Q.Show that the straight lines $\vec r=(5\hat i+7\hat j-3\hat k)+s(4\hat i+4\hat j-5\hat k)$ and $\vec r=(8\hat i+4\hat j+5\hat k)+t(7\hat i+\hat j+3\hat k)$ are coplanar. Find the vector equation of the plane in which they lie.

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.