A ( a → ) , B ( b → ) , C ( c → ) , D ( d → ) are four concyclic points, such that x a → + y b → + z c → + t d → = 0 → , x + y + z + t = 0 , where x , y , z , t are constants not all zero. If the chords A B and C D intersect at P , then

x y a → - b → 2 = z t c → - d → 2

x y a → + c → 2 = z t b → + d → 2

x t a → - c → 2 = y z b → - d → 2

x z b → + d → 2 = y t a → + c → 2

HARD

Earn 100

Find the length and the equation of the common perpendicular to the lines: $\frac{x + 3}{- 4} = \frac{y - 6}{3} = \frac{z}{2} and \frac{x + 2}{- 4} = \frac{y}{1} = \frac{z - 7}{1}$ .

Important Questions on Three Dimensional Geometry

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

A plane

P

meets the coordinate axes at

A, B

and

C

respectively. The centroid of

∆ ABC

is given to be

(1, 1, 2)

. Then the equation of the line through this centroid and perpendicular to the plane

P

is :

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

From a point

P (λ, λ, λ),

perpendiculars

P Q

and

P R

are drawn respectively on the lines

y = x, z = 1

and

y = - x, z = - 1

. If

P

is such that

∠ Q P R

is a right angle, then the possible value (s) of

λ

is (are)

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

Let

L_{1}

and

L_{2}

denotes the lines,

\vec{r} = \hat{i} + λ (- \hat{i} + 2 \hat{j} + 2 \hat{k}), λ \in R

and

\vec{r} = μ (2 \hat{i} - \hat{j} + 2 \hat{k}), μ \in R

, respectively. If

L_{3}

is a line which is perpendicular to both

L_{1}

and

L_{2}

and cuts both of them, then which of the following options describe(s)

L_{3}

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

A (1, 2, 3), B (3, 7, - 2), C (6, 7, 7)

and

D (- 1, 0, - 1)

are points in a plane, then the vector equation of the line passing through the centroids of

Δ A B D

and

Δ A C D

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

If the points

(- 1, 3, 2), (- 4, 3, - 2)

and

(5, l, m)

lie on a straight line then

l

and

m

are

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

A (\vec{a}), B (\vec{b}), C (\vec{c}), D (\vec{d})

are four concyclic points, such that

x \vec{a} + y \vec{b} + z \vec{c} + t \vec{d} = \vec{0}, x + y + z + t = 0

, where

x, y, z, t

are constants not all zero. If the chords

A B

and

C D

intersect at

P

, then

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

Three lines are given by

$\vec{r} = λ \hat{i}, λ \in R$ ,

$\vec{r} = μ (\hat{i} + \hat{j}), μ \in R$ and

$\vec{r} = ν (\hat{i} + \hat{j} + \hat{k}), ν \in R .$

Let the lines cut the plane $x + y + z = 1$ at the points $A, B$ and $C$ respectively. If the area of the triangle $A B C$ is $Δ$ then the value of ${(6 Δ)}^{2}$ equals_________

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

The equation of the line passing through the point

(2, 3, - 4)

and perpendicular to the

X O Z

plane is

(λ \in R)

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

The equation of line equally inclined to co-ordinate axes and passing through

(- 3, 2, - 5)

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

If the origin and the points

P (2, 3, 4), Q (1, 2, 3)

and

R (x, y, z)

are co-planar then

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

Let

3 \hat{i} + \hat{j} - \hat{k}

be the position vector of a point

B

. Let

A

be a point on the line which is passing through

B

and parallel to the vector

2 \hat{i} - \hat{j} + \hat{k}

. If

|\vec{B A}| = 18

, then the position vector of

A

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

The equation of the line passing through

(- 4, 3, 1),

parallel to the plane

x + 2 y - z - 5 = 0

and intersecting the line

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

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

If the Cartesian equation of the line is

x - 1 = 2 y + 3 = 3 - z,

then its vector equation is

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

If vector equation of the line

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

, is

\vec{r} = (2 \hat{i} + \frac{5}{2} \hat{j} - \hat{k}) + λ (2 \hat{i} - \frac{3}{2} \hat{j} + p \hat{k})

, then

p

is equal to

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

The parametric equations of the line passing through

A (3, 4, - 7), B (1, - 1, 6)

are

(λ \in R)

MEDIUM

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

A point

P

lies on a line through

Q (1, - 2, 3)

and is parallel to the line

\frac{x}{1} = \frac{y}{4} = \frac{z}{5} .

P

lies on the plane

2 x + 3 y - 4 z + 22 = 0,

then segment

P Q

equals

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

\vec{a}, \vec{b} & \vec{c}

are non-coplanar vectors, then the point of intersection of the line passing through the points

2 \vec{a} + 3 \vec{b} - \vec{c}, 3 \vec{a} + 4 \vec{b} - 2 \vec{c}

with the line joining the points

\vec{a} - 2 \vec{b} + 3 \vec{c}

\vec{a} - 6 \vec{b} + 6 \vec{c}

HARD

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

A line

l

passing through the origin is perpendicular to the lines

l_{1} : (3 + t) \hat{i} + (- 1 + 2 t) \hat{j} + (4 + 2 t) \hat{k}, - \infty < t < \infty

l_{2} : (3 + 2 s) \hat{i} + (3 + 2 s) \hat{j} + (2 + s) \hat{k}, - \infty < s < \infty

Then, the coordinate(s) of the point(s) on

l_{2}

at a distance of

\sqrt{17}

from the point of intersection of

l

and

l_{1}

is (are)

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

The equation of the line passing through the point

(5, 3, 2)

and perpendicular to the lines

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

and

\frac{x - 1}{2} = \frac{y - 1}{1} = \frac{z + 1}{0}

EASY

Mathematics>Coordinate Geometry>Three Dimensional Geometry>Equation of a Line in Space

The vector equation of the symmetrical form of equation of straight line

\frac{x - 5}{3} = \frac{y + 4}{7} = \frac{z - 6}{2}

Find the length and the equation of the common perpendicular to the lines: x+3-4=y-63=z2 and x+2-4=y1=z-71.

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Find the length and the equation of the common perpendicular to the lines: $\frac{x + 3}{- 4} = \frac{y - 6}{3} = \frac{z}{2} and \frac{x + 2}{- 4} = \frac{y}{1} = \frac{z - 7}{1}$ .