Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z =

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Q. Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.

West Bengal WbchseWest Bengal HS (WBCHSE) Board 2022Subjective· 4mImportance★★★★★

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Write the family of planes through the line of intersection using one parameter $\lambda$ , then pick $\lambda$ so this family's normal is perpendicular to the given plane's normal.

Given planes $x+y+z=1$ and $2x+3y+4z=5$ , and we need the member of their intersection-family that is perpendicular to $x-y+z=0$ .

Family of planes through the line of intersection:

$(x+y+z-1) + \lambda(2x+3y+4z-5) = 0$

$(1+2\lambda)x + (1+3\lambda)y + (1+4\lambda)z - (1+5\lambda) = 0$

This plane has normal vector $\vec n_1 = (1+2\lambda,\ 1+3\lambda,\ 1+4\lambda)$ .

Perpendicularity condition. Two planes are perpendicular iff their normals are perpendicular, i.e. $\vec n_1\cdot \vec n_2 = 0$ , where $\vec n_2=(1,-1,1)$ is the normal of $x-y+z=0$ .

$(1+2\lambda)(1) + (1+3\lambda)(-1) + (1+4\lambda)(1) = 0$

$1+2\lambda -1-3\lambda +1+4\lambda = 0$ …

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

Showing the 12 most recent of 28 on this concept.

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 equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.

Concept understanding — Equation of a Plane

Equation of a Plane

Point + normal form

Normal (perpendicular) form

Intercept form …

Q.Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to the plane x - y + z = 0.

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