Three Dimensional Geometry Questions

MathematicsThree Dimensional Geometry72 Questions Found

MathematicsEquation of a PlaneJEE Advanced 2005Moderate

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

Answer:

62 x + 29 y + 19 z - 105 = 0

MathematicsEquation of a PlaneJEE Advanced 2010Moderate

View in:English Hindi

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

Answer:6

MathematicsIntersection of a Line and a PlaneJEE Advanced 2003SEasy

View in:English Hindi

The value of $k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2 x - 4 y + z = 7$ , is

(A)

7

(B)

- 7

(C)

no real value

(D)

4

Answer:A

MathematicsShortest Distance Between Two Skew LinesJEE Advanced 2004SModerate

View in:English Hindi

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

(A)

3/2

(B)

9/2

(C)

- 2/9

(D)

- 3/2

Answer:B

MathematicsEquation of a PlaneJEE Advanced 2005SModerate

View in:English Hindi

A variable plane at a distance of the one unit from the origin cuts the coordinates axes at $A, B$ and $C$ . If the centroid $D (x, y, z)$ of triangle $A B C$ satisfies the relation $\frac{1}{x ^{2}} + \frac{1}{y ^{2}} + \frac{1}{z ^{2}} = k$ , then the value $k$ is

(A)

3

(B)

1

(C)

1/3

(D)

9

Answer:D

MathematicsEquation of a PlaneJEE Advanced 2006Moderate

View in:English Hindi

Let $A$ be vector parallel to line of intersection of planes $P_{1}$ and $P_{2}$ . Plane $P_{1}$ is parallel to the vectors $2 \hat{j} + 3 \hat{k}$ and $4 \hat{j} - 3 \hat{k}$ and that $P_{2}$ is parallel to $\hat{j} - \hat{k}$ and $3 \hat{i} + 3 \hat{j}$ , then the angle between vector $A$ and a given vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is

* Multiple Correct Options

(A)

\frac{π}{2}

(B)

\frac{π}{4}

(C)

\frac{π}{6}

(D)

\frac{3 π}{4}

Answer:B, D

MathematicsEquation of a PlaneJEE Advanced 2006Moderate

View in:English Hindi

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

(A)

0

(B)

1

(C)

2

(D)

22

Answer:D

MathematicsEquation of a Line in SpaceJEE Advanced 2007Moderate

View in:English Hindi

Consider the planes $3 x - 6 y - 2 z = 15$ and $2 x + y - 2 z = 5$ . STATEMENT-1 : The parametric equations of the line of intersection of the given planes are $x = 3 + 14 t, y = 1 + 2 t, z = 15 t$ . because STATEMENT-2 : The vector $14 \hat{i} + 2 \hat{j} + 15 \hat{k}$ is parallel to the line of intersection of given planes.

(A)

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1

(B)

Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1

(C)

Statement-1 is True, Statement-2 is False

(D)

Statement-1 is False, Statement-2 is True.

Answer:D

MathematicsEquation of a PlaneJEE Advanced 2008Moderate

View in:English Hindi

Consider three planes $P_{1} : x - y + z = 1$ , $P_{2} : x + y - z = 1$ , $P_{3} : x - 3 y + 3 z = 2$ . Let $L_{1}, L_{2}, L_{3}$ be the lines of intersection of the planes $P_{2}$ and $P_{3}$ , $P_{3}$ and $P_{1}$ , $P_{1}$ and $P_{2}$ , respectively. STATEMENT-1 : At least two of the lines $L_{1}, L_{2}$ and $L_{3}$ are non-parallel and STATEMENT-2 : The three planes do not have a common point.

(A)

STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

(B)

STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

(C)

STATEMENT - 1 is True, STATEMENT - 2 is False

(D)

STATEMENT - 1 is False, STATEMENT - 2 is True

Answer:D

MathematicsIntersection of a Line and a PlaneJEE Advanced 2009Moderate

View in:English Hindi

A line with positive direction cosines passes through the point $P (2, - 1, 2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x + y + z = 9$ at point $Q$ . The length of the line segment $P Q$ equals

(A)

(B)

2

(C)

3

(D)

Answer:C

...8

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the plane containing the line 2x−y+z−3=0,3x+y+z=5 and at a distance of 6​1​ from the point (2,1,−1).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of k such that 1x−4​=1y−2​=2z−k​ lies in the plane 2x−4y+z=7, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Find the equation of the plane containing the line 2x−y+z−3=0,3x+y+z=5 and at a distance of 6​1​ from the point (2,1,−1).

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the distance between the plane Ax−2y+z=d and the plane containing the lines 2x−1​=3y−2​=4z−3​ and 3x−2​=4y−3​=5z−4​ is 6​, then find ∣d∣.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of k such that 1x−4​=1y−2​=2z−k​ lies in the plane 2x−4y+z=7, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the lines 2x−1​=3y+1​=4z−1​ and 1x−3​=2y−k​=1z​ intersect, then the value of k is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A plane which is perpendicular to two planes 2x−2y+z=0 and x−y+2z=4, passes through (1,−2,1). The distance of the plane from the point (1,2,2) is

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The value of $k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2 x - 4 y + z = 7$ , is

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is

Find the equation of the plane containing the line $2 x - y + z - 3 = 0, 3 x + y + z = 5$ and at a distance of $\frac{1}{6}$ from the point $(2, 1, - 1)$ .

If the distance between the plane $A x - 2 y + z = d$ and the plane containing the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is $6$ , then find $∣ d ∣$ .

The value of $k$ such that $\frac{x - 4}{1} = \frac{y - 2}{1} = \frac{z - k}{2}$ lies in the plane $2 x - 4 y + z = 7$ , is

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ intersect, then the value of $k$ is

A plane which is perpendicular to two planes $2 x - 2 y + z = 0$ and $x - y + 2 z = 4$ , passes through $(1, - 2, 1)$ . The distance of the plane from the point $(1, 2, 2)$ is