Question

Find the direction cosines of the line x+2/2=2y-7/6=5-z/6 also find the vector equation of the line through the point a(-1,2,3)are parallel to given line

Answer 1

SOLUTION

TO DETERMINE

1. The direction cosines of the line

$\displaystyle \sf{ \frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6} }$

2. The vector equation of the line through the point A(-1,2,3) and parallel to given line

EVALUATION

1. Here the given equation of the line is

$\displaystyle \sf{ \frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6} }$

Which can be rewritten as

$\displaystyle \sf{ \frac{x + 2}{2} = \frac{y - \frac{7}{2} }{3} = \frac{z - 5}{ - 6} }$

So the direction ratios of the line is 2, 3 - 6

So the required direction cosines are

$\displaystyle \sf{ \frac{2}{ \sqrt{ {2}^{2} + {3}^{2} + {( - 6)}^{2} } } \: , \: \frac{3}{ \sqrt{ {2}^{2} + {3}^{2} + {( - 6)}^{2} } } \: , \frac{ - 6}{ \sqrt{ {2}^{2} + {3}^{2} + {( - 6)}^{2} } } }$

$\displaystyle \sf{i.e \: \: \: \: \frac{2}{ 7} \: , \: \frac{3}{ 7 } \: , \frac{ - 6}{ 7} \: \: \: }$

2. The cartesian equation of the line parallel to given line and passing through the point A(-1,2,3) is

$\displaystyle \sf{ \frac{x + 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{ - 6} }$

So the vector equation of the line is

$\vec{r} = ( \hat{ \imath} + 2 \hat{ \jmath} + 3 \hat{k}) + \lambda \: ( 2\hat{ \imath} + 3 \hat{ \jmath} - 6 \hat{k})$

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Answer 2

Answer

Step-by-step explanation:

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