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The Cartesian equations of a line are...

The Cartesian equations of a line are `6x-2=3y+1=2z-2.` Find its direction ratios and also find a vector equation of the line.

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The given line is `6x-2=3y+1=2z-2`
To put it in the symmetrical form, we must make the coeffcients `x, y and z` as 1. To do this, we divide each of the expressions in (i) by 6 and obtain `(x-(1//3))/(1)=(y+(1//3))/(2)=(z-1)/(3)`.
This shows that the given line passes through `(1//3, -1//3, 1)` and is parallel to the line whose direction ratios are 1, 2 and 3. Therefore, its vector equation is
`" "vecr=(1)/(3)hati-(1)/(3)hatj+hatk+lamda(hati+2hatj+3hatk)`
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