239. The unit vector to the points p1(0,1,0), p2(1,0,1), p3(0,0,1) is a) (-j – k)/1.414 b) (-i – k)/1.414 c) (-i – j)/1.414 d) (-i – j – k)/1.414
239. The unit vector to the points p1(0,1,0), p2(1,0,1), p3(0,0,1) is a) (-j – k)/1.414 b) (-i – k)/1.414 c) (-i – j)/1.414 d) (-i – j – k)/1.414
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Answer: a
Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
The unit normal vector is given by, (-j –
k)/1.414.
Answer: a
Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
The unit normal vector is given by, (-j –
k)/1.414
Answer: a
Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
The unit normal vector is given by, (-j –
k)/1.414
a) (-j – k)/1.414
Answer: a
Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
The unit normal vector is given by, (-j –
k)/1.414
Answer: a
Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
The unit normal vector is given by, (-j –
k)/1.414.
Answer: a
Explanation: The cross product of p1, p2, p3 is a X b = -j – k and its magnitude is 1.414.
The unit normal vector is given by, (-j –
k)/1.414.