Find the equation of the plane passing through the point $(1, 0, -2)$ and perpendicular to each of the planes $2x + y - z - 2 = 0$ and $x - y - z - 3 = 0$.

**OR**

Find the length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$.

Question

Find the equation of the plane passing through the point $(1, 0, -2)$ and perpendicular to each of the planes $2x + y - z - 2 = 0$ and $x - y - z - 3 = 0$.

**OR**

Find the length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$.

Accepted Answer

>

> [!TLDR]
> The required plane's normal must be perpendicular to the normals of both given planes, so it is their cross product; then use the point-normal form of a plane.
>
> **Step 1 — Identify the normals of the given planes.**
> $$2x+y-z-2=0 \implies \vec n_1=(2,1,-1),\qquad x-y-z-3=0 \implies \vec n_2=(1,-1,-1).$$
>
> **Step 2 — Find the normal of the required plane.**
> Since the required plane is perpendicular to both given planes, its normal $\vec N$ is perpendicular to both $\vec n_1$ and $\vec n_2$, i.e. $\vec N=\vec n_1	imes \vec n_2$:
> $$\vec N=\begin{vmatrix}\hat i & \hat j & \hat k\ 2 & 1 & -1\ 1 & -1 & -1\end{vmatrix}=\hat i\big[(1)(-1)-(-1)(-1)\big]-\hat j\big[(2)(-1)-(-1)(1)\big]+\hat k\big[(2)(-1)-(1)(1)\big]$$
> $$=\hat i[-1-1]-\hat j[-2+1]+\hat k[-2-1]=-2\hat i+\hat j-3\hat k.$$
> So $\vec N=(-2,1,-3)$.
>
> **Step 3 — Write the plane through $(1,0,-2)$ with normal $\vec N$.**
> $$-2(x-1)+1(y-0)-3(z-(-2))=0$$
> $$-2x+2+y-3z-6=0$$
> $$-2x+y-3z-4=0.$$
> Multiply by $-1$ for a cleaner form:
> $$2x-y+3z+4=0.$$
>
> **Step 4 — Verify.**
> Point $(1,0,-2)$: $2(1)-0+3(-2)+4=2-6+4=0$ ✓.
> $\vec N\cdot\vec n_1=(-2)(2)+(1)(1)+(-3)(-1)=-4+1+3=0$ ✓ (perpendicular to first plane).
> $\vec N\cdot\vec n_2=(-2)(1)+(1)(-1)+(-3)(-1)=-2-1+3=0$ ✓ (perpendicular to second plane).
> …

Q.Find the equation of the plane passing through the point $(1, 0, -2)$ and perpendicular to each of the planes $2x + y - z - 2 = 0$ and $x - y - z - 3 = 0$ . OR Find the length and foot of the perpendicular from the point $(7, 14, 5)$ to the plane $2x + 4y - z = 2$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

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