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Question: The Cartesian form of the plane \(r = (s - 2t)\hat i + (3 - t)\hat j + (2s + t)\hat k\) is A) 2x-5...

The Cartesian form of the plane r=(s−2t)i^+(3−t)j^+(2s+t)k^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

Explanation

Solution

First we’ll compare the given equation of the plane with the general of the equation of the plane. On comparing we’ll get the vectors parallel to the plane and a position vector lying on the plane.
Then using parallel vectors we’ll find the normal vector of the plane. After finding all these vectors we’ll substitute them in the Cartesian equation of any plane to get the answer.

Complete step by step solution: Given data: the vector form of plane i.e. r=(s−2t)i^+(3−t)j^+(2s+t)k^r = (s - 2t)\hat i + (3 - t)\hat j + (2s + t)\hat k
On simplifying the equation of the plane
⇒r=3j^+s(i^+2k^)+t(−2i^−j^+k^)..........(i)\Rightarrow r = 3\hat j + s(\hat i + 2\hat k) + t( - 2\hat i - \hat j + \hat k)..........(i)
We know that the general equation of a plane is r⃗=a+λb+μc\vec r = a + \lambda b + \mu cwhere the plane passes through a and is parallel to b and c
On comparing equation(i) and the general equation, we can say that
a=3j^a = 3\hat j
b⃗=i^+2k^\vec b = \hat i + 2\hat k
And c⃗=−2i^−j^+k^\vec c = - 2\hat i - \hat j + \hat k
Since b and c are parallel to the plane, we can say that their normal or the cross product is the normal of the plane
i.e.n⃗=b⃗×c⃗\vec n = \vec b \times \vec c
= \left| {\begin{array}{*{20}{c}} {\widehat i}&{\widehat j}&{\widehat k} \\\ 1&0&2 \\\ { - 2}&{ - 1}&1 \end{array}} \right|
=(0+2)i^−(1+4)j^+(−1+0)k^= (0 + 2)\hat i - (1 + 4)\hat j + ( - 1 + 0)\hat k
On simplification we get,
=2i^−5j^+−1k^= 2\hat i - 5\hat j + - 1\hat k
∴n⃗=2i^−5j^+−1k^\therefore \vec n = 2\hat i - 5\hat j + - 1\hat k
Now the equation of the plane is given by
⇒r⃗.n⃗=a⃗.n⃗\Rightarrow \vec r.\vec n = \vec a.\vec n
Now, substituting the value of a⃗\vec aand n⃗\vec n
⇒r⃗.(2i^−5j^+−1k^)=3j^.(2i^−5j^+−1k^)\Rightarrow \vec r.\left( {2\hat i - 5\hat j + - 1\hat k} \right) = 3\hat j.\left( {2\hat i - 5\hat j + - 1\hat k} \right)
Now using j^.i^=0\hat j.\hat i = 0 , j^.j^=1\hat j.\hat j = 1and j^.k^=0\hat j.\widehat k = 0, we get,
⇒r⃗.(2i^−5j^+−1k^)=−15\Rightarrow \vec r.\left( {2\hat i - 5\hat j + - 1\hat k} \right) = - 15
Now substituting r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k
⇒(xi^+yj^+zk^).(2i^−5j^+−1k^)=−15\Rightarrow \left( {x\hat i + y\hat j + z\hat k} \right).\left( {2\hat i - 5\hat j + - 1\hat k} \right) = - 15
Again using i^.j^=j^.i^=0\hat i.\hat j = \hat j.\hat i = 0,j^.k^=k^.j^=0\hat j.\hat k = \hat k.\hat j = 0,i^.k^=k^.i^=0\hat i.\hat k = \hat k.\hat i = 0 and, i^.i^=j^.j^=k^.k^=1\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1
⇒2x−5y−z=−15\Rightarrow 2x - 5y - z = - 15
⇒2x−5y−z+15=0\Rightarrow 2x - 5y - z + 15 = 0

Hence, Option (C) is correct.

Note: Here we’ve substitute the normal vector by the cross product of b and c as they are parallel to the plane and the cross product is always normal to the respective vectors. We can also prove that by the dot product of b or c with the n vector and it will result in zero as the dot product of two perpendicular vectors is zero.
b⃗=i^+2k^\vec b = \hat i + 2\hat k
n⃗=2i^−5j^+−1k^\vec n = 2\hat i - 5\hat j + - 1\hat k
⇒n⃗.b⃗=(2i^−5j^+−1k^).(i^+2k^)\Rightarrow \vec n.\vec b = (2\hat i - 5\hat j + - 1\hat k).(\hat i + 2\hat k)
using i^.j^=j^.i^=0\hat i.\hat j = \hat j.\hat i = 0,j^.k^=k^.j^=0\hat j.\hat k = \hat k.\hat j = 0,i^.k^=k^.i^=0\hat i.\hat k = \hat k.\hat i = 0 and, i^.i^=j^.j^=k^.k^=1\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1
⇒n⃗.b⃗=2(1)−1(2)\Rightarrow \vec n.\vec b = 2(1) - 1(2)
∴n⃗.b⃗=0\therefore \vec n.\vec b = 0 hence b vector and n vector are mutually perpendicular to each other.