+0

0
106
3

Find the distance between $$Q = (3, -7, -1)$$ and the line through $$A = (1, 1, 2)$$ and $$B = (2, 3, 4).$$ This distance is equal to $$\dfrac{\sqrt{d}}{3}$$
for some integer $$d$$. What is $$d$$?

Nov 8, 2019

#1
+1

The distance is sqrt(30)/3, so d = 30.

Nov 9, 2019
#2
+2833
+1

How did you get 30?

What are the steps involved?

CalculatorUser  Nov 9, 2019
#3
+109202
+1

We can find the  direction vector, AB  =  s ,  of the line thusly

( 2 - 1, 3 - 1, 4 - 2)   =  ( 1, 2,  2)  = s

Next....find the vector  QA  =  ( 1 - 3, 1  - - 7, 2 - - 1)  =  ( -2, 8, 3)

Next.....form the cross product  of    QA  x s  =

i      j       k       i      j

-2   8      3      - 2   8     =

1    2      2      1     2

[ i * 8 * 2  + j * 3 * 1 + k * -2 * 2]  -  [ k * 8 * 1 + i * 3 * 2  + j * - 2 * 2 ]  =

[ 16i  + 3j - 4k] - [ 8k + 6i - 4j]  =

10i + 7j  - 12 k

The distance   =

Length  of   [QA x s]

_________________   =

Length  of   [ s ]

√ [ 10^2 + 7^2 + (-12)^2  ]                 √ 293         √ 293

_______________________   =     _______  =  ______

√ [ 1^2 + 2^2 + 2^2 ]                           √9                 3

So  d  = 293

Nov 9, 2019