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junior college 2 | H2 Maths
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For part (i), why I cannot just cross OA and OB/ OA and OC/OB and OC since 2 points are needed to find the normal? I dont understand why the solution find the 2 possible direction vectors
Date Posted: 4 years ago
Views: 766
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To find the normal to the plane, you have to cross two vectors that are parallel to the plane (they can be contained in the plane or outside it), but not parallel to each other.
Now the question says the plane contains A, B and C. But it doesn't say that O is in that plane.
So you can't just cross OA&OB, OB&OC and OC&OA as we don't know if these vectors are parallel to the plane.
(They actually aren't since crossing them won't get you the required result)
In this question we know A, B and C are in the plane. And from their position vectors, we know they aren't collinear.
Since these 3 points are in the plane, then vectors AB, BC and AC are in the plane as well. None of them are parallel to each other since none of the points are collinear.
So we can cross any 2 of the 3 vectors to get the normal vector. The answer scheme finds AB and AC, then crosses them. It then uses OA for the position vector of the point on the plane to dot product with the normal vector. You can use the other 2 of position vectors as well.
If it was given that O was in the plane, then you can cross them directly.
Do note that you can also cross the vectors in the opposite direction (BA, CB, CB) as well. It's not only for 1 direction. Neither do the 2 vectors have to have the same starting point (eg. AB and CA is allowed, doesnt have to strictly be AB and AC with A as the common start point)
Now the question says the plane contains A, B and C. But it doesn't say that O is in that plane.
So you can't just cross OA&OB, OB&OC and OC&OA as we don't know if these vectors are parallel to the plane.
(They actually aren't since crossing them won't get you the required result)
In this question we know A, B and C are in the plane. And from their position vectors, we know they aren't collinear.
Since these 3 points are in the plane, then vectors AB, BC and AC are in the plane as well. None of them are parallel to each other since none of the points are collinear.
So we can cross any 2 of the 3 vectors to get the normal vector. The answer scheme finds AB and AC, then crosses them. It then uses OA for the position vector of the point on the plane to dot product with the normal vector. You can use the other 2 of position vectors as well.
If it was given that O was in the plane, then you can cross them directly.
Do note that you can also cross the vectors in the opposite direction (BA, CB, CB) as well. It's not only for 1 direction. Neither do the 2 vectors have to have the same starting point (eg. AB and CA is allowed, doesnt have to strictly be AB and AC with A as the common start point)
Typo in last paragraph. should be opposite direction (BA,CB,CA) instead of (BA, CB, CB)
Oh ok thanks a lot
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