The position vectors of points A and B are i | vedclass

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(C) The position vectors of the given points are $a = i - j + 3k$ and $b = 3i + 3j + 3k$.The equation of the plane is $r \cdot (5i + 2j - 7k) + 9 = 0$

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Are on the opposite side of the plane

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(C) The position vectors of the given points are $a = i - j + 3k$ and $b = 3i + 3j + 3k$.The equation of the plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. Let $f(r) = r \cdot (5i + 2j - 7k) + 9$.For point $A$,we calculate $f(a) = (i - j + 3k) \cdot (5i + 2j - 7k) + 9 = (1)(5) + (-1)(2) + (3)(-7) + 9 = 5 - 2 - 21 + 9 = -9$.Since $f(a) For point $B$,we calculate $f(b) = (3i + 3j + 3k) \cdot (5i + 2j - 7k) + 9 = (3)(5) + (3)(2) + (3)(-7) + 9 = 15 + 6 - 21 + 9 = 9$.Since $f(b) > 0$,point $B$ lies on the other side of the plane.Because $f(a)$ and $f(b)$ have opposite signs,the points $A$ and $B$ lie on the opposite sides of the plane.

The position vectors of points $A$ and $B$ are $i - j + 3k$ and $3i + 3j + 3k$ respectively. The equation of a plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. The points $A$ and $B$:

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