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Question
The equation of a line is 2x -2 = 3y +1 = 6z -2 find the direction ratios and also find the vector equation of the line.
Solution
The Cartesian equation of a line is
2x -2 =3y + 1 = 6z -2
2(x-1) = 3`("y" +1/3) = 6("z" -1/3)` ....dividing by 6 all side
`("x"-1)/3 = ("y"+1/3)/2 = ("z"-1/3)/1`
The direction ratios of the line are 3, 2, 1
Further, the line passes through the point `(1 , -1/3 , 1/3)`
Let A = `(1 , -1/3 , 1/3)`
Thus, the line passes through the point having the position vector.
`bar"a" = 1hat"i" -1/3hat"j" + 1/3hat"k"`
Let `hat"b" = 3hat"i" +2 hat"j" +hat"k" `
Then, the line is parallel to the vector `bar b`
Hence, the vector equation of the line is
`bar"r" = bar"a" + λbar"b"` , where λ ∈ R
i.e. `bar"r"=(1 hat"i"-1/3 hat"j" + 1/3 hat "k") +lambda (3 hat "i" +2 hat "j" +hat "k") `
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