Let L be the line with parametric equations
x = 1−2t
y = 3+3t
z = −1−3t
Find the shortest distance d from the point P0=(1, −4, −2) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where needed to give an exact
value for your answer.
Find
d=?
Q=(_,_,_)
Please Post a full solution
To find the shortest distance d from the point P0=(1, -4, -2) to the line L, we can use the formula for the distance between a point and a line in 3D space. The formula is:
d = |(P0 - P) · n| / |n|
Where P0 is the given point, P is a point on the line, n is the direction vector of the line. To find the point Q on the line that is closest to P0, we need to find the value of t that corresponds to Q.
Step 1: Find the direction vector of the line L.
The direction vector of a line is obtained by taking the coefficients of t in the parametric equations. In this case, the direction vector of L is < -2, 3, -3 >.
Step 2: Find a point P on the line L.
To find a point on the line, we set t to any value. Let's take t = 0.
Plugging t = 0 into the parametric equations, we get P = < 1, 3, -1 >.
Step 3: Find the vector from P0 to P.
The vector from P0 to P is obtained by subtracting the coordinates of P0 from the coordinates of P.
P0 - P = < 1, -4, -2 > - < 1, 3, -1 > = < 0, -7, -1 >.
Step 4: Calculate |(P0 - P) · n| / |n|.
To calculate the distance d, we need to find the dot product of the vector P0 - P with the direction vector n, and divide it by the magnitude of n.
|(P0 - P) · n| = |< 0, -7, -1 > · < -2, 3, -3 >| = |-6 + (-21) + 3| = | -24 |
|n| = |< -2, 3, -3 >| = √((-2)^2 + 3^2 + (-3)^2) = √(4 + 9 + 9) = √22
d = |-24| / √22 = 24 / √22.
So, the shortest distance d from the point P0=(1, -4, -2) to the line L is 24 / √22, and the point Q on the line that is closest to P0 is Q = (1 - 2t, 3 + 3t, -1 - 3t), where t is the parameter value corresponding to the point Q.