Question

The distance of the point i+2j+3k from the plane r (i+j+k)=5 which is measured parallel to a vector 2i+3j-6k is

Answer 1

Given that,

Point P = i+2j+3k = (1,2,3)

Plane =r i+rj+rk -5  

Vector = 2i+3j-6k = (2,3,-6)

We need to calculate the direction ratio from point to plane

Using formula of direction

$direction\ of\ ratio=(x_{2}-x_{1})+(y_{2}-y_{1})+(z_{2}-z_{1})$

Put the value into the formula

$direction\ of\ ratio=r-5-1, r-5-2, r-5-3$

$direction\ of\ ratio=r-6, r-7, r-8$...(I)

If we have perpendicular direction ratio.

Two direction is a₁,b₁,c₁ and a₂, b₂, c₂.

We need to find the value of r

Using the relation between two lines

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

Put the value of abc

$2(r-5)+3(r-7)-6(r-8)=0$

$2r-10+3r-21-6r+48=0$

$-r=-17$

$r=17$

We need to find the another point Q

Put the value of r in equation (I)

$Q=17-6, 17-7, 17-8$

$Q=11,10,9$

We need to calculate the distance between the points P and Q

Using formula of distance

$d=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}$

Put the value into the formula

$d=\sqrt{(11-1)^2+(10-2)^2+(9-3)^2}$

$d=14.1\ unit$

Hence, The distance between the points P and Q is 14.1 unit.