Obtain the projection of the vector P<1, 6, 5> in the vector Q<-3, -5, 5>

To find the projection of vector \(\mathbf{P}\) onto vector \(\mathbf{Q}\), you can use the formula:

\[
\text{Proj}_{\mathbf{Q}} \mathbf{P} = \frac{\mathbf{P} \cdot \mathbf{Q}}{|\mathbf{Q}|^2} \mathbf{Q}
\]

Here's how we can solve this step by step:

1. **Calculate the dot product** \(\mathbf{P} \cdot \mathbf{Q}\):

\[
\mathbf{P} \cdot \mathbf{Q} = (1)(-3) + (6)(-5) + (5)(5) = -3 - 30 + 25 = -8
\]

2. **Calculate the magnitude squared** of \(\mathbf{Q}\):

\[
|\mathbf{Q}|^2 = (-3)^2 + (-5)^2 + 5^2 = 9 + 25 + 25 = 59
\]

3. **Use the projection formula**:

\[
\text{Proj}_{\mathbf{Q}} \mathbf{P} = \frac{-8}{59} \mathbf{Q} = \frac{-8}{59} \langle -3, -5, 5 \rangle
\]

4. **Multiply the scalar by the vector** \(\mathbf{Q}\):

\[
\text{Proj}_{\mathbf{Q}} \mathbf{P} = \left\langle \frac{-8}{59} \times -3, \frac{-8}{59} \times -5, \frac{-8}{59} \times 5 \right\rangle = \left\langle \frac{24}{59}, \frac{40}{59}, \frac{-40}{59} \right\rangle
\]

So, the projection of vector \(\mathbf{P} = \langle 1, 6, 5 \rangle\) onto vector \(\mathbf{Q} = \langle -3, -5, 5 \rangle\) is:

\[
\text{Proj}_{\mathbf{Q}} \mathbf{P} = \left\langle \frac{24}{59}, \frac{40}{59}, \frac{-40}{59} \right\rangle
\]

Would you like any more details or have any other questions?

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Here are 5 related questions for further exploration:

1. How do you interpret the projection of one vector onto another?
2. What is the difference between the projection and the component of a vector?
3. How do you find the perpendicular component of a vector relative to another vector?
4. How does the projection formula change in two-dimensional space?
5. What are some applications of vector projections in physics or engineering?

**Tip:** The projection of a vector \(\mathbf{P}\) onto another vector \(\mathbf{Q}\) shows the shadow or footprint of \(\mathbf{P}\) along \(\mathbf{Q}\).

Learn how to calculate the vector projection of P<1, 6, 5> onto Q<-3, -5, 5> using the projection formula. Step-by-step solution provided with detailed calculations. Suitable for advanced high school students and above.

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