Let L be the line with parametric equations
x = −4+2t
y = 8+t
z = 8+3t
Find the vector equation for a line that passes through the point P=(−8, −6, 7) and intersects L at a point that is distance 3 from the point Q=(−4, 8, 8). Note that there are two possible correct answers. Use the square root symbol '√' where needed to give an exact value for your answer.
I have no idea how to do this question, please help
First I made sure that P was not on the given line, it isn't
So from Q we need a point on the given line which is 3 units away from
Q, let that point be R(x,y,z) or R(-4+2t, 8+t, 8+3t)
vector QR = <-4+2t + 4, 8+t - 8, 8+3t - 8>
= <2t, t, 3t>
|<2t, t, 3t>| = 3
√(4t^2 + t^2 + 9t^2) = 3
14t^2 = 9
t = ± √14/3
looks like a lot of tedious calculations ahead, would have expected
a "nicer" value for t
now that we have t, we can sub into the x,y, and z of the given line
to find the 2 points of R.
I will do the first one, you do the other one
if t = +√14/3
x = -4 + 2√14/3 = (2√14 - 12)/3
y = 8 + √14/3 = (√14 + 24)/3
z = 8 + 3√14/3 = 8+√14
Then vector PR = < (2√14+12)/3 + 8, (√14+24)/3 + 6, 8+√14-7>
= < (2√14 + 36)/3 , (√14 + 42)/3, √14 + 1>
= < 2√14+36, √14+42, 3√14 + 3>
so we now have the direction numbers and a point P(−8, −6, 7)
so the equation of one of the lines is
x = -8 + (2√14+36)t
y = -6 + (√14 + 42)t
z = 7 + (3√14+3)t
Repeat using t = -√14/3 to find the other point R
find the direction vector PR, as above,
then write the other line, perhaps use s as the parameter
Ohh btw, better check my arithmetic
To find the vector equation for a line that passes through the point P and intersects line L at a point that is distance 3 from point Q, we can follow these steps:
Step 1: Find the equations of the given line L
The parametric equations of line L are given as:
x = -4 + 2t
y = 8 + t
z = 8 + 3t
Step 2: Find the vector direction of line L
The vector direction of line L can be found by taking the coefficients of t:
L: <2, 1, 3>
Step 3: Find the vector from point Q to the point of intersection on line L
The vector from point Q to the point of intersection on line L can be found by subtracting the coordinates of Q from the coordinates of the point of intersection. Let's call this vector PQ.
PQ: <x₁ - x₂, y₁ - y₂, z₁ - z₂>
PQ: <(-4 + 2t) - (-8), (8 + t) - (-6), (8 + 3t) - 7>
PQ: <4 + 2t, 14 + t, 3t + 1>
Step 4: Set up the distance equation
We want the distance between points P and Q to be 3 units. We can use the distance formula:
√[(x₁ - x₂)² + (y₁ - y₂)² + (z₁ - z₂)²] = 3
Substitute the values from points P and Q into the distance equation:
√[(x₁ - (-8))² + (y₁ - (-6))² + (z₁ - 7)²] = 3
√[(-8 + 4 + 2t)² + (-6 - 14 - t)² + (1 + 3t - 7)²] = 3
√[(2t - 4)² + (-20 - t)² + (3t - 6)²] = 3
Step 5: Solve the distance equation
Squaring both sides to eliminate the square root:
[(2t - 4)² + (-20 - t)² + (3t - 6)²] = 3²
(2t - 4)² + (-20 - t)² + (3t - 6)² = 9
Expand and simplify:
4t² - 16t + 16 + 400 + 40t + t² + 9t² - 36t + 36 = 9
14t² - 73t + 421 = 9
Rearrange and simplify:
14t² - 73t + 412 = 0
Step 6: Solve the quadratic equation
Using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
For our quadratic equation 14t² - 73t + 412 = 0, the values of a, b, and c are:
a = 14
b = -73
c = 412
t = (-(-73) ± √((-73)² - 4(14)(412))) / (2(14))
t = (73 ± √(5329 - 23104)) / 28
t = (73 ± √(-17775)) / 28
Since we have a negative value under the square root (√(-17775)), this means there are no real solutions to the quadratic equation. Therefore, there is no point of intersection between line L and the line passing through point P.
As a result, there may be a mistake in the problem statement or additional information is needed to solve the problem correctly.
To find the vector equation for the line passing through point P and intersecting line L at a point that is a distance of 3 from point Q, we can follow these steps:
Step 1: Find the parametric equations for the line passing through point P and line L.
- For any point on the line passing through point P and intersecting line L, we can use the parametric form:
x = x1 + at
y = y1 + bt
z = z1 + ct
where (x1, y1, z1) is a known point on the line, and (a, b, c) are the direction ratios of the line.
Step 2: Use the distance formula to find the specific value of the parameter t.
- The distance between two points in 3D space can be calculated using the distance formula:
d = √[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]
In this case, we know that the distance between the point Q and the point on the line passing through P and intersecting L is 3. We can set up the equation:
3 = √[(x - (-4))^2 + (y - 8)^2 + (z - 8)^2]
Substitute the equations for x, y, and z from Step 1 and simplify.
Step 3: Solve for the values of a, b, and c.
- By comparing the coefficients of t in the equations obtained in Step 1 and Step 2, we can equate them and solve a system of three equations to find the values of a, b, and c.
Step 4: Write the vector equation.
- Once we have the direction ratios (a, b, c), we can write the vector equation for the line as:
r = P + t * <a, b, c>
where P is the point (x1, y1, z1) on line L and <a, b, c> are the direction ratios we found.
Note: Since the problem mentions there are two possible correct answers, you might get two different sets of direction ratios for the two possible lines.