Let L be the line through the point Q = (−1, 11, 11) and direction vector →d=[−1, 2, 2]T.
Find the two distinct points R1 and R2 on L at distance
3√10
to the point P = (1, 4, 1).
R1 = (0, 0, 0)
R2 = (0, 0, 0)
Well, it seems like there was a mix-up of coordinates here. Both R1 and R2 are actually the same point (0, 0, 0), which is not on the line L through point Q.
Let's try solving the problem correctly this time:
To find the two distinct points R1 and R2 on L at distance 3√10 from point P, we'll have to use some math.
First, let's find a point A on the line L that is closest to point P.
The vector from P to Q is given by →pq = Q - P = (-1, 11, 11) - (1, 4, 1) = (-2, 7, 10).
To find the projection vector of →pq onto →d, we can use the projection formula:
proj→d(→pq) = (→pq • →d / |→d|^2) • →d
where • denotes the dot product and |→d| represents the magnitude of →d.
Calculating further, we have:
proj→d(→pq) = (-2, 7, 10) • (-1, 2, 2) / ((-1)^2 + 2^2 + 2^2) • (-1, 2, 2)
= (-2)(-1) + (7)(2) + (10)(2) / 1^2 + 2^2 + 2^2 • (-1, 2, 2)
= 3 / 2 • (-1, 2, 2)
= (-3/2)(1, -2, -2)
= (-3/2, 3, 3).
Therefore, the point A on the line L closest to point P is given by A = P + proj→d(→pq) = (1, 4, 1) + (-3/2, 3, 3) = (-1/2, 7, 4).
Now, to find the points R1 and R2, we can use the vector equation of the line L:
→r = →a + t•→d
where →a is a point on the line (in this case, A), t is a parameter, and →d is the direction vector.
Substituting the values, we get:
→r = (-1/2, 7, 4) + t(-1, 2, 2)
We want to find the values of t that give us points R1 and R2 at a distance of 3√10 from point P. Using the distance formula:
distance = √[(r_x - p_x)^2 + (r_y - p_y)^2 + (r_z - p_z)^2] = 3√10
Substituting the values, we have:
√[(-1/2 - 1)^2 + (7 - 4)^2 + (4 - 1)^2] = 3√10
Simplifying this equation, we get:
√[(1/2)^2 + 3^2 + 3^2] = 3√10
√[(1/4) + 9 + 9] = 3√10
√(9 + 36/4) = 3√10
√(81/4) = 3√10
9/2 = 3√10
Multiplying both sides by 2/9, we get:
1 = 2/3√10
Square both sides to eliminate the square root:
1^2 = (2/3√10)^2
1 = (4/9)(10)
1 = 40/9
Oh boy, it seems like we've hit a contradiction. The equation is not true, which means there are no such points R1 and R2 on the line L at a distance of 3√10 from point P.
I apologize for the confusion, but it looks like this problem doesn't have a solution.
Apologies, but I made a mistake in the previous response. Let's try to find the correct solution.
To find the two distinct points R1 and R2 on L at a distance of 3√10 from P, we can follow these steps:
1. Find the parametric equations of the line L:
Let's start with the point Q = (-1, 11, 11) and the direction vector →d = [-1, 2, 2]ᵀ. The parametric equations of the line L can be expressed as:
x = -1 - t
y = 11 + 2t
z = 11 + 2t
where t is a parameter representing any real number.
2. Calculate the distance between the point P = (1, 4, 1) and the line L:
The distance between a point and a line can be calculated using the formula:
distance = |(P - Q) · →n| / |→d|
where |·| represents the magnitude, · represents the dot product, Q is a point on the line, →n is a vector orthogonal to the line, and →d is the direction vector of the line.
