QBManager

Three Dimensional Geometry

121275 In three dimensional space $x^{2} - 5 x + 6 = 0$ represents

1 two points

2 two parallel planes

3 two parallel lines

4 a pair of non parallel lines

Explanation:

B We have,
$x^{2} - 5 x + 6 = 0$
$(x - 3) (x - 2) = 0$
$x = 3, 2$
In 3D, these equation represents two parallel plane.

AMU-2015

Three Dimensional Geometry

121276 If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3$ , $4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + c z + 6 = 0$ . Then $a + b + c$ is equal to

1 14

2 12

3 13

4 15

Explanation:

A Given, the plane
$P_{1} : 2 x - y + z = 3$
$P_{2} : 4 x - 3 y + 5 z + 9 = 0$
Equation of plane formed by intersection of $P_{1}$ and $P_{2}$ are
$P : (2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$
$(2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z - (3 - 9 λ) = 0$
Now, this plane is parallel to the given line. Hence, dot product of normal of plane to line is zero.
$∴ (2 + 4 λ) (- 2) - (1 + 3 λ) 4 + (1 + 5 λ) 5 = 0$
$- 4 - 8 λ - 4 - 12 λ + 5 + 25 λ = 0$
$5 λ = 3$
$λ = \frac{3}{5}$
Then, equation of plane will be.
$(2 + 4 \times \frac{3}{5}) x - (1 + 3 \times \frac{3}{5}) y + (1 + \frac{5 \times 3}{5}) z - (3 - 9 \times \frac{3}{5}) = 0$
$\frac{22}{5} x - \frac{14}{5} y + 4 z + \frac{12}{5} = 0$
$Or 11 x - 7 y + 10 z + 6 = 0$
$Here,$
$a = 11, b = - 7 & c = 10$
$So, a + b + c = 11 - 7 + 10$
$a + b + c = 14$
Here,
So, $a + b + c = 11 - 7 + 10$

JEE Main-06.04.2023

Three Dimensional Geometry

121277 A plane $x$ passes through the point $(1, 1, 1)$ . If $b, c$ , a are the direction ratios of a normal to the plane, where $a, b, c (a < b < c)$ are the factors of 2001,then the equation of the plane is

1

29 x + 31 y + 3 z = 63

2

23 x + 29 y - 29 z = 23

3

23 x + 29 y + 3 z = 55

4

31 x + 37 y + 3 z = 71

Explanation:

C Given point, $A (1, 1, 1)$
The factor of 2001are 3, 23 and 29.
So, we can write as
$a = 3, b = 23, and c = 29$
Direction ratio of normal to plane is $(23, 29, 3)$
Plane passing through $(1, 1, 1)$ , here
$23 x + 29 y + 3 z = 55$

AP EAMCET-2002

Three Dimensional Geometry

121279 If the plane $56 x + 4 y + 9 z = 2016$ meets the coordinated axes in $A, B, C$ then the centroid of the triangle $ABC$ is

1

(12, 168, 224)

2

(12, 168, 112)

3

(12, 168, \frac{224}{3})

4

(12, - 168, \frac{224}{3})

Explanation:

C Given,
Plane : $56 x + 4 y + 9 z = 2016$
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$
So,
$(A, B, C) = (36, 504, 224)$
Now, centroid of $ABC$ is $= (\frac{A}{3}, \frac{B}{3}, \frac{C}{3})$
$= (\frac{36}{3}, \frac{504}{3}, \frac{224}{3})$
$= (12, 168, \frac{224}{3})$

AP EAMCET-2016

Three Dimensional Geometry

121280 The plane $3 x + 4 y + 6 z + 7 = 0$ is rotated about the line $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$ unit the plane passes through origin. The equation of the plane in the new position is

1

4 x - 5 y - 2 z = 0

2

x + 2 y + 4 z = 0

3

6 x + 3 y - 4 z = 0

4

x + y + z = 0

Explanation:

