The line frac{x-1}{2}=frac{y+2}{-1}=frac{z}{1 | vedclass

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Question

(A) The given line is $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1} = \lambda$. Any point on the line is $(2\lambda+1, -\lambda-2, \lambda)$. For the $XY$

Vedclass · Accepted Answer

$[\bar{r}-(\hat{i}-2 \hat{j}+0 \hat{k})] 	imes(-\hat{i}+\frac{1}{2} \hat{j}-\frac{1}{2} \hat{k})=\overline{0}$

Vedclass · Answer

(A) The given line is $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1} = \lambda$. Any point on the line is $(2\lambda+1, -\lambda-2, \lambda)$. For the $XY$ plane,$z=0$,so $\lambda=0$. Point $A$ is $(1, -2, 0)$. For the $YZ$ plane,$x=0$,so $2\lambda+1=0 \implies \lambda=-\frac{1}{2}$. Point $B$ is $(0, -\frac{3}{2}, -\frac{1}{2})$. The vector equation of the line passing through $A(1, -2, 0)$ and $B(0, -\frac{3}{2}, -\frac{1}{2})$ is $\bar{r} = \vec{a} + t(\vec{b}-\vec{a})$. Here $\vec{a} = \hat{i}-2\hat{j}$ and $\vec{b}-\vec{a} = (0-1)\hat{i} + (-\frac{3}{2}-(-2))\hat{j} + (-\frac{1}{2}-0)\hat{k} = -\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}$. The equation is $\bar{r} = (\hat{i}-2\hat{j}) + t(-\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k})$. This can be written in cross product form as $[\bar{r}-(\hat{i}-2\hat{j})] 	imes (-\hat{i} + \frac{1}{2}\hat{j} - \frac{1}{2}\hat{k}) = \overline{0}$.

The line $\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z}{1}$ intersects the $XY$ plane and the $YZ$ plane at points $A$ and $B$ respectively. The equation of the line passing through the points $A$ and $B$ is

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