Question

find the equation of the plane through the point (2,3,5) and perpendicular to the planes 2x-3y+z=2, 4x+y-3z+1=0​

Answer 1

Answer:

The required plane is $4x+5y+7z-58=0$

Step-by-step explanation:

The normal vector to the plane 2x-3y+z-2=0 is

$2\vec{i}-3\vec{j}+\vec{k}$

The normal vector to the plane 4x+y-3z+1=0 is $4\vec{i}+\vec{j}-3\vec{k}$

Hence the required plane passes through the point (2,3,.5) and parallel to the vectors $2\vec{i}-3\vec{j}+\vec{k}$ and $4\vec{i}+\vec{j}-3\vec{k}$

Equation of the required plane is

$\left|\begin{array}{ccc}x-x_1&y-y_1&z-z_1\\l_1&m_1&n_1\\l_2&m_2&n_2\end{array}\right|=0$

$\left|\begin{array}{ccc}x-2&y-3&z-5\\2&-3&1\\4&1&-3\end{array}\right|=0$

Expanding along $R_1$, we get

$(x-2)(8)-(y-3)(-10)+(z-5)(14)=0$

$8(x-2)+10(y-3)+14(z-5)=0$

Divide both sides by 2

$4(x-2)+5(y-3)+7(z-5)=0$

$4x-8+5y-15+7z-35=0$

$4x+5y+7z-58=0$