A Given,
Plane: $3 x + 4 y + 6 z + 7 = 0$
Line: $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$
Equation of required plane passing through this line and origin is
$(\vec{r} - 0) \cdot [(\hat{i} + 2 \hat{j} - 3 \hat{k}) \times (2 \hat{i} - 3 \hat{j} + 3 \hat{k})] = 0$
$\vec{r} [(2 - 9) \hat{i} - (1 + 6) \hat{j} + (- 3 - 4) \hat{k}] = 0$
$\vec{r} \cdot (- 7 \hat{i} - 7 \hat{j} - 7 \hat{k}) = 0$
$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$ or

AP EAMCET-20.04.2019

Three Dimensional Geometry

121275 In three dimensional space $x^{2} - 5 x + 6 = 0$ represents

1 two points

2 two parallel planes

3 two parallel lines

4 a pair of non parallel lines

Explanation:

B We have,
$x^{2} - 5 x + 6 = 0$
$(x - 3) (x - 2) = 0$
$x = 3, 2$
In 3D, these equation represents two parallel plane.

AMU-2015

Three Dimensional Geometry

121276 If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3$ , $4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + c z + 6 = 0$ . Then $a + b + c$ is equal to

1 14

2 12

3 13

4 15

Explanation:

A Given, the plane
$P_{1} : 2 x - y + z = 3$
$P_{2} : 4 x - 3 y + 5 z + 9 = 0$
Equation of plane formed by intersection of $P_{1}$ and $P_{2}$ are
$P : (2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$
$(2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z - (3 - 9 λ) = 0$
Now, this plane is parallel to the given line. Hence, dot product of normal of plane to line is zero.
$∴ (2 + 4 λ) (- 2) - (1 + 3 λ) 4 + (1 + 5 λ) 5 = 0$
$- 4 - 8 λ - 4 - 12 λ + 5 + 25 λ = 0$
$5 λ = 3$
$λ = \frac{3}{5}$
Then, equation of plane will be.
$(2 + 4 \times \frac{3}{5}) x - (1 + 3 \times \frac{3}{5}) y + (1 + \frac{5 \times 3}{5}) z - (3 - 9 \times \frac{3}{5}) = 0$
$\frac{22}{5} x - \frac{14}{5} y + 4 z + \frac{12}{5} = 0$
$Or 11 x - 7 y + 10 z + 6 = 0$
$Here,$
$a = 11, b = - 7 & c = 10$
$So, a + b + c = 11 - 7 + 10$
$a + b + c = 14$
Here,
So, $a + b + c = 11 - 7 + 10$

JEE Main-06.04.2023

Three Dimensional Geometry

121277 A plane $x$ passes through the point $(1, 1, 1)$ . If $b, c$ , a are the direction ratios of a normal to the plane, where $a, b, c (a < b < c)$ are the factors of 2001,then the equation of the plane is

1

29 x + 31 y + 3 z = 63

2

23 x + 29 y - 29 z = 23

3

23 x + 29 y + 3 z = 55

4

31 x + 37 y + 3 z = 71

Explanation:

C Given point, $A (1, 1, 1)$
The factor of 2001are 3, 23 and 29.
So, we can write as
$a = 3, b = 23, and c = 29$
Direction ratio of normal to plane is $(23, 29, 3)$
Plane passing through $(1, 1, 1)$ , here
$23 x + 29 y + 3 z = 55$

AP EAMCET-2002

Three Dimensional Geometry

121279 If the plane $56 x + 4 y + 9 z = 2016$ meets the coordinated axes in $A, B, C$ then the centroid of the triangle $ABC$ is

1

(12, 168, 224)

2

(12, 168, 112)

3

(12, 168, \frac{224}{3})

4

(12, - 168, \frac{224}{3})

Explanation:

