Question

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.)

Answer 1

Answer:

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 $d=\frac{|a\times b|}{|a|}$

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:

$a \times b= \begin{vmatrix} i & j & k\\ 4 &1&-1\\-5 &-5&-5\\ \end{vmatrix}$

= -6i+15j-15k

The distance is : $d=\frac{\sqrt{(-6)^{2}+(15)^{2}+(-15)^{2}}}{\sqrt{(4)^{2}+(1)^{2}+(-1)^{2}}}$

$=\frac{\sqrt{36+225+225}}{\sqrt{16+1+1}}$

$=\frac{\sqrt{36+225+225}}{\sqrt{16+1+1}}$

$d=\frac{\sqrt{486}}{\sqrt{18}}$

≈4.726

Therefore, the distance is  4.726