Question

If $2(\vec a \times \vec c)+3(\vec b \times \vec c)=0$, where $\vec a=2\hat i-5\hat j+5\hat k$, $\vec b=\hat i-\hat j+3\hat k$ and $(\vec a-\vec b)\cdot\vec c=-97$, find $|\vec c \times \vec k|^2$.

Hint:

If $\vec p\times\vec q=0$, then the vectors are always parallel.

Answer 1

Correct Answer: 218

Step 1: Use given vector equation.
\[ 2(\vec a \times \vec c)+3(\vec b \times \vec c)=0 \] \[ (2\vec a+3\vec b)\times\vec c=0 \] Step 2: Find $2\vec a+3\vec b$.
\[ 2\vec a+3\vec b=2(2,-5,5)+3(1,-1,3) \] \[ =(7,-13,19) \] Step 3: Write $\vec c$ proportional to this vector.
\[ \vec c=\lambda(7,-13,19) \] Step 4: Use dot product condition.
\[ (\vec a-\vec b)\cdot\vec c=-97 \] \[ (1,-4,2)\cdot\lambda(7,-13,19)=-97 \] \[ \lambda(7+52+38)=-97 \] \[ \lambda=-1 \] Step 5: Find $\vec c$.
\[ \vec c=(-7,13,-19) \] Step 6: Compute $|\vec c\times\vec k|^2$.
\[ \vec c\times\vec k=(-7,13,0) \] \[ |\vec c\times\vec k|^2=49+169=218 \] Final conclusion.
The required value is 218.