Let a=2 i+_1 j+3 k, b=4 i+(3-_2) j+6 k, c=3 i+6 j+(_3-1) k, with b=2 a and a c.

Question

Let $\vec a=2\hat i+\lambda_1\hat j+3\hat k$, $\vec b=4\hat i+(3-\lambda_2)\hat j+6\hat k$, $\vec c=3\hat i+6\hat j+(\lambda_3-1)\hat k$, with $\vec b=2\vec a$ and $\vec a\perp\vec c$. A possible $(\lambda_1,\lambda_2,\lambda_3)$ is:

Accepted Answer

$\vec b=2\vec a$: $3-\lambda_2=2\lambda_1$ \ldots  (1)

$\vec a\cdot\vec c=0$: $6+6\lambda_1+3(\lambda_3-1)=0 \implies \lambda_3=-2\lambda_1-1$ \ldots  (2)

Testing $\lambda_1=-\dfrac12$: from (1), $\lambda_2=3-2(-	frac12)=4$; from (2), $\lambda_3=-2(-	frac12)-1=0$. This matches option B: $\left(-\dfrac12,4,0ight)$.

Solutions · 1