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Point P(6,3,0) to the line determined by the points Q(3,1,5) and R(4,5,2)

Short Answer

Expert verified

The answer is9.64 units.

Step by step solution

01

Given information

A point P(-6,3,0) to the line determined by the points Q(3,-1,5) and R(4,5,-2)

02

Calculation

The goal is to calculate the distance between the point and the line.

First find the line equation for the points Q(3,-1,5)and R(4,5,-2)

The direction vector d=(4-3,5-(-1),-2-5)

d=(1,6,-7)

The formula to find the line Lequation is as follows,

r(t)=P0+tdWhere, P0is the point and dis the direction vector.

Then, for Q(3,-1,5)and d=(1,6,-7)

r(t)=(3,-1,5)+t(1,6,-7)r(t)=(3+t,-1+6t,5-7t)

The formula for the distance is d×P0Pd

Let the point Pis P(-6,3,0)

The point P0 on the line equation r(t)=(3+t,-1+6t,5-7t)is (3,-1,5)

The direction vector P0P=(-6-3,3-(-1),0-5)

Then P0P=(-9,4,-5)

Take the direction vector d=(1,6,-7)

Substitute the values d=(1,6,-7) and P0P=(-9,4,-5) in the formula d×P0Pd

Then the distance =(1,6,-7)×(-9,4,-5)(1,6,-7)

03

Calculation

The cross product of (1,6,-7)×(-9,4,-5)is calculated as follows,

(1,6,-7)×(-9,4,-5)=ijk16-7-94-5

=i(-30+28)-j(-5-63)+k(4+54)=-2i+68j+58k

Thus,

Distance=-2i+68j+58k(1,6,-7)Distance=4+4624+33641+36+49Distance=799286

Distance=799286

On further simplification,

Distance =399643

Distance =9.64 units

Thus, the distance between the point P(-6,3,0) and the line joining Q(3,-1,5) and R(4,5,-2) is 9.64 units.

Therefore, the answer is 9.64 units.R(4,5,-2)

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