and \(\overleftrightarrow{PQ}\)