Find the vector and Cartesian equations of the plane passing through the points with position vectors 3vec i +

Question 118 of 162

3\vec i + 4\vec j + 2\vec k

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2017Subjective· 10mImportance★★★★★

73% · 118/162 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Find two vectors lying in the plane from the three given points, take their cross product to get the normal vector, then write the plane equation through any one of the points.

Let $A=3\vec i+4\vec j+2\vec k$ , $B=2\vec i-2\vec j-\vec k$ , $C=7\vec i+0\vec j+\vec k$ be the given position vectors.
Vectors in the plane: $\vec{AB}=B-A=(2-3)\vec i+(-2-4)\vec j+(-1-2)\vec k = -\vec i-6\vec j-3\vec k$ $\vec{AC}=C-A=(7-3)\vec i+(0-4)\vec j+(1-2)\vec k = 4\vec i-4\vec j-\vec k$
Normal vector $\vec n = \vec{AB}\times\vec{AC} = \begin{vmatrix}\vec i&\vec j&\vec k\\-1&-6&-3\\4&-4&-1\end{vmatrix}$
$\vec n_i = (-6)(-1)-(-3)(-4) = 6-12=-6$ $\vec n_j = -[(-1)(-1)-(-3)(4)] = -[1+12]=-13$ $\vec n_k = (-1)(-4)-(-6)(4) = 4+24=28$ So $\vec{AB}\times\vec{AC} = -6\vec i-13\vec j+28\vec k$ ; taking the equivalent opposite normal $\vec n=6\vec i+13\vec j-28\vec k$ . …

Start your 14-day free trial to unlock the full solution →

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

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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

$3x - 5y + 4z = k$ …

Start your 14-day free trial to see the full solution →
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 $d = 2$ .

…

Start your 14-day free trial to see the full solution →
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 $\vec r=x\hat i+y\hat j+z\hat k$ :

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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.

…

Start your 14-day free trial to see the full solution →
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 $a=2,\ b=3,\ c=4$ :

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

…

Start your 14-day free trial to see the full solution →
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 $5$ :

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $(5,-2.5,10)$ .

…

Start your 14-day free trial to see the full solution →

←Previous questionQ19 — cos(A+B) = cos Acos B - sin Asin B : prove by vector method.→Next questionQ21 — vecr = sveci + tvecj is the equation of : (a) yoz plane (b) a straight…

Q.Find the vector and Cartesian equations of the plane passing through the points with position vectors $3\vec i + 4\vec j + 2\vec k$ , $2\vec i - 2\vec j - \vec k$ and $7\vec i + \vec k$ .

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 passing through the points with position vectors 3i⃗+4j⃗+2k⃗3\vec i + 4\vec j + 2\vec k3i+4j​+2k, 2i⃗−2j⃗−k⃗2\vec i - 2\vec j - \vec k2i−2j​−k and 7i⃗+k⃗7\vec i + \vec k7i+k.

🎓Unlock everything free for 14 days

Q.Find the vector and Cartesian equations of the plane passing through the points with position vectors $3\vec i + 4\vec j + 2\vec k$ , $2\vec i - 2\vec j - \vec k$ and $7\vec i + \vec k$ .

Unlock everything free for 14 days