Check whether the 3 vectors ar coplanar or not
A= t+j+k , B= i+3j+k , C= 2i+2j+2k
If three vectors are coplanar, their scalar triple product is zero. Give it a go, and someone will check your work :)
I assume you mean A = i+j+k
Is A X B in the same direction as A X C?
You can work it out but by inspection you can see that vector A is the same direction as vector C and therefore, A and B (or C and B) define a plane.
No it's A = t+j+k
well, we know they are coplanar if t = 1
not if not
Thanks .. thats what i was doubtful about !!!
well it is very difficult for A to be perpendicular to both B and C otherwise
because j + k and 3 j+k just do not work :)
To check if the three vectors A, B, and C are coplanar, we need to determine if they lie on the same plane.
One way to check coplanarity is to find the normal vector to the plane formed by the three vectors. If this normal vector is zero or parallel to any of the three vectors, then the vectors are coplanar.
Let's find the normal vector to the plane. We can do this by taking the cross product of any two vectors on the plane. Here, we choose vectors AB and AC.
1. Calculate vector AB by subtracting vector A from vector B:
AB = B - A = (i + 3j + k) - (t + j + k) = (i + 3j + k) - (t + 1j + 1k) = (i - t + 2j + 0k).
2. Calculate vector AC by subtracting vector A from vector C:
AC = C - A = (2i + 2j + 2k) - (t + j + k) = (2i + 2j + 2k) - (t + 1j + 1k) = (2i - t + 1j + 1k).
3. Take the cross product of vectors AB and AC:
N = AB x AC = (i - t + 2j + 0k) x (2i - t + 1j + 1k).
To find the cross product, we use the following formula:
(AB x AC) = det(i, j, k; i - t, 2, 0; 2 - t, 1, 1).
Expanding the determinant, we get:
(AB x AC) = (2 - t)j - ((2 - t)(2 - t) - 1(i - t))k
= (2 - t)j - (t^2 - 4t + 3)i + (t^2 - 4t + 3)k.
If the normal vector N is equal to zero or is parallel to any of the vectors A, B, or C, then A, B, and C are coplanar. Otherwise, they are not coplanar.
So compare the components of the normal vector N with zero and each of the vectors A, B, and C to check for coplanarity.