Visual SolutionStandard

Question

The equation of the plane passing through the points

(1, 1, 0)

(1, 2, 1)

, and

(-2, 2, -1)

is:

(A)

2x + 3y - 3z = 5

(B)

2x - 3y + 3z = 5

(C)

2x + 3y + 3z = 5

(D)

x + 3y - 3z = 4

Solution Path

Cross product of AB=(0,1,1) and AC=(-3,1,-1) gives normal (2,3,-3). Plane: 2x+3y-3z=5.

01Question Setup

1/4

Find the equation of the plane through

A(1,1,0)

B(1,2,1)

C(-2,2,-1)

Plane through 3 points

02Direction Vectors

2/4

\vec{AB} = (0,1,1)

and

\vec{AC} = (-3,1,-1)

. Normal

= \vec{AB} \times \vec{AC}

\vec{n} = \vec{AB} \times \vec{AC}

03Cross ProductKEY INSIGHT

3/4

\vec{n} = (-2,-3,3)

, equivalently

(2,3,-3)

. Using point

A(1,1,0)

2(x-1)+3(y-1)-3z = 0

\vec{n} = (2, 3, -3)

04Final Answer

4/4

Expanding:

2x+3y-3z = 5

. Verified:

A

gives

5

B

gives

5

\boxed{2x+3y-3z=5}

Concepts from this question1 concepts unlocked

★

Equation of a Plane

STANDARD

A plane with normal direction (a, b, c) passing through (x1, y1, z1) has the equation a(x-x1) + b(y-y1) + c(z-z1) = 0, which simplifies to ax + by + cz + d = 0. In vector form: (r - a).n = 0 or r.n = d.

ax + by + cz + d = 0 \quad \text{or} \quad \vec{r} \cdot \hat{n} = p

The plane equation is central to 3D geometry. It appears in distance problems, angle calculations, intersection of planes, and image/foot of perpendicular questions. JEE tests multiple ways of forming the equation: normal form, three-point form, and intercept form.

Plane through three pointsDistance from point to planeAngle between planesIntersection of planes

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