let OA,OB,OC be 3 mutually perpendicular lines through the origin whose direction cosines are<l1,m1,n1>,<l2,m2,n2>,<l3,m3,n3>.if OA=OB=OC=a find equation of plane ABC
To find the equation of the plane ABC, we need to find the normal vector to the plane. The normal vector will be perpendicular to both vectors AB and AC, which lie in the plane.
1. Find vectors AB and AC:
- Vector AB: Subtract the coordinates of point A from point B.
AB = OB - OA = (l2, m2, n2) - (l1, m1, n1) = (l2 - l1, m2 - m1, n2 - n1)
- Vector AC: Subtract the coordinates of point A from point C.
AC = OC - OA = (l3, m3, n3) - (l1, m1, n1) = (l3 - l1, m3 - m1, n3 - n1)
2. Find the cross product of vectors AB and AC:
The cross product of two vectors is a vector that is perpendicular to both of them.
Cross product AB × AC = ((m2 - m1)(n3 - n1) - (n2 - n1)(m3 - m1), (n2 - n1)(l3 - l1) - (l2 - l1)(n3 - n1), (l2 - l1)(m3 - m1) - (m2 - m1)(l3 - l1))
Expand and simplify the equation to get the cross product vector.
3. The cross product vector is the normal vector to the plane. So, the equation of the plane is given by:
(x - x1)(l4) + (y - y1)(m4) + (z - z1)(n4) = 0,
where (x1, y1, z1) are the coordinates of any point on the plane and (l4, m4, n4) are the components of the cross product vector.
This equation represents the plane ABC.