Let, , , . A vector coplanar to and has a projection along of magnitude , then the vector is
- A.
- B.
- C.
- D.None of these
Show answer and solution
Correct option: A
Solution
A vector coplanar to a and b has the form v = sa + tb = (s+t)î + (2s−t)ĵ + (s+t)k̂.
Note the î and k̂ components are always equal, so v = (s+t)î + (2s−t)ĵ + (s+t)k̂.
The projection along c = î + ĵ − k̂ (magnitude √3) is:
(v · c)/|c| = [(s+t) + (2s−t) − (s+t)]/√3 = (2s−t)/√3 = 1/√3
So 2s − t = 1, giving t = 2s − 1.
Now check option 1: 4î − ĵ + 4k̂. Here î and k̂ components are both 4 ✓ (consistent with coplanar form). Setting s+t = 4 and 2s−t = −1: adding gives 3s = 3, s = 1, t = 3. But 2s−t = 2−3 = −1 ≠ 1.
Re-examining: projection of magnitude 1/√3 means |(v·c)/|c|| = 1/√3, so |2s−t| = 1, allowing 2s−t = −1. With s=1, t=3: v = 4î + (2−3)ĵ + 4k̂
Note the î and k̂ components are always equal, so v = (s+t)î + (2s−t)ĵ + (s+t)k̂.
The projection along c = î + ĵ − k̂ (magnitude √3) is:
(v · c)/|c| = [(s+t) + (2s−t) − (s+t)]/√3 = (2s−t)/√3 = 1/√3
So 2s − t = 1, giving t = 2s − 1.
Now check option 1: 4î − ĵ + 4k̂. Here î and k̂ components are both 4 ✓ (consistent with coplanar form). Setting s+t = 4 and 2s−t = −1: adding gives 3s = 3, s = 1, t = 3. But 2s−t = 2−3 = −1 ≠ 1.
Re-examining: projection of magnitude 1/√3 means |(v·c)/|c|| = 1/√3, so |2s−t| = 1, allowing 2s−t = −1. With s=1, t=3: v = 4î + (2−3)ĵ + 4k̂
[AIEEE 2006]