Question

# 1. Find the distance between the point and the line given by the set of parametric...

1. Find the distance between the point and the line given by the set of parametric equations. (Round your answer to three decimal places.)

(8, -1, 4); x = 2t, y = t − 3, z = 2t + 2

I had put 6.87 but it shows as it being wrong.

2. Consider the following planes.

 −6x + y + z = 6 24x − 4y + 6z = 36

Find the angle between the two planes. (Round your answer to two decimal places.)

3. Verify that the points are the vertices of a parallelogram, and find its area.

A(1, 1, 3), B(6, −7, 7), C(8, −4, 0), D(3, 4, −4)

1.

P(8,-1,4)

Line,R(2t,t-3,2t+2)

=(0+2t,-3+t,2+2t)

Directional vector of line(2,1,2)

When t=0,point on line(0,-3,2)

Distance

=(0-8,-3+1,2-4)(2,1,2)/√(9)

=|(-8,-2,-2)(2,1,2)|/3

=22/3=7.33

2.

-6x+y+z=6

24x-4y+6z=36

Cos∅=((-6*24-4+6)/(√38)(√628))

Cos∅=0.919

∅=23.22⁰

3.

A(1,1,3)

B(6,-7,7)

C(8,-4,0)

D(3,4,-4)

AB=√(6-1)^2 +(-7-1)^2 +(7-3)^2

=√105

BC=√62

CD=√105

DA=√62

AB=CD

BC=DA

SO it is a parallelogram

AC=√83

Area of triangle ABC

s= (√105+√62+√83)/2=13.62

s-AB=(13.62-√105)/2=1.68

s-BC=2.87

s-AC=2.255

Area=√s(s-AB)(s-BC)(s-AC)

=12.17

area of parallelogram=2*12.17=24.34

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