Question

The vector projection of b on a , where a = 3$\hat {i}$ + 2$\hat {j}$ + 5$\hat {k}$ and b = 7$\hat {i}$ - 5$\hat {j}$ - $\hat {k}$ is

• A. $\frac {3(3\hat i + 2\hat j + 5\hat k)}{\sqrt {38}}$
• B. $\frac {9\hat i + 6\hat j + 15\hat k}{19}$
• C. $\frac {3(3\hat i + 2\hat j + 5\hat k)}{38}$
• D. $\frac {6(3\hat i + 2\hat j + 5\hat k)}{\sqrt {38}}$

Answer 1

Answer: (B)

$\frac {9\hat i + 6\hat j + 15\hat k}{19}$

Answer 2

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 $\frac {9\hat i + 6\hat j + 15\hat k}{19}$.