Question

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* = 2*t*, *y* = *t* −
3, *z* = 2*t* + 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)

Can you please help me solve these 3 questions

Answer #1

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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