Find the distance from the point to the line. (-1,-2,1); x=4+4t, y=3+t, z=6-t .the distance is typn exact answer, using radicals as needed.)

The distance is  4.726

Step-by-step explanation:

we need to find the distance from the point to the line

Given:- point (-1,-2,1) and line ; x=4+4t, y=3+t, z=6-t .

used formula

Let point P be (-1,-2,1)

using value t=0 and t=1

The point Q (4 , 3, 6) and R ( 8, 4, 5)

Let a be the vector from Q to R :   a = < 8 - 4, 4 - 3, 5 - 6 > = < 4, 1, -1 >

Let b be the vector from Q to P:    b = < -1 - 4, -2 - 3, 1 - 6> = < -5, -5, -5 >

The cross product of a and b is:

= -6i+15j-15k

The distance is :

≈4.726

Therefore, the distance is  4.726

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

not only in mathematics, scientific proof also are required in the other field of science. also maths proofs are of particular structure, containing steps and logics. so proving is like doing science, it is science. that's why it is important to prove. 

an inaccurate proof leads to errors, or stating results that are not accurate.  which wil mess the probelm up you dont want this