Question

Vector Question,See Attachment.

Answer 1

@cwrw238

Answer 2

What i did was i knew dot product was equal to 0 so P1.P2=0 and <3,2,4>.<a,1,1> =0
therfore a=-2

Answer 3

give a generic plane, namely:
\[ax+by+cz=d\]
that plane is perpendicular to the vector:
\[(a,b,c)\]
so our planes are perpendicular each other if and only if, the subsequent condition is checked:
\[3a+2+4=0\]
from which:
\[a=2\].
Now your vector:
\[j-k+\lambda (i+2j+2k)\]
is:
\[(\lambda, 1+2\lambda, -1+2\lambda)\]
sowe get coordinates of A when lamda=1:
\[A=(1,3,1)\]
and, we have coordinates of B when lamda=2:
\[B=(2,5,3)\]
form which their distance d is:
\[d=3\]

Answer 4

where I applied the formula of euclidean distance in the geometrical space

Answer 5

Well When lambda is 1 how is it A?

Answer 6

A=(1,3,1)

Answer 7

Oh Perfect! Thanks a ton :)

Answer 8

Thanks!