C Given,
Plane : $56 x + 4 y + 9 z = 2016$
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$
So,
$(A, B, C) = (36, 504, 224)$
Now, centroid of $ABC$ is $= (\frac{A}{3}, \frac{B}{3}, \frac{C}{3})$
$= (\frac{36}{3}, \frac{504}{3}, \frac{224}{3})$
$= (12, 168, \frac{224}{3})$

AP EAMCET-2016

Three Dimensional Geometry

121280 The plane $3 x + 4 y + 6 z + 7 = 0$ is rotated about the line $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$ unit the plane passes through origin. The equation of the plane in the new position is

1

4 x - 5 y - 2 z = 0

2

x + 2 y + 4 z = 0

3

6 x + 3 y - 4 z = 0

4

x + y + z = 0

Explanation:

A Given,
Plane: $3 x + 4 y + 6 z + 7 = 0$
Line: $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$
Equation of required plane passing through this line and origin is
$(\vec{r} - 0) \cdot [(\hat{i} + 2 \hat{j} - 3 \hat{k}) \times (2 \hat{i} - 3 \hat{j} + 3 \hat{k})] = 0$
$\vec{r} [(2 - 9) \hat{i} - (1 + 6) \hat{j} + (- 3 - 4) \hat{k}] = 0$
$\vec{r} \cdot (- 7 \hat{i} - 7 \hat{j} - 7 \hat{k}) = 0$
$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$ or

AP EAMCET-20.04.2019

Three Dimensional Geometry

121275 In three dimensional space $x^{2} - 5 x + 6 = 0$ represents

1 two points

2 two parallel planes

3 two parallel lines

4 a pair of non parallel lines

Explanation:

B We have,
$x^{2} - 5 x + 6 = 0$
$(x - 3) (x - 2) = 0$
$x = 3, 2$
In 3D, these equation represents two parallel plane.

AMU-2015

Three Dimensional Geometry

121276 If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3$ , $4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + c z + 6 = 0$ . Then $a + b + c$ is equal to

1 14

2 12

3 13

4 15

Explanation:

A Given, the plane
$P_{1} : 2 x - y + z = 3$
$P_{2} : 4 x - 3 y + 5 z + 9 = 0$
Equation of plane formed by intersection of $P_{1}$ and $P_{2}$ are
$P : (2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$
$(2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z - (3 - 9 λ) = 0$
Now, this plane is parallel to the given line. Hence, dot product of normal of plane to line is zero.
$∴ (2 + 4 λ) (- 2) - (1 + 3 λ) 4 + (1 + 5 λ) 5 = 0$
$- 4 - 8 λ - 4 - 12 λ + 5 + 25 λ = 0$
$5 λ = 3$
$λ = \frac{3}{5}$
Then, equation of plane will be.
$(2 + 4 \times \frac{3}{5}) x - (1 + 3 \times \frac{3}{5}) y + (1 + \frac{5 \times 3}{5}) z - (3 - 9 \times \frac{3}{5}) = 0$
$\frac{22}{5} x - \frac{14}{5} y + 4 z + \frac{12}{5} = 0$
$Or 11 x - 7 y + 10 z + 6 = 0$
$Here,$
$a = 11, b = - 7 & c = 10$
$So, a + b + c = 11 - 7 + 10$
$a + b + c = 14$
Here,
So, $a + b + c = 11 - 7 + 10$

JEE Main-06.04.2023

Three Dimensional Geometry

121277 A plane $x$ passes through the point $(1, 1, 1)$ . If $b, c$ , a are the direction ratios of a normal to the plane, where $a, b, c (a < b < c)$ are the factors of 2001,then the equation of the plane is

1

29 x + 31 y + 3 z = 63

2

23 x + 29 y - 29 z = 23

3

23 x + 29 y + 3 z = 55

4

31 x + 37 y + 3 z = 71

Explanation:

C Given point, $A (1, 1, 1)$
The factor of 2001are 3, 23 and 29.
So, we can write as
$a = 3, b = 23, and c = 29$
Direction ratio of normal to plane is $(23, 29, 3)$
Plane passing through $(1, 1, 1)$ , here
$23 x + 29 y + 3 z = 55$

