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determine whether the given pairs of lines are parallel, identical, intersecting, or skew. If the lines are parallel, compute the distance between them. If the lines intersect, find the point of intersection and the angle at which the lines intersect.

r1(t)=4+5t,6,7-2t

r2(t)=6-4t,-3+3t,-1+4t

Short Answer

Expert verified

The lines are not parallel andthe intersecting point of the lines is(-6,6,11)

Step by step solution

01

Step 1:Given information

The given pairs of lines are

r1(t)=4+5t,6,7-2t

r2(t)=6-4t,-3+3t,-1+4t

02

Step 2:Calculation

Consider the two lines,

r1(t)=(4+5t,6,7-2t)....eq(1)

r2(t)=(6-4t,-3+3t,-1+4t)....eq(2)

For the line equation

(1) that is for r1(t)=(4+5t,6,7-2t)the direction vector is,

d1=(5,0,-2)whered1is the direction vector

For the line equation

(2)that is for r2(t)=(6-4t,-3+3t,-1+4t)the direction vector is

d2=(-4,3,4). whered2is the direction vector

Here the direction vectors d1,d2are not scalar multiples of each other.

Thus, the lines are not parallel.

Thus, the lines r1(t)=(4+5t,6,7-2t)and r2(t)=(6-4t,-3+3t,-1+4t)are not parallel lines.

Now calculate the point of intersection of two lines.

To calculate the intersection point of two equations, replace the parameter tby uin the equation(2)that is r2(t)=(6-4t,-3+3t,-1+4t).

Then,

r2(u)=(6-4u,-3+3u,-1+4u)

Equate the values of equations,

r1(t)=(4+5t,6,7-2t)andr2(u)=(6-4u,-3+3u,-1+4u)

4+5t=6-4u...eq(3)

6=-3+3u...eq(4)

7-2t=-1+4u...eq(5)

Taketheequation(4)thatis6=-3+3u.

6+3=3u

u=3

Substitute u=3in the equation

(3) that is4+5t=6-4u.

4+5t=6-4·3

4+5t=-6

t=-2

Substitutet=-2inr1(t)=(4+5t,6,7-2t).

r1(-2)=(4+5·-2,6,7-2·-2)

r1(-2)=(-6,6,11)

Now substituteu=3in the equationr2(u)=(6-4u,-3+3u,-1+4u).

r2(3)=(6-4·3,-3+3·3,-1+4·3)

r2(3)=(-6,6,11)

The intersecting point of the lines is(-6,6,11).

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