How do you find the parametric equations for the line through the point P = (2, 2, 1) that is perpendicular to the plane 1x + 3y  2z = 1?
for the plane
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To find the parametric equations for the line through the point ( P = (2, 2, 1) ) that is perpendicular to the plane ( 1x + 3y  2z = 1 ), follow these steps:

Find the normal vector to the plane by taking the coefficients of ( x ), ( y ), and ( z ) in the plane equation. In this case, the normal vector is ( \langle 1, 3, 2 \rangle ).

Since the line is perpendicular to the plane, the direction vector of the line will be parallel to the normal vector of the plane. Therefore, the direction vector of the line is ( \langle 1, 3, 2 \rangle ).

Write the parametric equations for the line using the point ( P ) and the direction vector ( \langle 1, 3, 2 \rangle ). The parametric equations are:
[ x = 2 + t ] [ y = 2 + 3t ] [ z = 1  2t ]
where ( t ) is a parameter representing any real number.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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