AP EAMCET-2002

Three Dimensional Geometry

121279 If the plane $56 x + 4 y + 9 z = 2016$ meets the coordinated axes in $A, B, C$ then the centroid of the triangle $ABC$ is

1

(12, 168, 224)

2

(12, 168, 112)

3

(12, 168, \frac{224}{3})

4

(12, - 168, \frac{224}{3})

Explanation:

C Given,
Plane : $56 x + 4 y + 9 z = 2016$
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$
So,
$(A, B, C) = (36, 504, 224)$
Now, centroid of $ABC$ is $= (\frac{A}{3}, \frac{B}{3}, \frac{C}{3})$
$= (\frac{36}{3}, \frac{504}{3}, \frac{224}{3})$
$= (12, 168, \frac{224}{3})$

AP EAMCET-2016

Three Dimensional Geometry

121280 The plane $3 x + 4 y + 6 z + 7 = 0$ is rotated about the line $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$ unit the plane passes through origin. The equation of the plane in the new position is

1

4 x - 5 y - 2 z = 0

2

x + 2 y + 4 z = 0

3

6 x + 3 y - 4 z = 0

4

x + y + z = 0

Explanation:

A Given,
Plane: $3 x + 4 y + 6 z + 7 = 0$
Line: $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$
Equation of required plane passing through this line and origin is
$(\vec{r} - 0) \cdot [(\hat{i} + 2 \hat{j} - 3 \hat{k}) \times (2 \hat{i} - 3 \hat{j} + 3 \hat{k})] = 0$
$\vec{r} [(2 - 9) \hat{i} - (1 + 6) \hat{j} + (- 3 - 4) \hat{k}] = 0$
$\vec{r} \cdot (- 7 \hat{i} - 7 \hat{j} - 7 \hat{k}) = 0$
$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$ or

AP EAMCET-20.04.2019

Three Dimensional Geometry

121275 In three dimensional space $x^{2} - 5 x + 6 = 0$ represents

1 two points

2 two parallel planes

3 two parallel lines

4 a pair of non parallel lines

Explanation:

B We have,
$x^{2} - 5 x + 6 = 0$
$(x - 3) (x - 2) = 0$
$x = 3, 2$
In 3D, these equation represents two parallel plane.

AMU-2015

Three Dimensional Geometry

121276 If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3$ , $4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + c z + 6 = 0$ . Then $a + b + c$ is equal to

1 14

2 12

3 13

4 15

Explanation:

A Given, the plane
$P_{1} : 2 x - y + z = 3$
$P_{2} : 4 x - 3 y + 5 z + 9 = 0$
Equation of plane formed by intersection of $P_{1}$ and $P_{2}$ are
$P : (2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$
$(2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z - (3 - 9 λ) = 0$
Now, this plane is parallel to the given line. Hence, dot product of normal of plane to line is zero.
$∴ (2 + 4 λ) (- 2) - (1 + 3 λ) 4 + (1 + 5 λ) 5 = 0$
$- 4 - 8 λ - 4 - 12 λ + 5 + 25 λ = 0$
$5 λ = 3$
$λ = \frac{3}{5}$
Then, equation of plane will be.
$(2 + 4 \times \frac{3}{5}) x - (1 + 3 \times \frac{3}{5}) y + (1 + \frac{5 \times 3}{5}) z - (3 - 9 \times \frac{3}{5}) = 0$
$\frac{22}{5} x - \frac{14}{5} y + 4 z + \frac{12}{5} = 0$
$Or 11 x - 7 y + 10 z + 6 = 0$
$Here,$
$a = 11, b = - 7 & c = 10$
$So, a + b + c = 11 - 7 + 10$
$a + b + c = 14$
Here,
So, $a + b + c = 11 - 7 + 10$

