Question

Find the vector equation of the plane passing through points 4i - 3j - k, 3i + 7j -10k and 2i + 5j - 7k and show that the point i + 2j - 3k lies in the plane.

Answer 1

Answer:

Equation of the plane:

$\vec r.\hat i+\hat j+\hat k=0$

Step-by-step explanation:

let

$\vec a=4\hat i-3\hat j-\hat k\\\\\vec b=3\hat i+7\hat j-10\hat k\\\\\vec c=\hat i+2\hat j-3\hat k$

vector equation

$(\vec r-\vec a).(\vec {AB}\times\vec{AC} )\\\\\vec {AB}=-\hat i+10\hat j-9\hat k\\\\\vec {AC}=-3\hat i+5\hat j-2\hat k\\\\\\(\vec {AB}\times\vec{AC} )=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\-1&10&-9\\-3&5&-2\end{array}\right|\\\\=\hat i(-20+45)-\hat j(2-27)+\hat k(-5+30)\\\\\\=25\hat i+25\hat j+25\hat k$

So

$(\vec r-\vec a).(25\hat i+25\hat j+25\hat k)=0$

$\vec r-(4\hat i-3\hat j-\hat k).(25\hat i+25\hat j+25\hat k)=0\\\\\vec r.\hat i+\hat j+\hat k=0$

is the required equation of the plane.

2) If the points lie on the plane than it(1,2,-3) satisfies the plane

so equation of plane in cartesian form is x+y+z=0

put point(1,2,-3)

1+2-3=0

hence the point lie on the plane.