What is the equation of the plane which passes through the z-axis and is perpendicular to the line `(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/0?`a. `x+y

Question

What is the equation of the plane which passes through the z-axis and is perpendicular to the line `(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/0?`a. `x+y

What is the equation of the plane which passes through the z-axis and is perpendicular to the line `(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/0?`a. `x+yt a ntheta=0` b. `y+x t a ntheta=0` c. `xcostheta-ysintheta=0`d. `xsintheta-ycostheta=0`
A. `x+ytentheta=0`
B. `y+xtantheta=0`
C. `xcostheta-ysintheta=0`
D. `xsintheta-ycostheta=0`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Correct Answer - a
The plane is perpendicular to the line
`" "(x-a)/(costheta)=(y+2)/(sintheta)=(z-3)/(0)`
Hence, the direction ratios of the normal of the plane are `costheta, sintheta and 0`. `" "` (i)
Now, the required plane passes through the z-axis.
Hence, the point `(0, 0, 0)` lies on the plane.
From Eqs. (i) and (ii) , we get equation of the plane as
`" "costheta (x-0)+ sin theta(y-0) + 0(z-0) =0`
`" "costhetasx + sin theta y =0`
`" "x+ytantheta=0`