The distance of the point (7, $-$3, $-$4) from the plane passing through the points (2, $-$3, 1), ($-$1, 1, $-$2) and (3, $-$4, 2) is :
Solution
$A(2,-3,1), B(-1,1,-2), C(3,-4,2)$
<br/><br/>
$$
\begin{aligned}
& \overrightarrow{A B}=-3 \hat{i}+4 \hat{j}-3 \hat{k} \quad \overrightarrow{A C}=\hat{i}-\hat{j}+\hat{k} \\\\
& \vec{n}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\\\
-3 & 4 & -3 \\\\
1 & -1 & 1
\end{array}\right|=\hat{i}-\hat{k}
\end{aligned}
$$
<br/><br/>
Let equation of plane is $x-z+\lambda=0$ passes through point $A(2,-3,1) \Rightarrow \lambda=-1$
<br/><br/>
Equation of plane is $x-z-1=0$
<br/><br/>
Distance of point $(7,-3,-4)$ from the plane $x-z-$ $1=0$ is $5 \sqrt{2}$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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