Q.Find the non-parametric form of vector equation, and Cartesian equation of the plane passing through the point $(2,3,6)$ and parallel to the straight lines $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-3}{1}$ and $\dfrac{x+3}{2}=\dfrac{y-3}{-5}=\dfrac{z+1}{-3}$.

Showing the 12 most recent of 28 on this concept.

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

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

   Multiply through by $6$ (LCM of $2,3,1$):

   $3x - 2y + 6z = 6$

   ✓Final answer

   $3x-2y+6z=6$ — option (b).

2. CBSE 2025Set E1 markMCQ
   Q.Equation of a plane parallel to the plane $9x - 8y + 7z = 10$ is

   (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 $x, y, z$, differing only in the constant term. The only such option is $9x - 8y + 7z = 5$.

   ✓Final answer

   (B) $9x - 8y + 7z = 5$.

3. CBSE 2024Set D1 markMCQ
   Q.The equation of the $xy$-plane is

   (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 $z = 0$. (Similarly $x=0$ is the $yz$-plane and $y=0$ is the $zx$-plane.)

   ✓Final answer

   (c) $z = 0$.

4. CBSE 2024Set D1 markMCQ
   Q.The equation of the plane parallel to the plane $3x - 5y + 4z = 11$ is

   (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

   $3x - 5y + 4z = k$

   for some constant $k \neq 11$. Among the options, only (A) preserves all three coefficients $(3, -5, 4)$; options (B) and (C) change $-5y$ to $+5y$, so their normals differ.

   ✓Final answer

   (A) $3x - 5y + 4z = 21$.

5. CBSE 2024Set ANNUAL1 markMCQ
   Q.The Cartesian equation of the plane r⃗ · (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 $d = 2$.

   Substituting: $(x\hat\imath + y\hat\jmath + z\hat k)\cdot(\hat\imath + \hat\jmath - \hat k) = 2 \;\Rightarrow\; x + y - z = 2$.

   ✓Final answer

   x + y − z = 2 — option (b).

6. CBSE 2023Set ANNUAL1 markMCQ
   Q.The Cartesian equation of the plane $\vec{r}\cdot(\hat{i}+\hat{j}-\hat{k}) = 2$ is-

   (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 $\vec r=x\hat i+y\hat j+z\hat k$:

   $(x\hat i+y\hat j+z\hat k)\cdot(\hat i+\hat j-\hat k)=2$

   $x(1)+y(1)+z(-1)=2$

   $x+y-z=2$.

   ✓Final answer

   Option (b) $x+y-z=2$

7. CBSE 2023Set E1 markMCQ
   Q.Direction ratios of the normal to the plane $x+2y-3z+15=0$ are

   (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 coefficients $a,b,c$.

   For $x+2y-3z+15=0$ these are $1,2,-3$.

   ✓Final answer

   (c) $1,2,-3$.

8. CBSE 2023Set E1 markMCQ
   Q.Equation of a plane parallel to the plane $x-8y-9z=12$ is

   (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.

   Among the options, $x-8y-9z=2023$ has exactly the coefficients $1,-8,-9$, so it is parallel to $x-8y-9z=12$.

   ✓Final answer

   (b) $x-8y-9z=2023$.

9. CBSE 2023Set ANNUAL1 markMCQ
   Q.The equation of the plane with intercepts of 2, 3 and 4 on the $x, y$ and $z$-axes respectively is:

   (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 $a=2,\ b=3,\ c=4$:

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

   Multiply through by the LCM $12$:

   $6x+4y+3z=12$

   ✓Final answer

   (b) $6x+4y+3z=12$.

10. CBSE 2023Set ANNUAL1 mark
    Q.Find the intercepts cut off by the plane $2x+y-z=5$ on co-ordinate axes.

    ›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 $5$:

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

    So the intercepts are $a=\dfrac{5}{2}$ on the $x$-axis, $b=5$ on the $y$-axis, and $c=-5$ on the $z$-axis.

    ✓Final answer

    $x$-intercept $=\frac{5}{2}$, $y$-intercept $=5$, $z$-intercept $=-5$.

11. CBSE 2022Set HE2191 mark
    Q.Write true or false: Equation of a plane in normal form is $lx + my + nz = d$.

    ›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 the perpendicular distance of the plane from the origin. This matches the given statement.

    ✓Final answer

    True.

12. CBSE 2022Set HE2191 mark
    Q.Write true or false: The planes $2x - y + 4z = 5$ and $5x - 2.5y + 10z = 6$ are parallel.

    ›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 $(5,-2.5,10)$.

    Check proportionality: $\dfrac{5}{2} = 2.5,\ \dfrac{-2.5}{-1}=2.5,\ \dfrac{10}{4}=2.5$. All three ratios are equal, so the normals are parallel (scalar multiples of each other), meaning the planes are parallel. (Since $6\ne 2.5\times 5=12.5$, they are two distinct parallel planes, not the same plane.)

    ✓Final answer

    True — the planes are parallel.