If the foot of the perpendicular drawn from the origin to the plane is $(4, -2, -3)$, then find the equation of the plane in vector and Cartesian form. **OR** If $f(1) = 10$ and $f'(x) \geq 2$ for $1 \leq x \leq 4$, how small can $f(4)$ possibly be?

> [!TLDR] > Main part: the foot of the perpendicular gives both the normal direction and a point on the plane, so its position vector directly yields the plane equation. OR alternative: apply the Mean Value Theorem on [1,4] to bound f(4)−f(1) below. > > **Main part** > 1. Let $F=(4,-2,-3)$ be the foot of the perpendicular from the origin $O$ to the required plane. > 2. Since $OF$ is perpendicular to the plane, the direction of $\overrightarrow{OF}=4\vec i-2\vec j-3\vec k$ is a normal vector $\vec n$ to the plane. > 3. The plane passes through $F$, whose position vector is $\vec a=4\vec i-2\vec j-3\vec k$ — the same as $\vec n$ here. > 4. Vector equation of a plane through point $\vec a$ with normal $\vec n$: $\vec r\cdot\vec n=\vec a\cdot\vec n$. > 5. $\vec a\cdot\vec n=(4)(4)+(-2)(-2)+(-3)(-3)=16+4+9=29$. > 6. So the vector equation is $\vec r\cdot(4\vec i-2\vec j-3\vec k)=29$. > 7. Writing $\vec r=x\vec i+y\vec j+z\vec k$, the Cartesian form is $4x-2y-3z=29$. > > **OR — alternative** > 8. We are given $f(1)=10$ and $f'(x)\geq2$ for all $x\in[1,4]$; $f$ is differentiable (hence continuous) on $[1,4]$. …

If the foot of the perpendicular drawn from the origin to the plane is (4, -2, -3), then find the equation of

Question 126 of 162

(4, -2, -3)

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 6mImportance★★★★★

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Main part: the foot of the perpendicular gives both the normal direction and a point on the plane, so its position vector directly yields the plane equation. OR alternative: apply the Mean Value Theorem on [1,4] to bound f(4)−f(1) below.

Main part

Let $F=(4,-2,-3)$ be the foot of the perpendicular from the origin $O$ to the required plane.
Since $OF$ is perpendicular to the plane, the direction of $\overrightarrow{OF}=4\vec i-2\vec j-3\vec k$ is a normal vector $\vec n$ to the plane.
The plane passes through $F$ , whose position vector is $\vec a=4\vec i-2\vec j-3\vec k$ — the same as $\vec n$ here.
Vector equation of a plane through point $\vec a$ with normal $\vec n$ : $\vec r\cdot\vec n=\vec a\cdot\vec n$ .
$\vec a\cdot\vec n=(4)(4)+(-2)(-2)+(-3)(-3)=16+4+9=29$ .
So the vector equation is $\vec r\cdot(4\vec i-2\vec j-3\vec k)=29$ .
Writing $\vec r=x\vec i+y\vec j+z\vec k$ , the Cartesian form is $4x-2y-3z=29$ .

OR — alternative

8. We are given $f(1)=10$ and $f'(x)\geq2$ for all $x\in[1,4]$ ; $f$ is differentiable (hence continuous) on $[1,4]$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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CBSE 2026Set ANNUAL1 markMCQ
$Q. The equation of the plane having intercepts 2, -3, 1 on X, Y and Z axis respectively is ................. .$
(a) 2x − 3y + z = 0
(b) 3x − 2y + 6z = 6
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(d) x + 2y + 3z = 2
›Reveal solutionSolution

Use the intercept form of a plane: $x/a+y/b+z/c=1$ .

With $a=2,\,b=-3,\,c=1$ :

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

…

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CBSE 2025Set E1 markMCQ
$9x - 8y + 7z = 10$
(a) $9x - 8y - 7z = 5$
(b) $9x - 8y + 7z = 5$
(c) $9x + 8y + 7z = 5$
(d) $9x - y + 7z = 5$
›Reveal solutionSolution

Parallel planes have identical coefficients of $x,y,z$ (same normal), only the constant differs.

The plane $9x-8y+7z=10$ has normal direction $(9,-8,7)$ . Any parallel plane must have the same normal, i.e. the same coefficients of $x, y, z$ , differi …

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CBSE 2024Set D1 markMCQ
$xy$
(a) $x = 0$
(b) $y = 0$
(c) $z = 0$
(d) none of these
›Reveal solutionSolution

Points on the $xy$ -plane have $z=0$ .

The $xy$ -plane consists of all points $(x, y, 0)$ ; the defining condition is that the $z$ -coordinate vanishes. Hence its equation is $z = 0$ . (Simila …

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CBSE 2024Set D1 markMCQ
$3x - 5y + 4z = 11$
(a) $3x - 5y + 4z = 21$
(b) $3x + 5y + 4z = 25$
(c) $3x + 5y + 4z = 35$
(d) none of these
›Reveal solutionSolution

Parallel planes share the same normal, so the coefficients of $x, y, z$ must be identical: $3x - 5y + 4z = 21$ .

