Problem 4. In R2, find the point on the line x2 = 2x1+1 that is closest to the point (5, 2) |
Problem 6. In R3, find the equation of the plane that passes through the points (2,4,1), (4,3,5), (7,2,3)7.
Problem 4: To find the point on the line x2 = 2x1+1 that is closest to the point (5, 2), we can use the formula for the distance between a point and a line.
First, let's solve the equation x2 = 2x1+1 for x2 in terms of x1:
x2 = 2x1 + 1
Next, we can substitute x2 into the distance formula:
distance = √((x1 - 5)^2 + (2 - (2x1 + 1))^2)
Now, let's simplify the expression inside the square root:
distance = √((x1 - 5)^2 + (-2x1 - 1)^2)
distance = √(x1^2 - 10x1 + 25 + 4x1^2 + 4x1 + 1)
distance = √(5x1^2 - 6x1 + 26)
To find the point on the line that minimizes the distance to (5, 2), we can find the minimum of the distance function. We can do this by finding the derivative of the distance function with respect to x1, setting it equal to 0, and solving for x1.
Now, let's find the derivative of the distance equation:
d(distance)/dx1 = (1/2) * (10x1 - 6) / √(5x1^2 - 6x1 + 26)
Setting the derivative equal to 0, we get:
10x1 - 6 = 0
10x1 = 6
x1 = 6/10
x1 = 3/5
Now, let's substitute x1 back into the line equation to find x2:
x2 = 2(3/5) + 1
x2 = 6/5 + 1
x2 = 11/5
Therefore, the point on the line x2 = 2x1+1 that is closest to the point (5, 2) is (3/5, 11/5).
Problem 6: To find the equation of the plane that passes through the points (2,4,1), (4,3,5), and (7,2,3), we can use the formula for the equation of a plane in three-dimensional space.
Let's assume the equation of the plane is ax + by + cz + d = 0. We want to find the values of a, b, c, and d.
To find a, b, and c, we can take the cross product of two vectors formed by the given points. Let vector AB be the vector from (2,4,1) to (4,3,5) and vector AC be the vector from (2,4,1) to (7,2,3).
Vector AB = <4-2, 3-4, 5-1> = <2, -1, 4>
Vector AC = <7-2, 2-4, 3-1> = <5, -2, 2>
Taking the cross product of AB and AC, we get:
<2, -1, 4> x <5, -2, 2> = <-14, 18, 9>
So, a = -14, b = 18, and c = 9.
Now, we can substitute the coordinates of one of the given points into the equation ax + by + cz + d = 0 to solve for d. Let's use (2,4,1):
-14(2) + 18(4) + 9(1) + d = 0
-28 + 72 + 9 + d = 0
53 + d = 0
d = -53
Therefore, the equation of the plane that passes through the points (2,4,1), (4,3,5), and (7,2,3) is -14x + 18y + 9z - 53 = 0.
To solve problem 4, we need to find the point on the line x2 = 2x1 + 1 that is closest to the point (5, 2) in R2.
Step 1: Rewrite the given equation in slope-intercept form:
x2 = 2x1 + 1
=> x2 - 2x1 = 1
Step 2: Calculate the slope of the line:
The coefficient of x1 is the slope, which is 2.
Step 3: Calculate the y-intercept of the line:
To find the y-intercept, let x1 = 0 and solve for x2:
x2 - 2(0) = 1
=> x2 = 1
So, the y-intercept is (0, 1).
Step 4: Use the point-slope form to find the equation of the line:
Using the point-slope form, we have:
y - y1 = m(x - x1)
=> y - 1 = 2(x - 0)
Simplifying:
=> y - 1 = 2x
Therefore, the equation of the line x2 = 2x1 + 1 is y - 1 = 2x.
To solve problem 6, we need to find the equation of the plane that passes through the points (2,4,1), (4,3,5), and (7,2,3) in R3.
Step 1: Identify two vectors lying on the plane:
We can take two vectors formed by subtracting one point from the other. Let's choose (4,3,5) - (2,4,1) and (7,2,3) - (2,4,1):
Vector 1 = (4,3,5) - (2,4,1) = (2, -1, 4)
Vector 2 = (7,2,3) - (2,4,1) = (5, -2, 2)
Step 2: Find the normal vector of the plane:
To find the normal vector, we take the cross product of Vector 1 and Vector 2:
Normal vector = (2, -1, 4) × (5, -2, 2)
= [(2*(-2)-(-1)*4), (2*2-4*5), ((2)*(-2)-(-1)*5)]
= (-8, -16, -1)
Step 3: Use the normal vector to find the equation of the plane:
Using the normal vector, we have:
Ax + By + Cz = D
-8x - 16y - z = D
To find the value of D, we substitute one of the points into the equation:
Using (2,4,1):
-8(2) - 16(4) - (1) = D
-16 - 64 - 1 = D
D = -81
Therefore, the equation of the plane passing through the points (2,4,1), (4,3,5), and (7,2,3) is -8x - 16y - z = -81.