Find the cartesian equations of the following planes:
\(\overrightarrow r.(\hat i+\hat j-\hat k)=2\) (b) \(\overrightarrow r.(2\hat i+3\hat j-4\hat k)=1\)
(c) \(\overrightarrow r.[(s-2t)\hat i+(3-t)\hat j+(2s+t)\hat k)=15\)
(a)It is given that equation of the plane is
\(\overrightarrow r.(\hat i+\hat j-\hat k)=2\)...(1)
For any arbitrary point P(x,y,z) on the plane, position vector r→ is given by,
\(\overrightarrow r.(x\hat i+y\hat j-z\hat k)=z\hat k\)
Substituting the value of \(\overrightarrow r\) in equation(1), we obtain
\((x\hat i+y\hat j-z\hat k).(\hat i+\hat j-\hat k)=2\)
\(\Rightarrow \) x+y-z=2
This is the cartesian equation of the plane.
(b) \(\overrightarrow r.(2\hat i+3\hat j-4\hat k)=1\)...(1)
For any arbitrary point P(x,y,z) on the plane, position vector \(\overrightarrow r\) is given by,
\(\overrightarrow r.(x\hat i+y\hat j-z\hat k)\)
Substituting the value of \(\overrightarrow r\) in equation(1), we obtain
\((x\hat i+y\hat j+z\hat k)=z\hat k (2\hat i+3\hat j-4\hat k)=1\)
\(\Rightarrow\) 2x+3y-4z=1
This is the cartesian equation of the plane.
(c) \(\overrightarrow r.[(s-2t)\hat i+(3-t)\hat j+(2s+t)\hat k)=15\)...(1)
For any arbitrary point P(x,y,z) on the plane, position vector\(\overrightarrow r\) is given by,
\(\overrightarrow r.(x\hat i+y\hat j-z\hat k)\)
Substituting the value of r→ in equation(1), we obtain
\(\overrightarrow r.(x\hat i+y\hat j-z\hat k)\).\([(s-2t)\hat i+(3-t)\hat j+(2s+t)\hat k)=15\)
\(\Rightarrow \) (s-2t)x+(3-t)y+(2s+t)z=15
This is the cartesian equation of the given plane.
Show that the following lines intersect. Also, find their point of intersection:
Line 1: \[ \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} \]
Line 2: \[ \frac{x - 4}{5} = \frac{y - 1}{2} = z \]
A surface comprising all the straight lines that join any two points lying on it is called a plane in geometry. A plane is defined through any of the following uniquely: