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  • Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

9.  Find the vector and the cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).

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Let the line passing through the points A(3,-2,-5) and B(3,-2,6) is AB;

Then as AB passes through through A so, we can write its position vector as;

\vec{a} =3\widehat{i}-2\widehat{j}-5\widehat{k}

Then direction ratios of PQ are given by,

(3-3)= 0,\ (-2+2) = 0,\ (6+5)=11

Therefore the equation of the vector in the direction of AB is given by,

\vec{b} =0\widehat{i}-0\widehat{j}+11\widehat{k} = 11\widehat{k}

We have then the equation of line AB in  vector form is given by,

\vec{r} =\vec{a}+\lambda\vec{b},\ where\ \lambda \epsilon R

\Rightarrow \vec{r} = (3\widehat{i}-2\widehat{j}-5\widehat{k}) + 11\lambda\widehat{k}

So, the equation of AB in Cartesian form is;

\frac{x-x_{1}}{a} =\frac{y-y_{1}}{b} =\frac{z-z_{1}}{c}

or \frac{x-3}{0} =\frac{y+2}{0} =\frac{z+5}{11}

 

Posted by

Divya Prakash Singh

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