8.  Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).

Answers (1)
D Divya Prakash Singh

GIven that the line is passing through the (0,0,0) and (5,-2,3)

Thus the required line passes through the origin. 

\therefore its position vector is given by,

\vec{a} = \vec{0}

So, the direction ratios of the line through  (0,0,0) and (5,-2,3) are,

(5-0) = 5, (-2-0) = -2, (3-0) = 3

The line is parallel to the vector given by the equation, \vec{b} = 5\widehat{i}-2\widehat{j}+3\widehat{k}

Therefore the equation of the line passing through the point with position vector \vec{a} and parallel to \vec{b} is given by;

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

\Rightarrow\vec{r} = 0+\lambda (5\widehat{i}-2\widehat{j}+3\widehat{k})

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

Now, the equation of the line through the point (x_{1},y_{1},z_{1}) and the direction ratios a, b, c is given by;

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

Therefore the equation of the required line in the Cartesian form will be;

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

OR \frac{x}{5} = \frac{y}{-2} =\frac{z}{3}

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