Question

By vector methods, find the cosine of the angle between the lines (x - 1)/(3) = (y - 0.5)/(2) = z and x = y = z

Answer 1

The angle between the lines
`(x-1)/(3)= (y-0.5)/(2)=(z-0)/(1)` and
`(x-0)/(1)= (y-0)/(1)=(z-0)/(1)` is
`\sqrt{(6)/(7)}`

Explanation:

The equation of a line with direction vector
`\vec{d}=(l,m.n)` that passes through the point
`(x_(1),y_(1),z_(1))` is given by the formula

`(x-x_(1))/(l)= (y-x_(1))/(m)=(z-z_(1))/(n),` where l,m, and n are non-zero real numbers.

This is called the symmetric equations of the line.

The angle between two lines
`(x-x_(1))/(l_(1) )= (y-y_(1))/(m_(1) )=(z-z_(1))/(n_(1))` and
`(x-x_(2))/(l_(2) )= (y-y_(2))/(m_(2) )=(z-z_(2))/(n_(2))` equal the angle subtended by direction vectors,
`d_(1)` and
`d_(2)` of the lines

`cos (\theta)=\frac{\vec{d_(1)}\cdot\vec{d_(2)}}{|\vec{d_(1)}|\cdot|\vec{d_(2)}|}=\frac{l_(1) \cdot\l_(2)+m_(1) \cdot\ m_(2)+n_(1) \cdot\ n_(2)}{\sqrt{l_(1)^(2)+m_(1)^(2)+n_(1)^(2)} \cdot \sqrt{l_(2)^(2)+m_(2)^(2)+n_(2)^(2)}}`

Given that

`(x-1)/(3)= (y-0.5)/(2)=(z-0)/(1)` and
`(x-0)/(1)= (y-0)/(1)=(z-0)/(1)`

`l_(1)=3, m_(1)=2,n_(1)=1\\ l_(2)=1, m_(2)=1,n_(2)=1`

We can use the formula above to find the cosine of the angle between the lines

`cos(\theta)=\frac{3 \cdot 1+2 \cdot 1 +1 \cdot 1}{\sqrt{3^(2)+2^(2)+1^(2)} \cdot \sqrt{1^(2)+1^(2)+1^(2)}} = \sqrt{(6)/(7)}`