Question

The Cartesian form of the plane \[r = (s - 2t)\hat i + (3 - t)\hat j + (2s + t)\hat k\] is
A) \[2x - 5y - z - 15 = 0\]
B) \[2x - 5y + z - 15 = 0\]
C) \[2x - 5y - z + 15 = 0\]
D) \[2x + 5y - z + 15 = 0\]

Answer 1

Hint: First we have to solve the co-efficient of \[\hat i\], \[\hat j\] & \[\hat k\] using the matrix in order to get the values of coefficients . Then we have to use the Cartesian form of the plane in the form of \[Ax + By + Cz + D = 0\]. Find the values of A,B,C,D using formula of direction ratios.

Formula used: Equation of a plane in Cartesian form is and the unit vector of the plane is obtained by taking the cross product of their direction ratios as \[\hat n = {d_1} \times {d_2}\].

Complete step-by-step answer:
For writing the equation of a line in parametric form we need to find a point on that line and direction ratios of that line.
We have given with the parametric form of the equation as \[r = (s - 2t)\hat i + (3 - t)\hat j + (2s + t)\hat k\]
Cartesian form of the plane is in the form of \[Ax + By + Cz + D = 0\] on comparing this Cartesian form to the parametric form of the equation we get ,
\[\
  A = 0 + s - 2t \\
  B = 3 + 0 - t \\
  C = 0 + 2s + t \\
\]
And the value of \[D\] is obtained by satisfying the equation \[Ax + By + Cz + D = 0\].
In parametric form the constant term represents the coordinates of point ( P ) & co-efficient of ‘s’ & ‘t’ represent the direction ratio.
P is a point on the plane whose coordinates are \[P(0,3,0)\]and ,\[{\vec d_2}\]are the direction ratios of that plane.
\[P(0,3,0)\], P is a point that lies on the plane \[r = (s - 2t)\hat i + (3 - t)\hat j + (2s + t)\hat k\] and the direction ratios of this plane are coefficients of ‘s’ and coefficients of ‘t’ respectively
\[\
  {{\vec d}_1} = (1,0,2) \\
  {{\vec d}_2} = ( - 2, - 1,1) \\
\ \]
Now we have to find a unit vector along the plane so for this we have to take cross product of the direction ratios
\[\hat n(A,B,C) = (1,0,2) \times ( - 2, - 1,1)\]\[\hat n(A,B,C) = (1,0,2) \times ( - 2, - 1,1) \equiv (2, - 5, - 1)\]
\[\left| {\begin{array}{*{20}{c}}
  {\hat i}&{\hat j}&{\hat k} \\
  1&0&2 \\
  { - 2}&{ - 1}&1
\end{array}} \right| = \]\[\hat i(2) - \hat j(5) + \hat k( - 1)\]
Unit vector obtained here is =\[\hat i(2) - \hat j(5) + \hat k( - 1)\]
Unit vector is \[\hat i(2) - \hat j(5) + \hat k( - 1)\]= 0
Now putting these values in Cartesian for\[D = 15\]m of the plane \[Ax + By + Cz + D = 0\]
\[A = 2,B = - 5,C = - 1\]
the equation becomes\[A = 2,B = - 5,C = - 1\]
\[2x - 5y - 1.z + D = 0\]
Now putting the values of (x,y,z) =(0,3,0)
\[0 - 15 - 0 + D = 0\]
\[Ax + By + Cz + D = 0\] where & \[D = 15\]
\[2x - 5y - z + 15 = 0\]

Option ( C ) is the correct option.

Note: Direction ratios : Any three numbers a,b,c proportional to the direction cosines \[l,m,n\] are called direction ratios of the line.
Direction cosine of x-axis = (1,0,0)
Direction cosine of y-axis = (0,1,0)
Direction cosine of z-axis = (0,0,1)
Also \[{l^2} + {m^2} + {n^2} = 1\]
Relation between direction ratio and direction cosines
\[\dfrac{l}{a} = \dfrac{m}{b} = \dfrac{n}{c}\]