A ray of light is incident on a surface in a direction given by vector `A=2i-2j+k`. The normal to that surface passing through the point of incidence is along the vector `N=j-2k`. The unit vector in the direction of reflected ray is given by `R=aj+bj+ck`. Find three equations in terms of `a,b,c` using which we can find the values of `a,b,& c`

Answer

`a^(2)+b^(2) + c^(2) =1;3a + 4b +2c = ; b - 2c = 4//3`

Explanation

`vec(A)=2hat(i)-2hat(j)+hat(k), vec(N)-hat(j)-2hat(k) vec(R)=ahat(i)+bhat(j)+chat(k) , thereforevec(R)` is a unit vector
`thereforea^(2)+b^(2)+c^(2)=1 .......(1)`
For all these vectors to be in a plane.
`(vec(A)xxvec(N)).vec(R)=0rArr[3hat(i)+4hat(j)+2hat(k)].[ahat(i)+bhat(j)+chat(k)]=0`
`rArr 3a+4b+2c=0..............(2)`
Now `vec(A).vec(N)=|vec(A)||vec(N)| cos (pi-i)`
and `vec(R).vec(N)=|vec(R)||vec(N)| cos i`
`therefore(vec(A).vec(N))/(|vec(A)||vec(N)|)=(-vec(R).vec(N))/(|vec(R)||vec(N|))rArr(-2-2)/(3)=(-(b-2c))/(1)rArrb-2c=(4)/(3).........(3)`
The equation `(1),(2)` and `(3)` are the required relations.