Question

Find the equation of the plane passing through the points (-1,2,0),(2,2,-1)& parallel to the line (x-1)/1=(2y+1)/2=(z+1)/-1

Answer 1

$\textsf{The required plane passes through the points (-1,2,0) and (2,2,-1) and parallel to the vector }$$\mathsf{\vec{i}+\vec{j}-\vec{k}}$

$\textsf{The required plane is}$

$\mathsf{\left|\begin{array}{ccc}x-x_1&y-y_1&z-z_1\\x_2-x_1&y_2-y_1&z_2-z_1\\l&m&n\end{array}\right|=0}$

$\mathsf{\left|\begin{array}{ccc}x+1&y-2&z-0\\3&0&-1\\1&1&-1\end{array}\right|=0}$

$\textsf{Expanding along first row, we get}$

$\textsf{(x+1)(1)-(y-2)(-3+1)+z(3)=0}$

$\textsf{x+1+2(y-2)+3z=0}$

$\textsf{x+1+2y-4+3z=0}$

$\implies\boxed{\textsf{x+2y+3z-3=0}}$

Answer 2

Answer: The equation of plane would be x-4y+3x+9=0

Step-by-step explanation:

Since we have given that

Equation of line passing through (-1,2,0), we get that

$A(x+1)+B(y-2)+Cz=0$

Now, the equation passing through (2,2,-1), we get that

$A(2+1)+B(2-2)+C(-1-0)=0\\3A-C=0\\3A=C$

Now, it is parallel to the line :

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

So, our equation of plane becomes:

$A+B-C=0$

$A+B+3A=0\\\\4A+B=0\\\\B=-4A$

So, our equation becomes,

$A(x+1)-4A(y-2)+3Az=0\\\\(x+1)-4(y-2)+3z=0\\\\x-4y+3z+9=0$

Hence, the equation of plane would be x-4y+3x+9=0