Find a unit vector perpendicular to the plane ABC, where the coordinates of A, B and C are A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1).
Given points A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1)
Let position vectors of the points A, B and C be,
and
respectively.
We know position vector of a point (x, y, z) is given by, where
,
and
are unit vectors along X, Y and Z directions.
Similarly, we have and
Plane ABC contains the two vectors and
.
So, a vector perpendicular to this plane is also perpendicular to both of these vectors.
Recall the vector is given by
Similarly, the vector is given by
We need to find a unit vector perpendicular to and
.
Recall a vector that is perpendicular to two vectors and
is
Here, we have (a1, a2, a3) = (–2, 0, –5) and (b1, b2, b3) = (1, –2, –1)
Let the unit vector in the direction of be
.
We know unit vector in the direction of a vector is given by
.
Recall the magnitude of the vector is
Now, we find.
So, we have
Thus, the required unit vector that is perpendicular to plane ABC is.