Aspire Faculty ID #17304 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

The equation of plane passing through the point $(0,7,-7)$ and containing the line

$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$

is:

Solution


The line

$\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1}$

gives a point on the line

$(-1,3,-2)$

Direction vector of line

$\vec d=(-3,2,1)$


Vector joining given point and line point

$(0,7,-7)-(-1,3,-2)=(1,4,-5)$


Normal vector of plane

$\vec n = \vec d \times (1,4,-5)$

$\vec n=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-3 & 2 & 1 \\
1 & 4 & -5
\end{vmatrix}$

$= -14\mathbf{i}-14\mathbf{j}-14\mathbf{k}$

Normal vector $\propto (1,1,1)$


Plane equation

$x+y+z+d=0$

Substitute point $(0,7,-7)$

$0+7-7+d=0$

$d=0$

Plane

$x+y+z=0$
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