Question

Find the vector equation of the line whose Cartesian equations are y = 2 and 4x – 3z + 5 = 0.

Answer 1

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")`.