Question

Find the acute angle between the lines x = –y, z = 0 and x = 0, z = 0.

Answer 1

The equations x = – y, z = 0 can be written as `x/(1) = -y/(1), z` = 0.

∴ the direction ratios of the line are 1, -1, 0.
The direction ratios of the line x = 0, z = 0, i.e., Y-axis are 0, 1, 0.

∴ its direction ratios are 0, 1, 0.

Let `bar"a" and bar"b"` be the vectors in the direction of the lines x = -y, z = 0 and x = 0, z = 0.

Then `bar"a" = hat"i" + hat"j",  bar"b" = hat"j"`

∴ `bar"a".bar"b" = (hat"i" + hat"j").hat"j"`

= (1)(0) + (-1)(1) + (0)(0)

= -1

`|bar"a"| = sqrt((1^2) + (-1^2)) = sqrt(2)`

`|bar"b"| = |hat"j"|` = 1

If θ is the acute angle between the lines, then

cos θ = `|(bar"a".bar"b")/(|bar"a"|bar"b"|)| = |(1)/(sqrt(2) xx 1)| = (1)/sqrt(2)` = cos 45°

∴ θ = 45°