Linear Algebra question

Question

Find the shortest distance from the point P=(-3,-2,-3) to a point on the line given by 1: (x,y,z) = (-7t, 1t, 1t). Please help ASAP! Thanks!

Answer 1

[Tutor's NOTE: Whatever is in the square brackets is the Tutor's Notes. Not a part of the solution, but, often, a part of my teaching]

[I teach students to be aware of possibility of using more than one approach/strategy for solving a problem. Knowledge of more than one approach often leads to better understanding of each of them, and it certainly allows the student to go for the one s/he is the most comfortable with]

Observe: Although the title is "Linear Algebra Question", the text does not constrain solutions to the methods of Linear Algebra.

For a visualization, follow the link https://www.desmos.com/calculator/4zhs61ryco

To find the shortest distance we need

I. Find the distance formula, possibly using

A. Linear Algebra, OR

B. Geometry

Then

II. Find the minimum of the formula/function, possibly using

A. Algebra, OR

B. Calculus

I. Find the distance formula

A. Linear Algebra

I would not suggest going Linear Algebra way, because that approach is, obviously, much heavier for this problem.

The two points P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂) are treated as Position Vectors, v₁ and v₂.

Then several steps are required ending up with calculating |P₁P₂|, the distance d, as

d = sqrt((x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²)

which is what the 3D Geometry gives us immediately!

B. Geometry

The geometric 3D distance formula is an extension of the Pythagorean Theorem into 3D.

For two points P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂), |P₁P₂| is calculated from

d² = (x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²

II. Find the shortest distance d. We need to find the minimum of the formula/function.

A. Algebra

Given: P₁(-3, -2, -3)

Given: P₂ is a point on the line (x,y,z) = (-7t, 1t, 1t)

d² = (-7t + 3)² + (t + 2)² + (t +3)² = f(t)

Shorted distance d will be reached at the point P₂ for which function f(t) will reach its minimum.

f(t) = 49t² - 42t + 9 + t² +4t + 4 + t² + 6t + 9 = 51t² - 32t + 22

a = 51 > 0, so the parabola opens up, and

its minimum is reached at its vertex V with x-axis coordinate t_v = - (b/2a) = 32/2•51.

Plug 32/102 into f(t) and calculate with Desmos or TI-84

f(t_v) = f(32/102) ≈ 16.98 ≈ d²

d ≈ 4.12

B. Calculus

If you were comfortable with Calculus, go for it.

Extreme of a function is reached when its (first) derivative function equals 0.

Solution is overall the same as in the II.B. Algebra way, except when getting the t_v value.

1. Find the distance squared the same (Algebra) way:

d² = (-7t + 3)² + (t + 2)² + (t +3)² = f(t) = 51t² - 32t + 22

2. Find the t for which f(t) reaches minimum the Calculus way

Derivative f´(t) = 102t - 32 = 0

t = 32/102

3. Find shortest d same (Algebra) way, but after getting t = 32/102 a little bit quicker

d ≈ 4.12

Answer: The shortest distance from the point P=(-3,-2,-3) to a point on the line given by 1: (x,y,z) = (-7t, 1t, 1t) is approximately 4.12

Answer 2

A unit vector tangent to the line is: T = (-7, 1, 1) / √51

The projection of the vector P on T is found with dot product:

P . T = (-3,-2,-3) . (-7, 1, 1) / √51 = 16 / √51

The component vector of P along T is:

P_|| = (16 / √51) T

The component vector of P, that is perpendicular to the line is the remainder:

P_⊥ = P - P_||

The magnitude of P_⊥ is the shortest distance to the line.