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Three Dimensional Geometry

121100 The line segment joining the points $A (2, 3, 4)$ and $B (- 3, 5, - 4)$ intersects $y z$ plane at the point.

1

(0, \frac{19}{5}, \frac{4}{5})

2

(0, 4, 5)

3

(9, \frac{14}{5}, 4)

4

(0, 0, 0)

Explanation:

A Given,
$A = (2, 3, 4)$
$B = (- 3, 5, - 4)$
Let $P$ divides $AB$ in ratio $λ : 1$ .
Coordinates of $P$ will be -
$P = (\frac{- 3 λ + 2}{λ + 1}, \frac{5 λ + 3}{λ + 1}, \frac{- 4 λ + 4}{λ + 1})$
$∵$ P lies on $yz =$ plane so $x$ -coordinate will be zero.
$\frac{- 3 λ + 2}{λ + 1} = 0$
$λ = \frac{2}{3}$
Put $t = 2 / 3$ in equation (i)
$P = (0, \frac{19}{5}, \frac{4}{5})$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121101 The line $x - 2 y + 4 z + 4 = 0, x + y + z - 8 = 0$ intersect the plane $x - y + 2 z + 1 = 0$ at the point

1

(- 2, 5, 1)

2

(2, - 5, 1)

3

(2, 5, - 1)

4

(2, 5, 1)

Explanation:

D We have equation of line,
$x - 2 y + 4 z + 4 = 0$
$x + y + z - 8 = 0$
Subtracting equation (i) and (ii) we get -
$y - z = 4$
Given, equation of plane -
$x - y + 2 z + 1 = 0$
According to question on solving equation (iii) and (iv) we get $x + z = 3$
Hence, $y = 5, x = 2$ and $z = 1$ ,

WB JEE-2022

Three Dimensional Geometry

121102 What will be the coordinates of foot of perpendicular from the point $(1, 1.5, 2)$ to the plane $2 x - 2 y + 4 z + 5 = 0$ ?

1

(1, 1 / 2, 1)

2

(0, 5 / 2, 0)

3

(0, 0, - 5 / 4) /

4

(6, 1, - 15 / 4)

Explanation:

B Let A $(1, 1.5, 2)$ be the given point and let $B$ be the foot of perpendicular drawn from $A$ to plane
$2 x - 2 y + 4 z + 5 = 0$
Direction ratio of normal to the plane (i) is $(2, - 2, 4)$ equation of line $A B$ is
$\frac{x - 1}{2} = \frac{y - 1.5}{- 2} = \frac{z - 2}{4} = λ$
$B (1, 1.5, 2)$
$x, y, 2)$
original image
Any point on line $AB$ is $B (2 λ + 1, 1.5 - 2 λ, 4 λ + 2)$ This point lies in plane (i)
$2 (2 λ + 1) - 2 (1.5 - 2 λ) + 4 (4 λ + 2) + 5 = 0$
$24 λ = - 12$
$λ = \frac{- 1}{2}$
So, coordinate of $B = (0, 2.5, 0)$
$= (0, 5 / 2, 0)$

J and K CET-2018

Three Dimensional Geometry

121103 Two lines in three-dimensional system are given as $(x - 1) / 2 = (y - 2) / 3 = (z - 3) / 4$ and $(x -$ 4) $/ 5 = (y - 1) / 2 = z$ . Their point of intersection is

1

(1, 2, 3)

2

(- 1, - 1, - 1)

3

(4, 1, 0)

4

(1, 1, 1)

Explanation:

C Given line are -
$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$
And $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1}$
Let the point on the line (i) is $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ According to question,
line (i) and (ii) intersect each other-
So, the above point satisfies the line (ii),
$\frac{2 λ + 1 - 4}{5} = \frac{3 λ + 2 - 1}{2} = \frac{4 λ + 3 - 0}{1}$
$\Rightarrow \frac{2 λ - 3}{5} = \frac{3 λ + 1}{2} = \frac{4 λ + 3}{1}$
$\Rightarrow 4 λ - 6 = 15 λ + 5$
$\Rightarrow 11 λ = - 11$
$\Rightarrow λ = - 1$
So, point of intersection is,
$(2 (- 1) + 1), (3 (- 1) + 2), (4 (- 1) + 3)$
$= (- 1, - 1, - 1)$

