Let L be the line with parametric equations
x = −8+2t
y = 4+3t
z = −4
Find the vector equation for a line that passes through the point P=(−2, 1, 3) and intersects L at a point that is distance 2 from the point Q=(−8, 4, −4). Note that there are two possible correct answers. Use the square root symbol '√' where needed to give an exact value for your answer.
To find the vector equation for a line that passes through point P and intersects line L at a point that is a distance of 2 from point Q, we can use the following steps:
Step 1: Find the direction vector of the line L.
The direction vector of line L can be determined by taking the coefficients of t in the parametric equations. In this case, the direction vector of L is ⟨2, 3, 0⟩.
Step 2: Find the vector connecting points P and Q.
To find the vector connecting points P and Q, we subtract the coordinates of Q from the coordinates of P. The vector connecting P and Q is ⟨-2 - (-8), 1 - 4, 3 - (-4)⟩ = ⟨6, -3, 7⟩.
Step 3: Find the unit vector in the direction of the vector connecting P and Q.
To find the unit vector, we divide the vector connecting P and Q by its magnitude. The magnitude of the vector connecting P and Q is √(6^2 + (-3)^2 + 7^2) = √(36 + 9 + 49) = √94. Therefore, the unit vector in the direction of the vector connecting P and Q is ⟨6/√94, -3/√94, 7/√94⟩.
Step 4: Determine the two possible vectors that are distance 2 from point Q.
Since we want to find the points that are a distance of 2 from point Q, we multiply the unit vector by 2 and subtract/add it to the coordinates of point Q. This gives us the two possible vectors:
Option 1:
Point A: ⟨-8, 4, -4⟩ + 2 * ⟨6/√94, -3/√94, 7/√94⟩ = ⟨-8 + 12/√94, 4 - 6/√94, -4 + 14/√94⟩
Option 2:
Point B: ⟨-8, 4, -4⟩ - 2 * ⟨6/√94, -3/√94, 7/√94⟩ = ⟨-8 - 12/√94, 4 + 6/√94, -4 - 14/√94⟩
Therefore, the two possible vector equations for a line that passes through point P and intersects line L at a point that is a distance of 2 from point Q are:
Option 1: r = ⟨-8 + 12/√94, 4 - 6/√94, -4 + 14/√94⟩ + t * ⟨2, 3, 0⟩
Option 2: r = ⟨-8 - 12/√94, 4 + 6/√94, -4 - 14/√94⟩ + t * ⟨2, 3, 0⟩
To find the vector equation for the line passing through point P that intersects line L at a distance of 2 from point Q, we need to follow these steps:
Step 1: Find the direction vector of line L.
The direction vector of line L can be obtained by looking at the coefficients of the parameter 't' in the given parametric equations. In this case, the direction vector of line L is d = (2, 3, 0).
Step 2: Find the vector connecting points P and Q.
The vector connecting points P and Q can be found by subtracting the coordinates of point P from the coordinates of point Q. Let's call this vector v = (Q - P) = (-8 + 2, 4 + 3, -4 - 3) = (-6, 7, -7).
Step 3: Find the normalized direction vector of line L.
To find one possible direction vector for the line passing through point P and intersecting line L at a distance of 2 from point Q, we need to add or subtract a scalar multiple of vector v to the direction vector of line L. It's important to note that to get the second possible direction vector, we can simply change the sign of the scalar multiple used. Since the question asks for the exact value of the answer, we'll leave it in the form of fractions.
To normalize vector d, we need to divide it by its magnitude. The magnitude of a vector (a, b, c) is given by √(a^2 + b^2 + c^2).
The magnitudes of vector d and vector v are:
|d| = √(2^2 + 3^2 + 0^2) = √13
|v| = √((-6)^2 + 7^2 + (-7)^2) = √(36 + 49 + 49) = √134
The normalized direction vector of line L is:
d_norm = (2/√13, 3/√13, 0)
Step 4: Find the two possible direction vectors for the line passing through point P.
Using vector v and the normalized direction vector d_norm, we can find the two possible direction vectors for the line passing through point P by adding or subtracting scalar multiples of v from d_norm.
Let's call the two direction vectors d1 and d2. Then:
d1 = d_norm + k1 * v
d2 = d_norm - k2 * v
where k1 and k2 are scalar multiples.
Step 5: Write the vector equation for the line passing through point P.
The vector equation for the line passing through point P with a direction vector d1 is:
r1 = P + t * d1
The vector equation for the line passing through point P with a direction vector d2 is:
r2 = P + t * d2
where r1 and r2 represent points on the line, and t is a parameter.
So, the two possible vector equations for the line passing through point P and intersecting line L at a distance of 2 from point Q are:
r1(t) = (-2, 1, 3) + t * (2/√13, 3/√13, 0) + 2(v/|v|^2)
r2(t) = (-2, 1, 3) + t * (2/√13, 3/√13, 0) - 2(v/|v|^2)