Question

Find the Cartesian and vector equation of the line passing through the point having position vector `hat"i" + 2hat"j" + 3hat"k"` and perpendicular to vectors `hat"i" + hat"j" + hat"k"` and `2hat"i" - hat"j" + hat"k"`

Answer 1

Let `bar"a" = hat"i" + 2hat"j" + 3hat"k", bar"b"_1 = hat"i" + hat"j" + hat"k", bar"b"_2 = 2hat"i" - hat"j" + hat"k"`

The required line is perpendicular to `bar"b"_1 = hat"i" + hat"j" + hat"k"` and `bar"b"_2 = 2hat"i" - hat"j" + hat"k"`

∴ It is parallel to `bar"b" = bar"b"_1 xx bar"b"_2`

Now, `bar"b"_1 xx bar"b"_2 = |(hat"i", hat"j", hat"k"),(1, 1, 1),(2, -1, 1)|`

= `hat"i"(1 + 1) - hat"j"(1 - 2) + hat"k"(-1 - 2)`

= `2hat"i" + hat"j" - 3hat"k"`

The vector equation of a line passing through a point with position vector `bar"a"` and parallel to `bar"b"` is `bar"r" = bar"a" + lambdabar"b"`

∴ Vector equation of the required line is

`bar"r" = (hat"i" + 2hat"j" + 3hat"k") + lambda(2hat"i" + hat"j" - 3hat"k")`    .......(i)

Putting `bar"r" = xhat"i" + yhat"j" + zhat"k"` in (i), we get

`xhat"i" + yhat"j" + zhat"k" = (1 + 2lambda)hat"i" + 2(2 + lambda)hat"j" + (3 - 3lambda)hat"k"`

Equating the coefficients of `hat"i", hat"j"` and `hat"k"`, we get

x = 1 + 2λ, y = 2 + λ, z = 3 – 3λ

∴ λ = `(x - 1)/2`, λ = `(y - 2)/1`, λ = `(z - 3)/(-3)`

∴ `(x- 1)/2 = (y - 2)/1 = (z - 3)/(-3)`,

which is required cartesian equation.