# Find the acute angle between x-2/1=y+3/2=z/-2 and plane -12x+3y+4z=9

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You need to remember that the acute angle between a line and a plane is the angle between the direction vector of the line and the normal vector to the plane.

You need to put the equation of the line in parametric form such that:

x = 2 + t

y = -3 + 2t

z = -2t

You need to identify the vector equation of the line such that:

`bar v = lt1,2,-2gt`

You need to identify the normal vector to the given plane such that:

`bar n = <12,3,4>`

You need to find the acute angle between the vectors `bar v`  and `bar n`  using the dot product such that:

`bar n*bar v = |bar n|*|bar v|*cos alpha`

`cos alpha = (bar n*bar v)/(|bar n|*|bar v|)`

You need to evaluate `bar n*bar v`  such that:

`bar n*bar v =12*1 +3*2 + 4*(-2)`

`bar n*bar v = 12 + 6 - 8`

`bar n*bar v = 10`

You need to evaluate `|bar n|*|bar v|`  such that:

`|bar n|*|bar v| = (sqrt(12^2+3^2+4^2))*(sqrt(1^2+2^2+(-2)^2))`

`|bar n|*|bar v| = (sqrt(169))*(sqrt(9))`

`|bar n|*|bar v| = 13*3`

`|bar n|*|bar v| = 39`

`cos alpha = (bar n*bar v)/(|bar n|*|bar v|)`

`cos alpha = 10/39`

`cos alpha = 0.256 =gt alpha = cos^(-1)(0.256)`

`alpha ~~ 75^o`

Hence, evaluating the acute angle between the given line and plane yields `alpha ~~ 75^o` .

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