Let vec{a}=hat{i}+2 hat{j}+lambda hat{k},vec{ | vedclass

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(D) Given $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$.From $\vec{a} 	imes \vec{c}=\vec{b} 	imes \v

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$8$

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(D) Given $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$ and $\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$.From $\vec{a} 	imes \vec{c}=\vec{b} 	imes \vec{c}$,we have $(\vec{a}-\vec{b}) 	imes \vec{c} = \vec{0}$.This implies that $(\vec{a}-\vec{b})$ is parallel to $\vec{c}$,so $\vec{a}-\vec{b} = \mu \vec{c}$ for some scalar $\mu$.Calculating $\vec{a}-\vec{b} = (1-3)\hat{i} + (2-(-5))\hat{j} + (\lambda - (-\lambda))\hat{k} = -2\hat{i} + 7\hat{j} + 2\lambda\hat{k}$.Thus,$\mu \vec{c} = -2\hat{i} + 7\hat{j} + 2\lambda\hat{k}$.Given $\vec{a} \cdot \vec{c} = 7$,we have $\vec{a} \cdot (\frac{1}{\mu} (-2\hat{i} + 7\hat{j} + 2\lambda\hat{k})) = 7$,which gives $-2 + 14 + 2\lambda^2 = 7\mu$,so $12 + 2\lambda^2 = 7\mu$.Given $2\vec{b} \cdot \vec{c} = -43$,we have $\vec{b} \cdot (\frac{1}{\mu} (-2\hat{i} + 7\hat{j} + 2\lambda\hat{k})) = -\frac{43}{2}$,which gives $-6 - 35 - 2\lambda^2 = -\frac{43}{2}\mu$,so $41 + 2\lambda^2 = \frac{43}{2}\mu$.Solving the system $2\lambda^2 = 7\mu - 12$ and $2\lambda^2 = \frac{43}{2}\mu - 41$,we get $7\mu - 12 = 21.5\mu - 41$,so $14.5\mu = 29$,which gives $\mu = 2$.Then $2\lambda^2 = 7(2) - 12 = 2$,so $\lambda^2 = 1$.Finally,$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-5) + (\lambda)(-\lambda) = 3 - 10 - \lambda^2 = -7 - 1 = -8$.Thus,$|\vec{a} \cdot \vec{b}| = |-8| = 8$.

Let $\vec{a}=\hat{i}+2 \hat{j}+\lambda \hat{k}$,$\vec{b}=3 \hat{i}-5 \hat{j}-\lambda \hat{k}$,$\vec{a} \cdot \vec{c}=7$,$2 \vec{b} \cdot \vec{c}+43=0$,and $\vec{a} \times \vec{c}=\vec{b} \times \vec{c}$. Then $|\vec{a} \cdot \vec{b}|$ is equal to

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