Find the acute angle between the plane 5*x* − 4*y* + 7*z* − 13 = 0 and the *y*-axis.

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#### Solution

The equation of the *y*-axis is

`(x−0)/0=(y−0)/1=(z−0)/0`

The direction ratios of the *y*-axis are 0, 1, 0.

The equation of the given plane is 5*x* − 4*y* + 7*z* − 13 = 0.

So, the direction ratios of the normal to the plane are 5, −4, 7.

Let *θ* be the acute angle between the given plane and the *y*-axis.

`therefore sin theta=|(a_1a_2+b_1b_2+c_1c_2)/(sqrt(a_1^2+b_1^2+c_1^2)sqrt(a_2^2+b_2^2+c_2^2))|`

`=> sin theta=|(0xx5−4xx1+0xx7)/(sqrt(0+1+0)sqrt(25+16+49))|`

`=|-4/(3sqrt10)|`

`=>theta=sin^-1(4/3sqrt10)`

Hence, the acute angle between the given plane and the *y*-axis is `sin^-1(4/3sqrt10)`

Concept: Angle Between Line and a Plane

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