1.If a theatre holds 1,800 people, and sells ( 1)/(3) of its seats, how many tickets does it sell?
2.Find the vector parametrization for the line segment that begins
at P ( 2, 7, -1) and ends at Q ( 4, 2, 3 ).
The correct answer and explanation is:
Solution 1: Number of Tickets Sold
The theater holds 1,800 people. Since it sells 13\frac{1}{3} of its seats, the number of tickets sold is: 13×1800=600\frac{1}{3} \times 1800 = 600
So, the theater sells 600 tickets.
Solution 2: Vector Parametrization of a Line Segment
To parametrize the line segment from P(2, 7, -1) to Q(4, 2, 3), we use the formula: r(t)=P+t(Q−P),0≤t≤1\mathbf{r}(t) = \mathbf{P} + t (\mathbf{Q} – \mathbf{P}), \quad 0 \leq t \leq 1
Step 1: Find the Direction Vector
The direction vector PQ is: PQ=Q−P=(4−2,2−7,3−(−1))=(2,−5,4)\mathbf{PQ} = Q – P = (4 – 2, 2 – 7, 3 – (-1)) = (2, -5, 4)
Step 2: Write the Parametric Equations
Since a point on the line is given by: r(t)=(2,7,−1)+t(2,−5,4),0≤t≤1\mathbf{r}(t) = (2, 7, -1) + t(2, -5, 4), \quad 0 \leq t \leq 1
Breaking it into components: x=2+2tx = 2 + 2t y=7−5ty = 7 – 5t z=−1+4tz = -1 + 4t
where 0≤t≤10 \leq t \leq 1.
Thus, the vector parametrization of the line segment is: (x,y,z)=(2+2t,7−5t,−1+4t),0≤t≤1( x, y, z ) = ( 2 + 2t, 7 – 5t, -1 + 4t ), \quad 0 \leq t \leq 1
Explanation
A vector parametrization is a way of representing a line segment using a parameter tt, which varies within a specific interval. The line segment from point P(2,7,-1) to Q(4,2,3) can be defined using vector operations.
The general equation for a line passing through two points P and Q in three-dimensional space is: r(t)=P+t(Q−P)\mathbf{r}(t) = \mathbf{P} + t(\mathbf{Q} – \mathbf{P})
where tt is a real number that determines the position on the line.
To construct the equation, we first determine the direction vector PQ by subtracting the coordinates of P from Q: (4−2,2−7,3−(−1))=(2,−5,4)(4-2, 2-7, 3 – (-1)) = (2, -5, 4)
This vector gives the direction in which the line moves. Using this, we write the parametric equations: x=2+2t,y=7−5t,z=−1+4t,0≤t≤1x = 2 + 2t, \quad y = 7 – 5t, \quad z = -1 + 4t, \quad 0 \leq t \leq 1
The values of tt range from 0 to 1 because at t=0t = 0, we are at P(2,7,-1), and at t=1t = 1, we reach Q(4,2,3). As tt varies between 0 and 1, the point moves along the line segment.
This form of parametrization is useful in physics, computer graphics, and engineering, as it allows easy representation of motion, interpolation, and geometric calculations.
Now, I’ll generate an image to visualize the line segment.
Here is the visualization of the line segment from P(2,7,-1) to Q(4,2,3) in 3D space. The red dot represents point P, the green dot represents point Q, and the blue line shows the line segment connecting them. Let me know if you need any modifications!