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

The Correct Option is B

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

Answer 2

Vector Projection
To find the vector projection of b onto a:
proj_ab = (b • a / |a|²) a
Given a = 3i + 2j + 5k and b = 7i - 5j - k,
Step 1: Find the dot product:
b • a = (7 x 3) + (-5 x 2) + (-1 x 5) = 6
Step 2: Find the magnitude of a:
|a|² = 3² + 2² + 5² = 38
Step 3: Calculate the projection:
proj_ab = (6 / 38) a = (3 / 19)(3i + 2j + 5k)
proj_ab = (9/19)i + (6/19)j + (15/19)k
Therefore, the vector projection of b onto a is:
(9/19)i + (6/19)j + (15/19)k