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Answer: a
Explanation: Distance vector from p1 and p2 is a = i – j + k. Distance vector from p1 and p3 is b = –j + k. The vector normal to these points is a X b = -j – k.
To find a vector normal to a plane defined by three points ( p_1(0,1,0) ), ( p_2(1,0,1) ), and ( p_3(0,0,1) ), we can use the following steps:
1. First, find two vectors that lie in the plane by subtracting the coordinates of the points:
[
vec{v_1} = p_2 – p_1 = (1,0,1) – (0,1,0) = (1, -1, 1)
]
[
vec{v_2} = p_3 – p_1 = (0,0,1) – (0,1,0) = (0, -1, 1)
]
2. Next, we need to find the cross product of these two vectors to determine a normal vector to the plane:
[
vec{n} = vec{v_1} times vec{v_2}
]
The cross product can be calculated using the determinant of the matrix:
[
vec{n} = begin{vmatrix}
hat{i} & hat{j} & hat{k} \
1 & -1 & 1 \
0 & -1 & 1