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 determine three non-collinear points on the plane. Since OA, OB, and OC are mutually perpendicular lines through the origin, we can use their direction cosines to find three points on the plane.
The coordinates of point A can be found using the direction cosines <l1,m1,n1> as:
A = (a * l1, a * m1, a * n1)
Similarly, the coordinates of points B and C can be found using the respective direction cosines:
B = (a * l2, a * m2, a * n2)
C = (a * l3, a * m3, a * n3)
Now, we have three points A, B, and C on the plane. To find the equation of the plane, we can use the cross product of the vectors AB and AC.
Vector AB = B - A, which is (a * l2 - a * l1, a * m2 - a * m1, a * n2 - a * n1)
Vector AC = C - A, which is (a * l3 - a * l1, a * m3 - a * m1, a * n3 - a * n1)
Taking the cross product of AB and AC, we get:
N = (a * (m2*n3 - m3*n2), a * (n2*l3 - n3*l2), a * (l2*m3 - l3*m2))
The normal vector N gives us the coefficients of the equation of the plane ABC.
So, the equation of the plane ABC is:
(a * (m2*n3 - m3*n2)) * (x - a * l1) + (a * (n2*l3 - n3*l2)) * (y - a * m1) + (a * (l2*m3 - l3*m2)) * (z - a * n1) = 0
Simplifying this equation, we get the final form:
(l2*n3 - l3*n2)*x + (n2*l3 - n3*l2)*y + (l3*m2 - l2*m3)*z - (a^2 * (m2*n3 - m3*n2) * l1 + a^2 * (n2*l3 - n3*l2) * m1 + a^2 * (l2*m3 - l3*m2) * n1) = 0
This is the equation of the plane ABC.