Let L be the line passing through the point P(−2, −2, 2) with direction vector d=[1, −2, 3]T. Find the shortest distance d from the point P0(−4, −4, −5) 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.
d=?
Q=(?,?,?)
how do you find Q?
To find the shortest distance d from the point P0(-4, -4, -5) to the line L, we can use the formula for the distance between a point and a line in 3D space.
First, let's find the equation of the line L in vector form. Since we have a point P(-2, -2, 2) on the line and a direction vector d = [1, -2, 3]𝑇, we can use the parametric form of the line:
𝑟 = 𝑟0 + 𝑡𝑑
where 𝑟 is a general point on the line, 𝑟0 is the point P(-2, -2, 2), 𝑡 is a parameter, and 𝑑 is the direction vector [1, -2, 3]𝑇.
So, the vector equation of the line L is 𝑟 = [-2, -2, 2] + 𝑡[1, -2, 3]𝑇.
Next, we need to find the point Q on the line L that is closest to P0(-4, -4, -5). To do this, we can find the projection of the vector between P0 and P onto the direction vector d.
Let 𝒗 be the vector between P0 and P, which is 𝒗 = P - P0 = [-2, -2, 2] - [-4, -4, -5] = [2, 2, 7].
We can find the projection of 𝒗 onto d using the dot product:
projd(𝒗) = (𝒗 · d/|d|^2) * d
= ([2, 2, 7] · [1, -2, 3]𝑇 / |[1, -2, 3]𝑇|^2) * [1, -2, 3]𝑇
= ((2 * 1) + (2 * -2) + (7 * 3) / (1^2 + (-2)^2 + 3^2)) * [1, -2, 3]𝑇
= (2 - 4 + 21 / 14) * [1, -2, 3]𝑇
= (19/14) * [1, -2, 3]𝑇
= [19/14, -19/7, 57/14]𝑇.
Now, we can find the point Q by adding the projection vector to the initial point P:
Q = P + projd(𝒗)
= [-2, -2, 2] + [19/14, -19/7, 57/14]𝑇
= [-2 + (19/14), -2 - (19/7), 2 + (57/14)]
= [-28/14 + 19/14, -14/7 - 38/14, 28/14 + 57/14]
= [-9/14, -36/14, 85/14]
= [-9/14, -18/7, 85/14].
Hence, the point Q on line L that is closest to P0 is Q = [-9/14, -18/7, 85/14].
Finally, to find the shortest distance d between P0 and line L, we can calculate the length of the vector between P0 and Q:
d = |P0 - Q|
= |[-4, -4, -5] - [-9/14, -18/7, 85/14]|
= |[-56/14, -56/14, -70/14] - [-81/14, -216/14, 425/14]|
= |[25/14, -160/14, 355/14]|
= √((25/14)^2 + (-160/14)^2 + (355/14)^2)
= √(625/196 + 25600/196 + 126025/196)
= √(187250/196)
= √(953125/196)
= (√953125) / (√196)
= (√953125) / 14
= 275√5 / 14.
Therefore, the shortest distance d from the point P0(-4, -4, -5) to the line L is d = 275√5 / 14, and the point Q on line L that is closest to P0 is Q = [-9/14, -18/7, 85/14].
so you want the line through P0 perpendicular to L. That is, it lies in the plane perpendicular to L.
That plane is (1)(x+2) + (-2)(y+2) + (3)(z-2) = 0
Or, x-2y+3z = 8
The distance is thus |(1)(-4)+(-2)(-4)+(3)(-5)|/√(1^2+2^2+3^2)
Now just find Q.