To find the vector projection of b onto a:
proj(a, b) = \(\frac {b\ dot \ a}{|a|^2}\) x a
Given a = 3\(\hat {i}\) + 2\(\hat {j}\) + 5\(\hat {k}\) and b = 7\(\hat {i}\) - 5\(\hat {i}\) - \(\hat {k}\)
We can calculate the vector projection.First, let's find the dot product of b and a:
b dot a = (7 x 3) + (-5 x 2) + (-1 x 5) = 21 - 10 - 5 = 6.
Now calculate the magnitude of a:
|a| = \(\sqrt{(3^2) + (2^2) + (5^2)}\)= \(\sqrt{9 + 4 + 25}\) = \(\sqrt{38}\)
Now we can substitute these values into the formula to find the vector projection:
proj(a, b) = \(\frac {b \ dot\ a }{|a|^2}\) x a
proj(a, b) = \(\frac {6}{(\sqrt{(38)^2}}\) x (3\(\hat i\) + 2\(\hat j\) + 5\(\hat k\))
proj(a, b) = \(\frac {6}{38}\) x (3\(\hat {i}\) + 2\(\hat {j}\) + 5\(\hat k\))
proj(a, b) = \(\frac {3}{19}\) x (3\(\hat {i}\) + 2\(\hat {j}\) + 5\(\hat k\))
proj(a, b) = \(\frac {9}{19}\)\(\hat {i}\) + \(\frac {6}{19}\)\(\hat {j}\) + \(\frac {15}{19}\)\(\hat {k}\)
Therefore, the vector projection of b onto a is \(\frac {9\hat i + 6\hat j + 15\hat k}{19}\).
Let $\vec{u}=\hat{i}-\hat{j}-2 \hat{k}, \vec{v}=2 \hat{i}+\hat{j}-\hat{k}, \vec{v} \cdot \vec{w}=2$ and $\vec{v} \times \vec{w}=\vec{u}+\lambda \vec{v}$. Then $\vec{u} \cdot \vec{w}$ is equal to