1.

Find the shortest distancebetween the z-axis and the line, `x+y+2z-3=0,2x+3y+4z-4=0.`

Answer» let `p_1` and `p_2` be the points on the given planes having coordinates (1,1,2) & (2,3,4) respectively
R = `|(i,j,k),(1,1,2),(2,3,4)|`
`=i(-2)-j(0) + k(1) = -2i+k`
`(-2,0,1)`
if z=0 `x+y-3 = 0 & 2x+3y-4 = 0`
we get `2(3-y) + 3y-4 = 0`
then, `y=-2 & x = 5`
so,intersection point `(5,-2,0)`
eqn of line is
`(x-5)/-2 = (y+2)/0 = z/1`
let direction ratios of this line is `(a_1,b_1,c_1)`
eqn of z axis is
`x/0 = y/0 = z/0`
let direction ratios of this line is `(a_2,b_2,c_2)`
S.D = `||(x_2-x_1,y_2-y_1,z_2-z_1),(a_1,b_1,c_1),(a_2,b_2,c_2)||/||(i,j,k),(a_1,b_1,c_1),(a_2,b_2,c_2)||`
`= ||(-5,2,0),(-2,0,1),(0,0,1)||/(||(i,j,k),(-2,0,1),(0,0,1)||)`
`|(-4)|/(|j(2)|)= 2`


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