J and K CET-2018

Three Dimensional Geometry

121100 The line segment joining the points $A (2, 3, 4)$ and $B (- 3, 5, - 4)$ intersects $y z$ plane at the point.

1

(0, \frac{19}{5}, \frac{4}{5})

2

(0, 4, 5)

3

(9, \frac{14}{5}, 4)

4

(0, 0, 0)

Explanation:

A Given,
$A = (2, 3, 4)$
$B = (- 3, 5, - 4)$
Let $P$ divides $AB$ in ratio $λ : 1$ .
Coordinates of $P$ will be -
$P = (\frac{- 3 λ + 2}{λ + 1}, \frac{5 λ + 3}{λ + 1}, \frac{- 4 λ + 4}{λ + 1})$
$∵$ P lies on $yz =$ plane so $x$ -coordinate will be zero.
$\frac{- 3 λ + 2}{λ + 1} = 0$
$λ = \frac{2}{3}$
Put $t = 2 / 3$ in equation (i)
$P = (0, \frac{19}{5}, \frac{4}{5})$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121101 The line $x - 2 y + 4 z + 4 = 0, x + y + z - 8 = 0$ intersect the plane $x - y + 2 z + 1 = 0$ at the point

1

(- 2, 5, 1)

2

(2, - 5, 1)

3

(2, 5, - 1)

4

(2, 5, 1)

Explanation:

D We have equation of line,
$x - 2 y + 4 z + 4 = 0$
$x + y + z - 8 = 0$
Subtracting equation (i) and (ii) we get -
$y - z = 4$
Given, equation of plane -
$x - y + 2 z + 1 = 0$
According to question on solving equation (iii) and (iv) we get $x + z = 3$
Hence, $y = 5, x = 2$ and $z = 1$ ,

WB JEE-2022

Three Dimensional Geometry

121102 What will be the coordinates of foot of perpendicular from the point $(1, 1.5, 2)$ to the plane $2 x - 2 y + 4 z + 5 = 0$ ?

1

(1, 1 / 2, 1)

2

(0, 5 / 2, 0)

3

(0, 0, - 5 / 4) /

4

(6, 1, - 15 / 4)

Explanation:

B Let A $(1, 1.5, 2)$ be the given point and let $B$ be the foot of perpendicular drawn from $A$ to plane
$2 x - 2 y + 4 z + 5 = 0$
Direction ratio of normal to the plane (i) is $(2, - 2, 4)$ equation of line $A B$ is
$\frac{x - 1}{2} = \frac{y - 1.5}{- 2} = \frac{z - 2}{4} = λ$
$B (1, 1.5, 2)$
$x, y, 2)$
original image
Any point on line $AB$ is $B (2 λ + 1, 1.5 - 2 λ, 4 λ + 2)$ This point lies in plane (i)
$2 (2 λ + 1) - 2 (1.5 - 2 λ) + 4 (4 λ + 2) + 5 = 0$
$24 λ = - 12$
$λ = \frac{- 1}{2}$
So, coordinate of $B = (0, 2.5, 0)$
$= (0, 5 / 2, 0)$

J and K CET-2018

Three Dimensional Geometry

121103 Two lines in three-dimensional system are given as $(x - 1) / 2 = (y - 2) / 3 = (z - 3) / 4$ and $(x -$ 4) $/ 5 = (y - 1) / 2 = z$ . Their point of intersection is

1

(1, 2, 3)

2

(- 1, - 1, - 1)

3

(4, 1, 0)

4

(1, 1, 1)

Explanation:

