# What is the projection of #<6,5,3 ># onto #<2,-1,8 >#?

The vector projection is

We can start by taking the dot product of the two vectors.

You can distribute the coefficient to each component of the vector and write as:

Hope that helps!

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To find the projection of vector ( \mathbf{v} = \langle 6, 5, 3 \rangle ) onto vector ( \mathbf{u} = \langle 2, -1, 8 \rangle ), we use the formula:

[ \text{proj}_{\mathbf{u}}(\mathbf{v}) = \left( \frac{\mathbf{v} \cdot \mathbf{u}}{| \mathbf{u} | ^2} \right) \mathbf{u} ]

where ( \cdot ) represents the dot product and ( | \mathbf{u} | ) represents the magnitude of vector ( \mathbf{u} ).

First, calculate the dot product:

[ \mathbf{v} \cdot \mathbf{u} = (6 \cdot 2) + (5 \cdot -1) + (3 \cdot 8) = 12 - 5 + 24 = 31 ]

Then, find the magnitude of ( \mathbf{u} ):

[ | \mathbf{u} | = \sqrt{2^2 + (-1)^2 + 8^2} = \sqrt{4 + 1 + 64} = \sqrt{69} ]

Now, substitute these values into the formula:

[ \text{proj}_{\mathbf{u}}(\mathbf{v}) = \left( \frac{31}{69} \right) \langle 2, -1, 8 \rangle = \langle \frac{62}{69}, -\frac{31}{69}, \frac{248}{69} \rangle ]

So, the projection of ( \langle 6, 5, 3 \rangle ) onto ( \langle 2, -1, 8 \rangle ) is approximately ( \langle \frac{62}{69}, -\frac{31}{69}, \frac{248}{69} \rangle ).

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