Question

The equation of the plane perpendicular to the yz-plane and passing through the point (1, -2, 4) and (3, -4, 5) is,
A. \[y+2z=5\]
B. \[2y+z=5\]
C. \[y+2z=6\]
D. \[2y+z=6\]

Answer 1

Hint: Plane perpendicular to the yz-plane is also parallel to the x-axis. So find the equation of the vector lying in the plane and the vector representing the normal vector. Thus take their cross product and find the equation of the plane using the point and normal.

Complete step-by-step answer:

It is said that a plane is perpendicular to the yz-plane, then the plane will be parallel to the x-axis. If the planes are perpendicular, then their normals are also perpendicular. Thus the normal vector, can be said as (1, 0, 0), as the normal of the x-axis, parallel to plane and yz are perpendicular.

Now we need to find the vector which is lying in the plane by using the 2 given points (1, -2, 4) and (3, -4, 5). Let us mark that vector as \[\overrightarrow{s}.\]

\[\begin{align}
  & \therefore ({{x}_{1}},{{y}_{1}},{{z}_{1}})=(1,-2,4) \\
 & \therefore ({{x}_{2}},{{y}_{2}},{{z}_{2}})=(3,-4,5) \\
 & \therefore \overrightarrow{s}=({{x}_{2}}-{{x}_{1}})\hat{i}+({{y}_{2}}-{{y}_{1}})\hat{j}+({{z}_{2}}-{{z}_{1}})\hat{k} \\
 & \overrightarrow{s}=(3-1)\hat{i}+(-4-(-2))\hat{j}+(5-4)\hat{k} \\
 & \overrightarrow{s}=2\hat{i}-2\hat{j}+\hat{k}. \\
\end{align}\]

Similarly, the parallel vector \[\overrightarrow{p}\] can be represented from point (1, 0, 0) as
\[\overrightarrow{p}=1\hat{i}+0\hat{j}+0\hat{k}.\]

So the normal of the plane is given as,

\[\overrightarrow{N}=\overrightarrow{S}\times \overrightarrow{P}\]

Let us take the cross product.

\[\begin{align}
  & \overrightarrow{N}=\left( \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & -2 & 1 \\
   1 & 0 & 0 \\
\end{matrix} \right) \\

 & \overrightarrow{N}=(-2\times 0-1\times 0)\hat{i}-(2\times 0-1\times 1)\hat{j}+(2\times

0-(-2)\times 1) \\

 & \overrightarrow{N}=0\hat{i}-\hat{j}(-1)+2\hat{k}=\hat{j}+2\hat{k}. \\

\end{align}\]
Therefore the equation of plane using the point (1, -2, 4) and normal \[\hat{j}+2\hat{k}\] is,

\[\begin{align}

  & (x-{{x}_{1}})\hat{i}+(y-{{y}_{1}})\hat{j}+(z-{{z}_{1}})\hat{k}=0 \\

 & \Rightarrow (x-1)\times 0+(y-(-2)\times 1+(z-4)\times 2=0 \\

 & (y+2)+(z-4)\times 2=0 \\

 & (y+2)(2z-8)=0 \\

 & \Rightarrow y+2+2z-8=0 \\

 & y+2z-6=0 \\

 & y+2z=6. \\

\end{align}\]

Hence we got the equation of plane perpendicular to the yz-plane and passing through the point (1, -2, 4).

Note: The equation of plane using the point (3, -4, 5) and normal \[\hat{j}+2\hat{k}\] is,

\[\begin{align}

  & (x-{{x}_{1}})\hat{i}+(y-{{y}_{1}})\hat{j}+(z-{{z}_{1}})\hat{k}=0 \\

 & (x-3)\hat{i}+(y-(-4))\times 1+(z-5)\times 2=0 \\

 & (y+4)+4(z-5)=0 \\

 & y+4+2z-10=0 \\

 & y+2z=6. \\

\end{align}\]
Hence we got the same equation of plane which is perpendicular to the plane and perpendicular to the yz-plane and through point (3, -4, 5).