Question

Find the vector and cartesian equations of the line that passes through the origin and (5,- 2, 3).

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Solution

The line passes through the points A(0, 0, 0) and B(5, – 2, 3)

therefore space space space space straight a with rightwards arrow on top space equals space 0 with rightwards arrow on top comma space space straight b with rightwards arrow on top space equals space 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top

where

straight a with rightwards arrow on top comma space straight b with rightwards arrow on top

are position vectors of A, B respectively.

therefore space space space space space straight b with rightwards arrow on top space minus space straight a with rightwards arrow on top space equals space 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top

Let

straight r with rightwards arrow on top

be the position vector of any point on the given line.
∴ the vector equation of line is

straight r with rightwards arrow on top space equals space straight a with rightwards arrow on top space plus space straight lambda left parenthesis straight b with rightwards arrow on top minus straight a with rightwards arrow on top right parenthesis

straight r with rightwards arrow on top space equals space 0 with rightwards arrow on top plus straight lambda left parenthesis 5 straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis

straight r with rightwards arrow on top space equals space straight lambda left parenthesis 5 space straight i with hat on top space minus space 2 space straight j with hat on top space plus space 3 space straight k with hat on top right parenthesis

Taking

straight r with rightwards arrow on top space equals space straight x space straight i with hat on top space plus space space straight y space straight j with hat on top space plus space straight z space straight k with hat on top comma space

the equation of line is

straight x space straight i with hat on top space plus space straight y space straight j with hat on top space space plus straight z space straight k with hat on top space equals space 5 space straight lambda space straight i with hat on top space minus space space 2 space straight lambda space straight j with hat on top space plus space 3 space straight lambda space straight k with hat on top

Comparing the coefficients of

straight i with hat on top comma space straight j with hat on top comma space straight k with hat on top comma space we space get comma

straight x space equals space 5 straight lambda comma space space space straight y space equals space minus 2 straight lambda comma space space straight z space equals space 3 straight lambda

fraction numerator straight x minus 0 over denominator 5 end fraction space equals space fraction numerator straight y minus 0 over denominator negative 2 end fraction space equals space fraction numerator straight z minus 0 over denominator 3 end fraction space left parenthesis space equals space straight lambda right parenthesis

which is cartesian equation of line.

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