A line makes angles of measures `45^(@)` and `60^(@)` with positive directions of Y-and Z-axes respectively. Find the d.c.s. of the line and also find the vector of magnitude 5 along the direction of line.

Explanation

Let `alpha, beta, gamma` be the direction angles of the line.
Then `beta = 45^(@) and gamma = 60^(@)`
We know that
`cos^(2)alpha+cos^(2)beta+cos^(2)gamma=1`
`therefore cos^(2) alpha + cos^(2) 45^(@) + cos^(2)60^(@) =1`
`therefore cos^(2)alpha + ((1)/(sqrt(2)))^(2) + ((1)/(2))^(2) =1`
`therefore cos^(2) alpha + (1)/(2) + (1)/(4) =1`
`therefore cos^(2)alpha = 1-(3)/(4)=(1)/(4)`
`therefore cos alpha = pm (1)/(2)`
Let l,m, n be the direction cosines of the line.
Then `l = cos alpha = pm (1)/(2)`
`m = cos beta = cos 45^(@) = (1)/(sqrt(2))`
`n = cos gamma = cos 60^(@) = (1)/(2)`
`therefore` direction cosines of the line are
` pm (1)/(2), (1)/(sqrt(2)), (1)/(2)`
Let `bare` be the unit vector along the line.
Then `bare = lhati + mhatj + nhatk = pm (1)/(2) hati + (1)/(sqrt(2)) hatj + (1)/(2) hatk`
`therefore` the required vectors of magnitude 5 along the direction of the lines are
`5(pm(1)/(2)hati +(1)/(sqrt(2))hatj+(1)/(2)hatk)`
i.e., `5((1)/(2)hati +(1)/(sqrt(2))hatj+(1)/(2)hatk)and 5(-(1)/(2)hati+(1)/(sqrt(2))hatj+(1)/(2)hatk)`