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

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 $$\dfrac{l}{a} = \dfrac{m}{b} = \dfrac{n}{c}$$