Find the vector and cartesian equations of the plane through the points (1, 2, 3) and (2, 3, 1) and perpendicu

Question 128 of 162

(1, 2, 3)

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★

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Find the plane's normal as the cross product of (the line joining the two given points) and (the normal of the given perpendicular plane), then fit it through one of the points.

Let the required plane have equation $a(x-1)+b(y-2)+c(z-3)=0$ (it passes through $(1,2,3)$ , with normal $\vec n=(a,b,c)$ ).
Since the plane also passes through $(2,3,1)$ : $a(2-1)+b(3-2)+c(1-3)=0\Rightarrow a+b-2c=0$ — so $\vec n$ is perpendicular to the direction $\vec d_1=(1,1,-2)$ joining the two points.
Since the required plane is perpendicular to $3x-2y+4z-5=0$ (normal $\vec d_2=(3,-2,4)$ ), the two planes' normals are themselves perpendicular: $\vec n\cdot\vec d_2=0\Rightarrow3a-2b+4c=0$ .
Since $\vec n$ is perpendicular to both $\vec d_1=(1,1,-2)$ and $\vec d_2=(3,-2,4)$ , take $\vec n=\vec d_1\times\vec d_2$ .
$\vec d_1\times\vec d_2=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&-2\\3&-2&4\end{vmatrix}=\vec i[(1)(4)-(-2)(-2)]-\vec j[(1)(4)-(-2)(3)]+\vec k[(1)(-2)-(1)(3)]$ . …

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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
(c) 3x + 2y + z = 3
(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.Find the vector and cartesian equations of the plane through the points $(1, 2, 3)$ and $(2, 3, 1)$ and perpendicular to the plane $3x - 2y + 4z - 5 = 0$ .

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.Find the vector and cartesian equations of the plane through the points (1,2,3)(1, 2, 3)(1,2,3) and (2,3,1)(2, 3, 1)(2,3,1) and perpendicular to the plane 3x−2y+4z−5=03x - 2y + 4z - 5 = 03x−2y+4z−5=0.

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Q.Find the vector and cartesian equations of the plane through the points $(1, 2, 3)$ and $(2, 3, 1)$ and perpendicular to the plane $3x - 2y + 4z - 5 = 0$ .

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