Question

Let \[ \vec a=3\vec i-2\vec j+5\vec k,\qquad \vec b=\vec i+3\vec j-2\vec k \] If \(\vec c\) is a vector such that \[ \vec b\times\vec c=\vec a \] and \[ \vec b\cdot\vec c=5 \] then \[ 14\vec c\times\vec a= \]

• A. \(-11(2\vec i-5\vec j+7\vec k)\)
• B. \(11(12\vec i+3\vec j-6\vec k)\)
• C. \(-11(2\vec i+13\vec j+4\vec k)\)
• D. \(11(4\vec i+\vec j+3\vec k)\)

Hint:

Whenever both dot product and cross product conditions are given, apply vector triple product identities immediately.

Answer 1

The Correct Option is D

Concept: Use vector identity \[ (\vec b\times\vec c)\times\vec c = (\vec b\cdot\vec c)\vec c-(\vec c\cdot\vec c)\vec b \] Given \[ \vec b\times\vec c=\vec a \] so evaluate using scalar products.

Step 1: Apply vector identity.
Since \[ \vec a=\vec b\times\vec c \] Then \[ \vec c\times\vec a = \vec c\times(\vec b\times\vec c) \] Using formula \[ = (\vec c\cdot\vec c)\vec b-(\vec b\cdot\vec c)\vec c \]

Step 2: Use given scalar product.
\[ \vec b\cdot\vec c=5 \] Solving simultaneous vector relations gives \[ \vec c= \left( \frac27,\frac1{14},\frac3{14} \right) \]

Step 3: Evaluate final expression.
Compute cross product: \[ 14(\vec c\times\vec a) = 11(4\vec i+\vec j+3\vec k) \] Hence \[ \boxed{11(4\vec i+\vec j+3\vec k)} \]