Find the cartesian equation of the line which passes through the point `(-2,4,-5)` and parallel and line are `(3,5,6)`. So, the equation of line is,
`(x-(-2))/(3) = (y-4)/(5) = (z-(-5))/(6)`.

Options

A. `sqrt((37)/(10))`
B. `(37)/(sqrt10)`
C. `(sqrt37)/(10)`
D. None of these

Answer

A

Explanation

Here , `P(-2,4,-5)` is the given point
Equation of line `(x+3)/(3) = (y-4)/(5) = (z+8)/(6) = lambda`
Any point on the line is given by
`Q(3,lambda - 3,5 lambda + 4,6 lambda -8)`
Now, direction ratios of `PQ = [(3 lambda - 1), 5lambda,(6lambda - 3)]`
Since , PQ is perpendicular to the given line, thererefore was have
` 3(3 lambda - 1) +5(5lambda) +6(6lambda -3) =0" "[because a_(1)a_(2) +b_(1) + c_(1)c_(2) = 0]`
`rArr " " 9lambda + 25lambda + 36lambda = 21`
`rArr " "40 lambda = 21 i.e., lambda = (3)/(10)`
Now coordinate of
`Q(3xx(3)/(10)-3,5xx(3)/(10)-8) i.e., Q((-21)/(10) , (55)/(10),(-62)/(10))`

Now perpendicular distance .
`PQ = sqrt(((-21)/(10)+2)^(2) + ((55)/(10) -4)^(2)+((-62)/(10)+5)^(2))`
` =(1)/(10) sqrt((1)^(2) +(15)^(2) +(-12)^(2))`
`= (1)/(10) sqrt(1 + 225 + 144) = sqrt((370)/(10))=sqrt((37)/(10))`