2004 (2nd)
course2-3.2
Let P and Q be points on the line y=x+4
where the x-coordinates of P and Q are p and p+1,respectively.Let O denote the origin.
(1)For 2 vectors OP(vector)and OQ(vector),the inner product OP(vector)*OQ(vector) takes its minimum when p is JK/L
(2)For P and Q satisfying (1),the value of cosPOQ is MN/OP
(3)For P and Q satisfying (1),consider the triangle PQR whose center of gravity is O.The coordinates of the vertex R are (Q,RS).
Please help me.Can you solve this problem?How to solve this?I don't understand well.So I need a help.
To solve this problem, let's break it down step by step:
(1) Finding the minimum value of OP(vector) * OQ(vector):
To find the minimum value of the inner product, OP(vector) * OQ(vector), we need to minimize the magnitudes of OP(vector) and OQ(vector) as well as the angle between them.
Given that P and Q are points on the line y = x + 4, we can write their coordinates as:
P: (p, p + 4)
Q: (p + 1, p + 5)
Now, we can compute the vectors OP(vector) and OQ(vector) as follows:
OP(vector) = (p, p + 4)
OQ(vector) = (p + 1, p + 5)
The inner product of two vectors can be calculated as the sum of the products of their corresponding components:
OP(vector) * OQ(vector) = (p)(p + 1) + (p + 4)(p + 5)
To find the minimum value of this expression, we can take the derivative with respect to p, set it equal to zero, and solve for p. Once we have p, we can substitute it back into the expression to find the minimum value.
(2) Finding the value of cosPOQ:
Once we have found the value of p in part (1), we can substitute it back into the vectors OP(vector) and OQ(vector) to calculate their magnitudes.
The magnitude of a vector (a, b) can be calculated using the formula: √(a^2 + b^2).
Next, we can use the dot product formula to find the value of cosPOQ:
cosPOQ = (OP(vector) * OQ(vector)) / (|OP(vector)| * |OQ(vector)|)
Substitute the computed values of OP(vector), OQ(vector), and their magnitudes, and simplify the expression to find the value of cosPOQ.
(3) Finding the coordinates of vertex R:
Given that the center of gravity (centroid) of the triangle PQR is at the origin O, we can determine the coordinates of R by averaging the x-coordinates and y-coordinates of P and Q, respectively.
The x-coordinate of R, QR (or Q - R), is the average of the x-coordinates of P and Q:
QR = (p + (p + 1)) / 2 = (2p + 1) / 2 = p + 1/2
The y-coordinate of R, SR (or S - R), is the average of the y-coordinates of P and Q:
SR = ((p + 4) + (p + 5)) / 2 = (2p + 9) / 2 = p + 9/2
Therefore, the coordinates of R are (Q, RS) = (p + 1/2, p + 9/2).
By following these steps, you should be able to solve the problem and find the answers for each part.