If [tex]$\overrightarrow{PQ}$[/tex] and [tex]$\overrightarrow{RS}$[/tex] intersect to form four right angles, which statement is true?

A. [tex]$\overrightarrow{PQ} \perp \overrightarrow{RS}$[/tex]

B. [tex]$\overrightarrow{PQ}$[/tex] and [tex]$\overrightarrow{QS}$[/tex] are skew

C. [tex]$\overline{PQ}$[/tex] and [tex]$\overline{RS}$[/tex] are parallel

D. [tex]$\overrightarrow{PQ} = \overrightarrow{RS}$[/tex]



Answer :

To determine the correct statement about the vectors [tex]\( \overrightarrow{P Q} \)[/tex] and [tex]\( \overrightarrow{R S} \)[/tex], given that they intersect to form four right angles, let's analyze the geometrical context.

When two vectors intersect and form four right angles, it essentially means that the vectors are perpendicular to each other. Perpendicular vectors, by definition, intersect at an angle of 90 degrees. Since a right angle is 90 degrees, and the fact that they form four right angles suggests that wherever they intersect, they maintain perpendicularity.

Let's evaluate each of the given statements:

- Option A: [tex]\( \overrightarrow{P Q} \perp \overrightarrow{R S} \)[/tex]
- This statement indicates that the vectors [tex]\( \overrightarrow{P Q} \)[/tex] and [tex]\( \overrightarrow{R S} \)[/tex] are perpendicular. Given that they form four right angles upon intersection, this statement is true.

- Option B: [tex]\( \overrightarrow{P Q} \)[/tex] and [tex]\( \overrightarrow{Q S} \)[/tex] are skew
- Skew vectors are non-coplanar vectors that do not intersect and are not parallel. Since [tex]\( \overrightarrow{P Q} \)[/tex] and [tex]\( \overrightarrow{R S} \)[/tex] do indeed intersect to form right angles, they cannot be skew. Therefore, this statement is false.

- Option C: [tex]\( \overline{P Q} \)[/tex] and [tex]\( \overline{R S} \)[/tex] are parallel
- Parallel vectors never intersect and always remain equidistant from each other. Since it is given that [tex]\( \overrightarrow{P Q} \)[/tex] and [tex]\( \overrightarrow{R S} \)[/tex] intersect, they cannot be parallel. Therefore, this statement is false.

- Option D: [tex]\( \overrightarrow{P Q} = \overrightarrow{R S} \)[/tex]
- This statement indicates that the vectors are identical in both magnitude and direction. Since the vectors intersect perpendicularly and form right angles rather than overlapping completely, this statement is false.

After evaluating all options, the correct statement is:

A. [tex]\( \overrightarrow{P Q} \perp \overrightarrow{R S} \)[/tex].

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