Question

The miller indices of a plane passing through the three points having coordinates (0, 0, 1) 4 , 1 2 , 1 2 1,0,0 1 are

Answer 1

The equation of any plane passing through (a,b,c) is

A(x-a) +B(y-b) +C(z-c)=0, where A,B and C are direction ratios of the line.

Also, it is given that plane passes through (0, 0, 1), (4 , 1 2) , (1 ,2 ,1),(0,0 1) respectively.

1.

A(0-a)+B(0-b)+C(1-c)=0

-a A-b B +C -Cc=0

a A + b B +c C= C------(1)

2.

A (4-a) +B(1-b) +C(2-c)=0

4 A +B +2 C=  a A + b B +c C--------(2)

Equating (1) and (2)

4 A +B +2 C=C

4 A +B + C=0-----------(a)

3.

A (1-a) +B(2-b) +C(1-c)=0

A +2 B +C=a A + b B +c C------(3)

Equating (1) and (3)

A +2 B +C=C

A +2 B=0--------------(b)

Equating (a) and (b), gives

$\frac{A}{0-2}=\frac{B}{1}=\frac{C}{8-1}=k\\\\ \frac{A}{-2}=\frac{B}{1}=\frac{C}{7}=k$

A= -2 k, B= k, C=7 k

So, equation of line having direction cosines , -2 k, k and 7 k is obtained by putting these values in equation (1),

→ -2 k a+k b + 7 k c= 7 k

→ -2 a +b+7 c=7, which is the required equation of plane.