# What is the unit vector that is orthogonal to the plane containing # (-2i- 3j + 2k) # and # (3i – 4j + 4k) #?

Take the cross product of the 2 vectors

Compute

The

The magnitude of this new vector is:

Now to find the unit vector normalize our new vector

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To find the unit vector orthogonal to the plane containing the given vectors, first find the cross product of the two vectors. Then, normalize the resulting vector to obtain the unit vector.

Cross product: [ \vec{v}_1 = (-2i- 3j + 2k) ] [ \vec{v}_2 = (3i - 4j + 4k) ] [ \vec{n} = \vec{v}_1 \times \vec{v}_2 ]

Normalize the resulting vector: [ |\vec{n}| = \sqrt{a^2 + b^2 + c^2} ] [ \vec{u} = \frac{\vec{n}}{|\vec{n}|} ]

Substitute the values of ( \vec{v}_1 ) and ( \vec{v}_2 ) into the cross product formula: [ \vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -2 & -3 & 2 \ 3 & -4 & 4 \end{vmatrix} ]

Solve for ( \vec{n} ), then normalize it to find the unit vector orthogonal to the given plane.

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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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