Question

Let \[ \vec F=2\hat i+2\hat j+5\hat k,\quad A=(1,2,5),\quad B=(-1,-2,-3) \] and \[ \overrightarrow{BA}\times \vec F=4\hat i+6\hat j+2\lambda\hat k, \] then \[ \lambda= \]

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(-2\)

Hint:

For two points \(A\) and \(B\), \[ \overrightarrow{BA}=A-B. \] Then use the determinant method to calculate the cross product of two vectors.

Answer 1

The Correct Option is D

Step 1: Find \(\overrightarrow{BA}\).
Given, \[ A=(1,2,5) \] and \[ B=(-1,-2,-3) \] Therefore, \[ \overrightarrow{BA}=A-B \] \[ =(1-(-1),2-(-2),5-(-3)) \] \[ =(2,4,8) \] So, \[ \overrightarrow{BA}=2\hat i+4\hat j+8\hat k \]

Step 2: Write the given vector \(\vec F\).
\[ \vec F=2\hat i+2\hat j+5\hat k \]

Step 3: Compute \(\overrightarrow{BA}\times \vec F\).
\[ \overrightarrow{BA}\times \vec F = \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 4 & 8 \\ 2 & 2 & 5 \end{vmatrix} \] \[ = \hat i(4\cdot5-8\cdot2) -\hat j(2\cdot5-8\cdot2) +\hat k(2\cdot2-4\cdot2) \] \[ = \hat i(20-16)-\hat j(10-16)+\hat k(4-8) \] \[ = 4\hat i+6\hat j-4\hat k \]

Step 4: Compare with the given expression.
Given, \[ \overrightarrow{BA}\times \vec F=4\hat i+6\hat j+2\lambda\hat k \] But we found, \[ \overrightarrow{BA}\times \vec F=4\hat i+6\hat j-4\hat k \] Comparing the coefficient of \(\hat k\), \[ 2\lambda=-4 \]

Step 5: Find \(\lambda\).
\[ \lambda=-2 \]

Step 6: Final conclusion.
Therefore, \[ \boxed{-2} \]