What is the unit vector that is normal to the plane containing # ( i - 2 j + 3 k) # and # ( i + 7 j + 4 k) #?

1 Answer
Sep 18, 2017

Answer:

#1/sqrt(923)(-29i-j+9k)#

Explanation:

The cross product of these two vectors will be in a suitable direction, so to find a unit vector we can take the cross product then divide by the length...

#(i-2j+3k) xx (i+7j+4k) = abs((i, j, k), (1, -2, 3), (1, 7, 4))#

#color(white)((i-2j+3k) xx (i+7j+4k)) = abs((-2, 3),(7, 4))i + abs((3,1),(4,1))j + abs((1, -2),(1, 7))k#

#color(white)((i-2j+3k) xx (i+7j+4k)) = -29i-j+9k#

Then:

#abs(abs(-29i-j+9k)) = sqrt(29^2+1^2+9^2) = sqrt(841+1+81) = sqrt(923)#

So a suitable unit vector is:

#1/sqrt(923)(-29i-j+9k)#