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