JEE Main-06.04.2023

Three Dimensional Geometry

121277 A plane $x$ passes through the point $(1, 1, 1)$ . If $b, c$ , a are the direction ratios of a normal to the plane, where $a, b, c (a < b < c)$ are the factors of 2001,then the equation of the plane is

1

29 x + 31 y + 3 z = 63

2

23 x + 29 y - 29 z = 23

3

23 x + 29 y + 3 z = 55

4

31 x + 37 y + 3 z = 71

Explanation:

C Given point, $A (1, 1, 1)$
The factor of 2001are 3, 23 and 29.
So, we can write as
$a = 3, b = 23, and c = 29$
Direction ratio of normal to plane is $(23, 29, 3)$
Plane passing through $(1, 1, 1)$ , here
$23 x + 29 y + 3 z = 55$

AP EAMCET-2002

Three Dimensional Geometry

121279 If the plane $56 x + 4 y + 9 z = 2016$ meets the coordinated axes in $A, B, C$ then the centroid of the triangle $ABC$ is

1

(12, 168, 224)

2

(12, 168, 112)

3

(12, 168, \frac{224}{3})

4

(12, - 168, \frac{224}{3})

Explanation:

C Given,
Plane : $56 x + 4 y + 9 z = 2016$
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$
So,
$(A, B, C) = (36, 504, 224)$
Now, centroid of $ABC$ is $= (\frac{A}{3}, \frac{B}{3}, \frac{C}{3})$
$= (\frac{36}{3}, \frac{504}{3}, \frac{224}{3})$
$= (12, 168, \frac{224}{3})$

AP EAMCET-2016

Three Dimensional Geometry

121280 The plane $3 x + 4 y + 6 z + 7 = 0$ is rotated about the line $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$ unit the plane passes through origin. The equation of the plane in the new position is

1

4 x - 5 y - 2 z = 0

2

x + 2 y + 4 z = 0

3

6 x + 3 y - 4 z = 0

4

x + y + z = 0

Explanation:

A Given,
Plane: $3 x + 4 y + 6 z + 7 = 0$
Line: $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$
Equation of required plane passing through this line and origin is
$(\vec{r} - 0) \cdot [(\hat{i} + 2 \hat{j} - 3 \hat{k}) \times (2 \hat{i} - 3 \hat{j} + 3 \hat{k})] = 0$
$\vec{r} [(2 - 9) \hat{i} - (1 + 6) \hat{j} + (- 3 - 4) \hat{k}] = 0$
$\vec{r} \cdot (- 7 \hat{i} - 7 \hat{j} - 7 \hat{k}) = 0$
$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$ or

AP EAMCET-20.04.2019

Three Dimensional Geometry

121275 In three dimensional space $x^{2} - 5 x + 6 = 0$ represents

1 two points

2 two parallel planes

3 two parallel lines

4 a pair of non parallel lines

Explanation:

B We have,
$x^{2} - 5 x + 6 = 0$
$(x - 3) (x - 2) = 0$
$x = 3, 2$
In 3D, these equation represents two parallel plane.

AMU-2015

Three Dimensional Geometry

121276 If the equation of the plane passing through the line of intersection of the planes $2 x - y + z = 3$ , $4 x - 3 y + 5 z + 9 = 0$ and parallel to the line $\frac{x + 1}{- 2} = \frac{y + 3}{4} = \frac{z - 2}{5}$ is $a x + b y + c z + 6 = 0$ . Then $a + b + c$ is equal to

1 14

2 12

3 13

4 15

Explanation:

