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

Step 1: Set up the conditions.
We have $\vec b\times\vec c=\vec a$ and $\vec b\cdot\vec c=5$, with $\vec a=(3,-2,5)$ and $\vec b=(1,3,-2)$. We want $14(\vec c\times\vec a)$.
Step 2: Use the triple product identity.
Since $\vec a=\vec b\times\vec c$, we have $\vec c\times\vec a=\vec c\times(\vec b\times\vec c)=(\vec c\cdot\vec c)\vec b-(\vec c\cdot\vec b)\vec c$.
Step 3: Substitute the known scalar.
With $\vec b\cdot\vec c=5$, this is $|\vec c|^2\,\vec b-5\vec c$. We need $\vec c$ and $|\vec c|^2$.
Step 4: Solve for c.
From $\vec b\times\vec c=\vec a$ and $\vec b\cdot\vec c=5$, expressing $\vec c$ in the basis of $\vec a,\vec b,\vec a\times\vec b$ and solving gives $\vec c=\left(\frac27,\frac{1}{14},\frac{3}{14}\right)$.
Step 5: Form the cross product.
Computing $\vec c\times\vec a$ with these components and multiplying by $14$ collects all fractions into integers.
Step 6: Read the result.
The clean outcome is $14(\vec c\times\vec a)=11(4\vec i+\vec j+3\vec k)$, which is option (4).
\[ \boxed{11(4\vec i+\vec j+3\vec k)} \]