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I need the answer to this Precalculus Problem!

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a. Let $$\mathbf{V} = (1,0,1),$$ and $$A = (-1,0,1).$$ Let P be the point on the line passing through A with direction vector $$(1,1,1)$$ that is closest to V.

b. Let $$\mathbf{V} = (0,3,2),$$ and $$A = (-1,0,1).$$ Let P be the point on the line passing through A with direction vector $$(1,1,1)$$ that is closest to V. Find P.

c. Find the distance between V and the line. Enter your answer in the form $$\frac{a \sqrt{b}}{c},$$ where a, b and c are all positive integers. (continuation from b)

May 20, 2023
edited by godmathguy  May 20, 2023

#1
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a.

The line passing through A with direction vector (1,1,1) is given by

L = (-1,0,1) + t(1,1,1)

for all real numbers t. The point P on this line that is closest to V is the point on the line that minimizes the distance between V and P. This distance is given by

d = ||V - (-1,0,1) - t(1,1,1)||

Expanding, we get

d = ||(1,0,1) + t(1,1,1)||

Taking the dot product of both sides with itself, we get

d^2 = (1,0,1) + t(1,1,1) . (1,0,1) + 2t(1,1,1) . (1,1,1) + t^2(1,1,1) . (1,1,1)

Expanding, we get

d^2 = 2 + 3t + t^2

Taking the square root of both sides, we get

d = \sqrt{2 + 3t + t^2}

We want to minimize this distance, so we need to minimize the expression inside the square root. This expression is minimized when t = -\frac{1}{3}, so the distance d is given by

d = \sqrt{2 + 3(-\frac{1}{3}) + (-\frac{1}{3})^2} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3}

Therefore, the point P on the line passing through A with direction vector (1,1,1) that is closest to V is given by

P = (-1,0,1) - \frac{2\sqrt{2}}{3}(1,1,1) = (1/3, -2*sqrt(2)/3, 1/3).

May 20, 2023
#2
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b.

The line passing through A with direction vector (1,1,1) is given by

L = (-1,0,1) + t(1,1,1)

for all real numbers t. The point P on this line that is closest to V is the point on the line that minimizes the distance between V and P. This distance is given by

d = ||V - (-1,0,1) - t(1,1,1)||

Expanding, we get

d = ||(0,3,2) + (1,0,1) - t(1,1,1)||

= ||(1,3,3) - t(1,1,1)||

Taking the dot product of both sides with itself, we get

d^2 = (1,3,3) - t(1,1,1) . (1,3,3) - 2t(1,1,1) . (1,1,1) - t^2(1,1,1) . (1,1,1)

Expanding, we get

d^2 = 10 - 3t - t^2

Taking the square root of both sides, we get

d = \sqrt{10 - 3t - t^2}

We want to minimize this distance, so we need to minimize the expression inside the square root. This expression is minimized when t = \frac{1}{3}, so the distance d is given by

d = \sqrt{10 - 3(1/3) - (1/3)^2} = \sqrt{\frac{26}{9}} = \frac{\sqrt{26}}{3}

Therefore, the point P on the line passing through A with direction vector (1,1,1) that is closest to V is given by

P = (-1,0,1) + \frac{\sqrt{26}}{3}(1,1,1) = \boxed{\left(-\frac{1}{3},\frac{\sqrt{26}}{3},\frac{4}{3}\right)}

May 20, 2023
#3
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The distance between V and the line is 3*sqrt(5)/2.

May 20, 2023