A Given, the plane
$P_{1} : 2 x - y + z = 3$
$P_{2} : 4 x - 3 y + 5 z + 9 = 0$
Equation of plane formed by intersection of $P_{1}$ and $P_{2}$ are
$P : (2 x - y + z - 3) + λ (4 x - 3 y + 5 z + 9) = 0$
$(2 + 4 λ) x - (1 + 3 λ) y + (1 + 5 λ) z - (3 - 9 λ) = 0$
Now, this plane is parallel to the given line. Hence, dot product of normal of plane to line is zero.
$∴ (2 + 4 λ) (- 2) - (1 + 3 λ) 4 + (1 + 5 λ) 5 = 0$
$- 4 - 8 λ - 4 - 12 λ + 5 + 25 λ = 0$
$5 λ = 3$
$λ = \frac{3}{5}$
Then, equation of plane will be.
$(2 + 4 \times \frac{3}{5}) x - (1 + 3 \times \frac{3}{5}) y + (1 + \frac{5 \times 3}{5}) z - (3 - 9 \times \frac{3}{5}) = 0$
$\frac{22}{5} x - \frac{14}{5} y + 4 z + \frac{12}{5} = 0$
$Or 11 x - 7 y + 10 z + 6 = 0$
$Here,$
$a = 11, b = - 7 & c = 10$
$So, a + b + c = 11 - 7 + 10$
$a + b + c = 14$
Here,
So, $a + b + c = 11 - 7 + 10$

JEE Main-06.04.2023

Three Dimensional Geometry

121277 A plane $x$ passes through the point $(1, 1, 1)$ . If $b, c$ , a are the direction ratios of a normal to the plane, where $a, b, c (a < b < c)$ are the factors of 2001,then the equation of the plane is

1

29 x + 31 y + 3 z = 63

2

23 x + 29 y - 29 z = 23

3

23 x + 29 y + 3 z = 55

4

31 x + 37 y + 3 z = 71

Explanation:

C Given point, $A (1, 1, 1)$
The factor of 2001are 3, 23 and 29.
So, we can write as
$a = 3, b = 23, and c = 29$
Direction ratio of normal to plane is $(23, 29, 3)$
Plane passing through $(1, 1, 1)$ , here
$23 x + 29 y + 3 z = 55$

AP EAMCET-2002

Three Dimensional Geometry

121279 If the plane $56 x + 4 y + 9 z = 2016$ meets the coordinated axes in $A, B, C$ then the centroid of the triangle $ABC$ is

1

(12, 168, 224)

2

(12, 168, 112)

3

(12, 168, \frac{224}{3})

4

(12, - 168, \frac{224}{3})

Explanation:

C Given,
Plane : $56 x + 4 y + 9 z = 2016$
$\frac{x}{36} + \frac{y}{504} + \frac{z}{224} = 1$
So,
$(A, B, C) = (36, 504, 224)$
Now, centroid of $ABC$ is $= (\frac{A}{3}, \frac{B}{3}, \frac{C}{3})$
$= (\frac{36}{3}, \frac{504}{3}, \frac{224}{3})$
$= (12, 168, \frac{224}{3})$

AP EAMCET-2016

Three Dimensional Geometry

121280 The plane $3 x + 4 y + 6 z + 7 = 0$ is rotated about the line $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$ unit the plane passes through origin. The equation of the plane in the new position is

1

4 x - 5 y - 2 z = 0

2

x + 2 y + 4 z = 0

3

6 x + 3 y - 4 z = 0

4

x + y + z = 0

Explanation:

A Given,
Plane: $3 x + 4 y + 6 z + 7 = 0$
Line: $\vec{r} = (\hat{i} + 2 \hat{j} - 3 \hat{k}) + (2 \hat{i} - 3 \hat{j} + \hat{k})$
Equation of required plane passing through this line and origin is
$(\vec{r} - 0) \cdot [(\hat{i} + 2 \hat{j} - 3 \hat{k}) \times (2 \hat{i} - 3 \hat{j} + 3 \hat{k})] = 0$
$\vec{r} [(2 - 9) \hat{i} - (1 + 6) \hat{j} + (- 3 - 4) \hat{k}] = 0$
$\vec{r} \cdot (- 7 \hat{i} - 7 \hat{j} - 7 \hat{k}) = 0$
$\vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 0$ or

AP EAMCET-20.04.2019