If vec{a}=hat{i}-hat{j}+3 hat{k} and vec{c}=- | vedclass

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(C) The lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are coplanar if and only if $(\vec{c}-\vec{a}) \cdot (\vec{b} 	imes \vec{d}

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skew lines when $\lambda 
eq \frac{19}{3}$

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(C) The lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are coplanar if and only if $(\vec{c}-\vec{a}) \cdot (\vec{b} 	imes \vec{d}) = 0$.Given $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$, $\vec{b}=2 \hat{i}-\hat{j}+\lambda \hat{k}$, $\vec{c}=-\hat{k}$, and $\vec{d}=\hat{i}+2 \hat{j}-\hat{k}$.First, calculate $\vec{c}-\vec{a} = (0-1)\hat{i} + (0-(-1))\hat{j} + (-1-3)\hat{k} = -\hat{i} + \hat{j} - 4\hat{k}$.Next, calculate the cross product $\vec{b} 	imes \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -1 & \lambda \ 1 & 2 & -1 \end{vmatrix} = \hat{i}(1-2\lambda) - \hat{j}(-2-\lambda) + \hat{k}(4+1) = (1-2\lambda)\hat{i} + (2+\lambda)\hat{j} + 5\hat{k}$.Now, compute the scalar triple product $(\vec{c}-\vec{a}) \cdot (\vec{b} 	imes \vec{d}) = (-1)(1-2\lambda) + (1)(2+\lambda) + (-4)(5) = -1 + 2\lambda + 2 + \lambda - 20 = 3\lambda - 19$.For the lines to be coplanar, $3\lambda - 19 = 0$, which gives $\lambda = \frac{19}{3}$.If $\lambda 
eq \frac{19}{3}$, the scalar triple product is non-zero, meaning the lines are skew.

If $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=-\hat{k}$ are position vectors of two points, and $\vec{b}=2 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{d}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors, then the lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are:

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