# What is the unit vector that is orthogonal to the plane containing # (2i + 3j – 7k) # and # (3i + 2j - 6k) #?

A vector which is orthogonal to a plane containing two vectors is also orthogonal to the given vectors. We can find a vector which is orthogonal to both of the given vectors by taking their cross product. We can then find a unit vector in the same direction as that vector.

Given

For the

#(3*-6)-(2*-7)=-18+ 14=-4#

For the

#-[(2*-6)-(3*-7)]=-[-12+21]=-9#

For the

#(2*2)-(3*3)=4-9=-5#

Our vector is

Now, to make this a unit vector, we divide the vector by its magnitude. The magnitude is given by:

#|vecn|=sqrt((n_x)^2+(n_y)^2+(n_z)^2)#

#|vecn|=sqrt((-4)^2+(-9)^2+(-5)^2)#

#|vecn|=sqrt(16+81+25)=sqrt(122)#

The unit vector is then given by:

#vecu=(vecaxxvecb)/(|vecaxxvecb|)#

#vecu=(< -4,-9,-5 >)/(sqrt(122))#

#vecu=1/(sqrt(122))< -4,-9,-5 >#

or equivalently,

#vecu=< -4/(sqrt(122)),-9/(sqrt(122)),-5/(sqrt(122))>#

You may also choose to rationalize the denominator.

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To find the unit vector that is orthogonal to the plane containing the vectors (2i + 3j - 7k) and (3i + 2j - 6k), you can first find the cross product of these two vectors. Then, divide the resulting vector by its magnitude to obtain the unit vector.

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When evaluating a one-sided 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 one-sided 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 one-sided 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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