Two planes are parallel iff their normal vectors are proportional. The given plane $3x - 5y + 4z = 11$ has normal $(3, -5, 4)$ . A parallel plane must therefore have the form

$3x - 5y + 4z = k$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. The Cartesian equation of the plane r⃗ \cdot (î + ĵ - k̂) = 2 is :$
(a) x − y − z = 2
(b) x + y − z = 2
(c) x + y + z = 2
(d) x + y − z = −2
›Reveal solutionSolution

Writing r⃗ = xî + yĵ + zk̂ and dotting with (î + ĵ − k̂) directly gives the Cartesian form x + y − z = 2.

The vector equation of a plane is $\vec r \cdot \vec n = d$ , where $\vec r = x\hat\imath + y\hat\jmath + z\hat k$ is the position vector of a general point and $\vec n$ is the plane's normal.

Here $\vec n = \hat\imath + \hat\jmath - \hat k$ and $d = 2$ .

…

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CBSE 2023Set ANNUAL1 markMCQ
$\vec{r}\cdot(\hat{i}+\hat{j}-\hat{k}) = 2$
(a) $x+y-z=0$
(b) $x+y-z=2$
(c) $x+y-z=1$
(d) $x+y+z+2=0$
›Reveal solutionSolution

Substitute $\vec r=x\hat i+y\hat j+z\hat k$ into the vector equation of the plane and take the dot product.

Given $\vec r\cdot(\hat i+\hat j-\hat k)=2$ , with $\vec r=x\hat i+y\hat j+z\hat k$ :

…

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CBSE 2023Set E1 markMCQ
$x+2y-3z+15=0$
(a) $1,2,3$
(b) $1,-2,3$
(c) $1,2,-3$
(d) $1,2,15$
›Reveal solutionSolution

The normal to $ax+by+cz+d=0$ has direction ratios $a,b,c$ , i.e. $1,2,-3$ .

For a plane written as $ax+by+cz+d=0$ , the normal vector is $a\vec{i}+b\vec{j}+c\vec{k}$ , so its direction ratios are the coe …

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CBSE 2023Set E1 markMCQ
$x-8y-9z=12$
(a) $x+8y+9z=12$
(b) $x-8y-9z=2023$
(c) $8x-y-9z=12$
(d) $x-9y-8z=12$
›Reveal solutionSolution

A plane parallel to $x-8y-9z=12$ keeps the coefficients $1,-8,-9$ ; only the constant changes, e.g. $x-8y-9z=2023$ .

Two planes are parallel iff their normal vectors are proportional, i.e. the coefficients of $x,y,z$ are the same (up to a common factor). Only the constant term may differ.

…

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CBSE 2023Set ANNUAL1 markMCQ
$x, y$
(a) $4x+6y+3z=12$
(b) $6x+4y+3z=12$
(c) $3x+4y+6z=12$
(d) $5x+4y+3z=0$
›Reveal solutionSolution

Use the intercept form of a plane, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ , then clear denominators.

With intercepts $a=2,\ b=3,\ c=4$ :

$\dfrac{x}{2}+\dfrac{y}{3}+\dfrac{z}{4}=1$

…

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CBSE 2023Set ANNUAL1 mark
$2x+y-z=5$
›Reveal solutionSolution

Rewrite the plane equation in intercept form $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ by dividing through so the RHS becomes 1.

$2x+y-z=5$

Divide both sides by $5$ :

$\dfrac{x}{5/2}+\dfrac{y}{5}+\dfrac{z}{-5}=1$

…

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CBSE 2022Set HE2191 mark
$lx + my + nz = d$
›Reveal solutionSolution

This is exactly the standard normal (or perpendicular) form of the equation of a plane.

The equation of a plane in normal form is $lx+my+nz=d$ , where $(l,m,n)$ are the direction cosines of the normal to the plane from the origin, and $d\ (\ge 0)$ is th …

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CBSE 2022Set HE2191 mark
$2x - y + 4z = 5$
›Reveal solutionSolution

Two planes are parallel if their normal vectors are proportional; check the ratio of coefficients.

Plane 1: $2x-y+4z=5$ , normal $(2,-1,4)$ .

Plane 2: $5x-2.5y+10z=6$ , normal $(5,-2.5,10)$ .

…

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Q.If the foot of the perpendicular drawn from the origin to the plane is $(4, -2, -3)$ , then find the equation of the plane in vector and Cartesian form. OR If $f(1) = 10$ and $f'(x) \geq 2$ for $1 \leq x \leq 4$ , how small can $f(4)$ possibly be?

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

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