Answer with explanation:
⇒u= P Q
Coordinates of P = (0,0)
Coordinates of Q= (9,12)
There is also a scalar,c such that, c<0.
Now, we will first find ,→ c P=(0,0)
→c Q= (9 c , 12 c), where c, is less than 0.
→c u=c × P Q
Both , 9 c, and 12 c, will be negative, So, this point will lie in third Quadrant.
Option A: The terminal point of vector cu lies in Quadrant III.