Practicing Success
Direction ratios of the line perpendicular to the lines $\frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1}$ and $\frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2}$ are : |
<4, -5, 7> <-4, 5, 7> <4, 5, -7> <4, 5, 7> |
<4, 5, 7> |
lines $l_1, : \frac{x-3}{2}=\frac{y+7}{-3}=\frac{z-2}{1}$ $l_2: \frac{x+2}{1}=\frac{y+3}{2}=\frac{z-5}{-2}$ so vector $\vec{v}_1||l_1$ $\Rightarrow \vec{v}_1=2 \hat{i}-3 \hat{j}+\hat{k}$ vector $\vec{v}_2 || l_2$ $\Rightarrow \vec{v}_2=\hat{i}+2 \hat{j}-2 \hat{k}$ so $\vec p=\vec{v}_1×\vec{v}_2$ (perpendicular to both $l_1,l_2$) $\vec{p}=\begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\\2 & -3 & 1\\1 & 2 & -2\end{vmatrix}$ $\vec{p} = 4\hat{i} + 5\hat{j}+7\hat{k}$ (Upon solving) so DR's(p) → $<4, 5, 7>$ |