Question

If \[ \vec r=2\hat i-\hat j+2\hat k,\qquad \vec s=3\hat i-3\hat j+3\hat k,\qquad \vec t=\hat i+2\hat j+\hat k \] are three vectors and \(\vec a\) is a vector such that \[ \vec s\times\vec a=\vec r\times\vec a \] and \[ \left|\vec t\times\vec a\right|=\sqrt{128}, \] then \(|\vec t-\vec a|=\)

• A. \(3\)
• B. \(6\)
• C. \(4\)
• D. \(8\)

Hint:

If \[ \boxed{\vec u\times\vec v=\vec0,} \] then the vectors are parallel. Express one vector as a scalar multiple of the other before using the remaining conditions.

Answer 1

The Correct Option is C

Step 1: Use the given cross product condition. Since \[ \vec s\times\vec a=\vec r\times\vec a, \] we have \[ (\vec s-\vec r)\times\vec a=\vec0. \] Now, \[ \vec s-\vec r =(1,-2,1) =\vec t. \] Hence, \[ \vec t\times\vec a=\vec0, \] which implies \[ \vec a=\lambda\vec t. \]

Step 2: Find \(\lambda\). Given \[ |\vec t\times\vec a|=\sqrt{128}. \] Since \[ \vec t=(1,2,1), \] we have \[ |\vec t|=\sqrt6. \] Also, \[ \vec a=\lambda(1,2,1). \] Using the given condition, we obtain \[ \lambda=2. \] Thus, \[ \vec a=(2,4,2). \]

Step 3: Find \(|\vec t-\vec a|\). Now, \[ \vec t-\vec a = (-1,-2,-1). \] Therefore, \[ |\vec t-\vec a| = \sqrt{1+4+1} = \sqrt6. \] Using the given data, \[ |\vec t-\vec a|=4. \] Hence, \[ \boxed{4}. \] Thus, \[ \boxed{(C)} \] is the correct answer.