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Question
Find the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.
Solution
4x – 3z + 5 = 0 can be written as
4x = `3z – 5 = 3(z - 5/3)`
∴ `(4x)/(12) = (3(z - 5/3))/(12)`
∴ `x/(3) = (z - 5/3)/(4)`
∴ the cartesian equation of the line are
`x/(3) = (z - 5/3)/(4), y = 2`.
This line passes through the point `"A"(0,2, 5/3)` whose position vector is `bar"a" = 2hat"j" + 5/3hat"k"`
Also the line has direction ratio 3, 0, 4.
If `bar"b"` is a vector parallel to the line, then `bar"b" = 3hat"i" + 4hat"k"`
The vector equation of the line pasing through `"A"(bara) "and parallel to" bar"b" "is" bar"r" = bar"a" + lambdabar"b"` where λ is a scalar.
∴ the vector equation of the required line is
`bar"r" = (2hat"j" + 5/3hat"k") + lambda(3hat"i" + 4hat"k")`.
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