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`        Find the equation of the plane containing the straight line (x-1)/2=(y+2)/-3=z/5 and perpendicular to the plane x - y + z + 2 = 0`
one year ago

Samyak Jain
333 Points
```							Equation of line : (x – 1) / 2  =  (y + 2) / – 3  =  (z – 0) / 5.The line passes through (1, –2, 0) and its direction ratios (DRs) are 2, – 3,  5. The required plane contains the line and hence, the point (1, –2, 0).So, a vector along this line is b = 2 i – 3 j + 5 kGiven equation of plane is  x – y + z + 2 = 0. DRs of normal to this plane are  1, –1, 1.A vector along this normal is c = i – j + k.Required plane is perpendicular to the given plane i.e. normal of required plane is perpendicularto that of given plane. It is also perpendicular to the straight line. Let a vector along the normal be n.$\dpi{100} \therefore$ n = b x c = (2 i – 3 j + 5 k) x (i – j + k) = 2 i + 3 j + kWe know that equation of a plane is r . n = d , where r = x i + y j + z k , d = constant, n = vector along normal of the plane whose equation is to be determined.So, (x i + y j + z k) . (2 i + 3 j + k) = d  $\dpi{120} \Rightarrow$ 2x + 3y + z = d         ...(1)For constant d, put (1, –2, 0) in the above equation as the plne passes through the point.$\dpi{100} \therefore$ 2(1) + 3(–2) + 1(0) = d = 2 – 6 = – 4. Substitue value of d in (1). 2x + 3y + z = – 4  $\dpi{120} \Rightarrow$  2x + 3y + z + 4 = 0 , which is the required equation of the plane.
```
9 months ago
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