Question

Let \( \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \) and \( \vec{b}=3 \hat{i}+2 \hat{j}+4 \hat{k} \). Let for some \( \lambda, \mu \in \mathbf{R} \), a vector \( \vec{c}=\lambda \vec{a}+\mu \vec{b} \). If \( \vec{a} \cdot \vec{c}=16 \) and \( \vec{b} \times \vec{c}=-12 \hat{i}+22 \hat{j}-2 \hat{k} \), then \( |\vec{c}|^{2} \) equals ________.

Hint:

\( \vec b\times\vec c=\lambda(\vec b\times\vec a) \) gives \( \lambda \); then \( \vec a\cdot\vec c=16 \) gives \( \mu \).

Answer 1

Correct Answer: 69

Since \( \vec{c}=\lambda\vec a+\mu\vec b \), \( \vec b\times\vec c=\lambda(\vec b\times\vec a) \).
\( \vec a\times\vec b=(1\cdot4-(-1)\cdot2)\hat i-(2\cdot4-(-1)\cdot3)\hat j+(2\cdot2-1\cdot3)\hat k=6\hat i-11\hat j+\hat k \), so \( \vec b\times\vec a=-6\hat i+11\hat j-\hat k \).
Comparing with \( -12\hat i+22\hat j-2\hat k \): \( \lambda=2 \).
\( \vec a\cdot\vec c=\lambda|\vec a|^2+\mu(\vec a\cdot\vec b)=2(6)+\mu(6+2-4)=12+4\mu=16 \Rightarrow \mu=1 \).
\( \vec c=2\vec a+\vec b=7\hat i+4\hat j+2\hat k \), so \( |\vec c|^2=49+16+4=69 \).
Answer: 69