Question
Let the line $L_{1}$ be parallel to the vector $- 3\widehat{i} + 2\widehat{j} + 4\widehat{k}$ and pass through the point $(2,6,7)$ and the line $L_{2}$ be parallel to the vector $2\widehat{i} + \widehat{j} + 3\widehat{k}$ and pass through the point $(4,3,5)$. If the line $L_{3}$ is parallel to the vector $- 3\widehat{i} + 5\widehat{j} + 16\widehat{k}$ and intersects the lines $L_{1}$ and $L_{2}$ at the points C and D , respectively, then $|\overrightarrow{CD}|^{2}$ is equal to :
Options
Solution
$\ L_{1}:\frac{x - 2}{- 3} = \frac{y - 6}{2} = \frac{z - 7}{4}$
Point C on $L_{1}:\left( - 3\lambda_{1} + 2,2\lambda_{1} + 6,4\lambda_{1} + 7 \right)$
$$L_{2}:\frac{x - 4}{2} = \frac{y - 3}{1} = \frac{z - 5}{3}$$
Point D on $L_{2}:\left( 2\lambda_{2} + 4,\lambda_{2} + 3,3\lambda_{2} + 5 \right)$
Dr's of line $L_{3}$ :
$${L_{3}:\frac{2\lambda_{2} + 3\lambda_{1} + 2}{- 3} = \frac{\lambda_{2} - 2\lambda_{1} - 3}{5} = \frac{3\lambda_{2} - 4\lambda_{1} - 2}{16} }{\lambda_{1} = - 3,\lambda_{2} = 2 }$$C (11, 0, -5)
D (8, 5, 11)
$$|\overrightarrow{CD}|^{2} = 3^{2} + 5^{2} + 16^{2} = 290$$
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