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If the equation of the plane containing the line of intersection of the plane $x + y + z - 6 = 0$ , $2 x + 3 y + 4 z - 5 = 0$ and passing through $(1, 1, 1)$ is $P x + 13 y + 16 z - 39 = 0$ , then find $P$ .

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Important Questions on Three Dimensional Geometry

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

Perpendiculars are drawn from points on the line

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

to the plane

x + y + z = 3

. The feet of perpendiculars lie on the line

MEDIUM

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

If the points

(1, 1, λ) & (- 3, 0, 1)

, are equidistant from the plane,

3 x + 4 y - 12 z + 13 = 0,

then

λ

satisfies the equation:

EASY

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The equation of the plane through

(- 1, 1, 2),

whose normal makes equal acute angle with co-ordinate axes is

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

Equation of the plane which passes through the point of intersection of lines

\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}

and

\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}

and has the largest distance from the origin is:

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The equation of the plane containing the straight line

\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

and perpendicular to the plane containing the straight lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2}

and

\frac{x}{4} = \frac{y}{2} = \frac{z}{3}

is:

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The coordinates of the foot of the perpendicular from the point

(1, - 2, 1)

on the plane containing the lines

\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}

and

\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7},

is:

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The distance of the point

(1, 3, - 7)

from the plane passing through the point

(1, - 1, - 1)

, having normal perpendicular to both the lines

\frac{x - 1}{1} = \frac{y + 2}{- 2} = \frac{z - 4}{3}

and

\frac{x - 2}{2} = \frac{y + 1}{- 1} = \frac{z + 7}{- 1}

, is:

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

R^{3}

, consider the planes

P_{1} : y = 0

and

P_{2} : x + z = 1 .

Let

P_{3}

, be a plane, different from

P_{1}

and

P_{2}

, which passes through the intersection of

P_{1}

and

P_{2}

. If the distance of the point

(0, 1, 0)

, from

P_{3}

1

and the distance of a point

(α, β, γ)

, from

P_{3}

2

, then which of the following relations is(are) true ?

MEDIUM

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

If an angle between the line,

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

and the plane,

x - 2 y - k z = 3

\cos^{- 1} (\frac{2 \sqrt{2}}{3})

, then a value of

k

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

Let

P_{1} : 2 x + y - z = 3 a n d P_{2} : x + 2 y + z = 2

be two planes. Then, which of the following statement(s) is (are) TRUE?

EASY

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

If the angle between the line

2 (x + 1) = y = z + 4

and the plane

2 x - y + \sqrt{λ} z + 4 = 0

\frac{π}{6}

, then the value of

λ

MEDIUM

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The plane passing through the point

(4, - 1, 2)

and parallel to the lines

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

and

\frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3}

also passes through the point

MEDIUM

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The number of distinct real values of

λ

, for which the lines

\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{λ^{2}}

and

\frac{x - 3}{1} = \frac{y - 2}{λ^{2}} = \frac{z - 1}{2}

, are coplanar is

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

Two lines

L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}

and

L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}

are coplanar. Then

α

can take value(s)

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The distance of the point

(1, - 2, 4)

from the plane passing through the point

(1, 2, 2)

and perpendicular to the planes

x - y + 2 z = 3

and

2 x - 2 y + z + 12 = 0,

is :

EASY

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The planes

2 x - y + 4 z = 5

and

5 x - 2.5 y + 10 z = 6

are

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The equation of the plane passing through the point

(1,1, 1)

and perpendicular to the planes

2 x + y - 2 z = 5

and

3 x - 6 y - 2 z = 7

EASY

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

If the lines

\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}

and

\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}

are coplanar, then

k

can have

HARD

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}

, is

MEDIUM

Mathematics>Vectors and 3D Geometry>Three Dimensional Geometry>Family of Planes

The equation of the plane containing the line of intersection of

2 x - 5 y + z = 3; x + y + 4 z = 5,

and parallel to the plane,

x + 3 y + 6 z = 1,

If the equation of the plane containing the line of intersection of the plane x+y+z-6=0, 2x+3y+4z-5=0 and passing through (1, 1, 1) is Px+13y+16z-39=0, then find P.

Important Questions on Three Dimensional Geometry

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If the equation of the plane containing the line of intersection of the plane $x + y + z - 6 = 0$ , $2 x + 3 y + 4 z - 5 = 0$ and passing through $(1, 1, 1)$ is $P x + 13 y + 16 z - 39 = 0$ , then find $P$ .