In this case, we have:
Q = (-1, 11, 11)
→d = [-1, 2, 2]ᵀ
(P - Q) = (1 - (-1), 4 - 11, 1 - 11) = (2, -7, -10)
|(P - Q)| = √(2² + (-7)² + (-10)²) ≈ √153
|→d| = √((-1)² + 2² + 2²) = √9 = 3
Therefore, the distance between P and L is:
distance = |(P - Q) · →n| / |→d| = √153 / 3
3. Set up an equation to find the value of t:
Since we know the distance between P and L is 3√10, we can set up an equation based on the distance formula:
√153 / 3 = (|(P - Q)·→n|) / 3
Squaring both sides:
153 / 9 = |(P - Q)·→n|² / 9
Simplifying:
17 = |(P - Q)·→n|²
4. Find the orthogonal vector →n:
An orthogonal vector →n to →d = [-1, 2, 2]ᵀ can be found by taking the cross product of →d with any non-parallel vector. Let's choose →u = [1, 0, 0]ᵀ as a non-parallel vector:
→n = →d x →u
→n = [-1, 2, 2]ᵀ x [1, 0, 0]ᵀ
→n = [0, 2, -2]ᵀ
5. Substitute the values into the equation:
We can substitute (P - Q)·→n = 2(0) + (-7)(2) + (-10)(-2) = 0 + (-14) + 20 = 6 into the equation:
17 = |(P - Q)·→n|²
17 = 6²
17 = 36
Since 17 is not equal to 36, there are no points R1 and R2 on L at a distance of 3√10 from P.
Apologies for the incorrect previous response. It seems that there was an error in the calculations.
Based on the given information, we are tasked with finding two distinct points, R1 and R2, on the line L that are at a distance of 3√10 from point P = (1, 4, 1).
To find these points, we can use the parametric equation of a line. Let's assume that R1 = (x1, y1, z1) and R2 = (x2, y2, z2) are points on the line L.
The parametric equation of a line passing through point Q and with direction vector →d can be written as follows:
x = -1 - t
y = 11 + 2t
z = 11 + 2t
where t is a scalar parameter.
Now, let's find the points R1 and R2 by substituting them into the parametric equations.
For R1:
x1 = -1 - t1
y1 = 11 + 2t1
z1 = 11 + 2t1
For R2:
x2 = -1 - t2
y2 = 11 + 2t2
z2 = 11 + 2t2
We are given that the distance between P = (1, 4, 1) and R1 or R2 is 3√10. This means we can use the distance formula to solve for t1 and t2.
The distance between two points (x1, y1, z1) and (x2, y2, z2) is given by the formula:
√[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2] = 3√10
Substituting the values of R1 and P into the distance formula:
√[(-1 - t1 - 1)^2 + (4 - (11 + 2t1))^2 + (1 - (11 + 2t1))^2] = 3√10
Simplifying and squaring both sides of the equation:
(-2 - t1)^2 + (4 - 11 - 2t1)^2 + (1 - 11 - 2t1)^2 = 90
Do the same for R2.
By solving the above equations for t1 and t2, we should be able to find the two distinct points, R1 and R2, on the line L that are at a distance of 3√10 from point P.
A general point R on our line L is
R(-1-t, 11+2t, 11+2t)
We want RP = 3/√10
Let's use our distance between two points formula
P(1,4,1) and R(-1-t, 11+2t, 11+2t)
√( (1+1+t)^2 + (4-11-2t)^2 + (1-11-2t)^2 ) = 3/√10
√(2+t)^2 + (-7-2t)^2 + (-10-2t)^2 ) = 3/√10
square both sides and expand
4 + 4t + t^2 + 49 + 28t + 4t^2 + 100 + 40t + 4t^2 = 9/10
9t^2 + 72t + 153 = 9/10
90t^2 + 720t + 1530 - 9 = 0
90t^2 + 720t + 1521 = 0
Argghhhhh, supposed to get 2 real values of t,
getting complex roots, can't find my error!!
Can somebody find where my error is ?
Found out what the problem is.
It is in the question itself, not my calculations.
Using projection, I just calculated that the perpendicular distance from
(1,4,1) to the given line is 3 units, which would be the closest to the line
we could get from the given point.
We are asked to find where the distance between (1,4,1) and a point on
the line is 3/√10.
of course 3/√10 < 3, so there are no points that close.