Question

Let \[ \vec a=\hat i+\hat j+\hat k \] \[ \vec b=\hat i-\hat j+2\hat k \] If a vector \( \vec c \) is coplanar with \( \vec a \) and \( \vec b \) such that \[ \vec c\cdot \vec a=1 \] and \[ \vec c\cdot \vec b=2 \] then \( \vec c \) is:

• A. \( \dfrac13(-\hat i+5\hat j+3\hat k) \)
• B. \( \dfrac13(5\hat i-\hat j+3\hat k) \)
• C. \( \hat i-\hat j+\hat k \)
• D. \( \dfrac12(\hat j+\hat k) \)

Hint:

Whenever vectors are coplanar, express one vector as linear combination of the other two.

Answer 1

The Correct Option is A

Concept: If vectors are coplanar: \[ \vec c=\lambda \vec a+\mu \vec b \] Use dot product conditions to determine constants.

Step 1: Express \( \vec c \). Let: \[ \vec c=\lambda(\hat i+\hat j+\hat k)+\mu(\hat i-\hat j+2\hat k) \] Then: \[ \vec c= (\lambda+\mu)\hat i + (\lambda-\mu)\hat j + (\lambda+2\mu)\hat k \]

Step 2: Use dot product conditions. Given: \[ \vec c\cdot\vec a=1 \] Thus: \[ (\lambda+\mu)+(\lambda-\mu)+(\lambda+2\mu)=1 \] \[ 3\lambda+2\mu=1 \] Also: \[ \vec c\cdot\vec b=2 \] \[ (\lambda+\mu)-(\lambda-\mu)+2(\lambda+2\mu)=2 \] \[ 2\lambda+6\mu=2 \]

Step 3: Solve equations. Solving: \[ 3\lambda+2\mu=1 \] \[ 2\lambda+6\mu=2 \] we get: \[ \lambda=\frac13, \qquad \mu=-\frac23 \] Substituting: \[ \vec c = \frac13(-\hat i+5\hat j+3\hat k) \]

Step 4: Final answer. \[ \boxed{ \frac13(-\hat i+5\hat j+3\hat k) } \]