Question

[a b c] QUESTION NO:25 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

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Answer 1

\underline{\sf Solution:-}

$\sf\vec a=4\hat i-3\hat j-\hat k\\\\\ \sf \: vec \: b=3\hat i+7\hat j-10\hat k\\\\\ \sf \: vec \: C=\hat i+2\hat j-3\hat k\\\\\sf vector \: \: \: equation\\ \\ ▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬ \\ \\ \longrightarrow\sf (\vec r-\vec a).(\vec {AB}\times\vec{AC} )\\\\ \sf \: \vec {AB}=-\hat i+10\hat j-9\hat k\\\\\longrightarrow \sf \: \vec {AC}=-3\hat i+5\hat j-2\hat k\\\\\\ \longrightarrow\sf(\vec {AB}\times\vec{AC} )\longrightarrow\left|\begin{array}{ccc} \sf\hat i&\hat j& \sf \: \hat k\\-1&10&-9\\-3&5&-2\end{array}\right|\\\\= \sf\hat i(-20+45)-\hat j(2-27)+\hat k(-5+30)\\\\\\ \underline{ \boxed{\longrightarrow \red{\sf25\hat i+25\hat j+25\hat k}}}\\\\\sf \longrightarrow (\vec r-\vec a).(25\hat i+25\hat j+25\hat k) 0 \vec r-(4\hat i-3\hat j-\hat k).(25\hat i+25\hat j+25\hat k)=0\\\\ \underline{ \boxed{\purple{\sf\vec r.\hat i+\hat j+\hat k=0}}}$

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.