Let $\vec a=\hat i+2\hat j+3\hat k,\ \vec b=3\hat i+\hat j-\hat k$ and $\vec c$ be three vectors such that $\vec c$ is coplanar with $\vec a$ and $\vec b$. If the vector $\vec c$ is perpendicular to $\vec b$ and $\vec a\cdot\vec c=5$, then $|\vec c|$ is equal to

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] Let $\vec c=\lambda\vec a+\mu\vec b$. $\vec c\cdot\vec b=0$: $2\lambda+11\mu=0$. $\vec a\cdot\vec c=5$: $14\lambda+2\mu=5$. So $\mu=-\frac1{15}$, $\lambda=\frac{11}{30}$ and $\vec c=\left(\frac16,\frac23,\frac76\right)$. $|\vec c|^2=\frac{1+16+49}{36}=\frac{11}6$. Answer (1).

Q. Let $\vec a=\hat i+2\hat j+3\hat k,\ \vec b=3\hat i+\hat j-\hat k$ and $\vec c$ be three vectors such that $\vec c$ is coplanar with $\vec a$ and $\vec b$. If the vector $\vec c$ is perpendicular to $\vec b$ and $\vec a\cdot\vec c=5$, then $|\vec c|$ is equal to

Explanation

Reviewed by Dr. Abubacker

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