C Given line are -
$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$
And $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1}$
Let the point on the line (i) is $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ According to question,
line (i) and (ii) intersect each other-
So, the above point satisfies the line (ii),
$\frac{2 λ + 1 - 4}{5} = \frac{3 λ + 2 - 1}{2} = \frac{4 λ + 3 - 0}{1}$
$\Rightarrow \frac{2 λ - 3}{5} = \frac{3 λ + 1}{2} = \frac{4 λ + 3}{1}$
$\Rightarrow 4 λ - 6 = 15 λ + 5$
$\Rightarrow 11 λ = - 11$
$\Rightarrow λ = - 1$
So, point of intersection is,
$(2 (- 1) + 1), (3 (- 1) + 2), (4 (- 1) + 3)$
$= (- 1, - 1, - 1)$

J and K CET-2018

Three Dimensional Geometry

121100 The line segment joining the points $A (2, 3, 4)$ and $B (- 3, 5, - 4)$ intersects $y z$ plane at the point.

1

(0, \frac{19}{5}, \frac{4}{5})

2

(0, 4, 5)

3

(9, \frac{14}{5}, 4)

4

(0, 0, 0)

Explanation:

A Given,
$A = (2, 3, 4)$
$B = (- 3, 5, - 4)$
Let $P$ divides $AB$ in ratio $λ : 1$ .
Coordinates of $P$ will be -
$P = (\frac{- 3 λ + 2}{λ + 1}, \frac{5 λ + 3}{λ + 1}, \frac{- 4 λ + 4}{λ + 1})$
$∵$ P lies on $yz =$ plane so $x$ -coordinate will be zero.
$\frac{- 3 λ + 2}{λ + 1} = 0$
$λ = \frac{2}{3}$
Put $t = 2 / 3$ in equation (i)
$P = (0, \frac{19}{5}, \frac{4}{5})$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121101 The line $x - 2 y + 4 z + 4 = 0, x + y + z - 8 = 0$ intersect the plane $x - y + 2 z + 1 = 0$ at the point

1

(- 2, 5, 1)

2

(2, - 5, 1)

3

(2, 5, - 1)

4

(2, 5, 1)

Explanation:

D We have equation of line,
$x - 2 y + 4 z + 4 = 0$
$x + y + z - 8 = 0$
Subtracting equation (i) and (ii) we get -
$y - z = 4$
Given, equation of plane -
$x - y + 2 z + 1 = 0$
According to question on solving equation (iii) and (iv) we get $x + z = 3$
Hence, $y = 5, x = 2$ and $z = 1$ ,

WB JEE-2022

Three Dimensional Geometry

121102 What will be the coordinates of foot of perpendicular from the point $(1, 1.5, 2)$ to the plane $2 x - 2 y + 4 z + 5 = 0$ ?

1

(1, 1 / 2, 1)

2

(0, 5 / 2, 0)

3

(0, 0, - 5 / 4) /

4

(6, 1, - 15 / 4)

Explanation:

B Let A $(1, 1.5, 2)$ be the given point and let $B$ be the foot of perpendicular drawn from $A$ to plane
$2 x - 2 y + 4 z + 5 = 0$
Direction ratio of normal to the plane (i) is $(2, - 2, 4)$ equation of line $A B$ is
$\frac{x - 1}{2} = \frac{y - 1.5}{- 2} = \frac{z - 2}{4} = λ$
$B (1, 1.5, 2)$
$x, y, 2)$
original image
Any point on line $AB$ is $B (2 λ + 1, 1.5 - 2 λ, 4 λ + 2)$ This point lies in plane (i)
$2 (2 λ + 1) - 2 (1.5 - 2 λ) + 4 (4 λ + 2) + 5 = 0$
$24 λ = - 12$
$λ = \frac{- 1}{2}$
So, coordinate of $B = (0, 2.5, 0)$
$= (0, 5 / 2, 0)$

J and K CET-2018

Three Dimensional Geometry

121103 Two lines in three-dimensional system are given as $(x - 1) / 2 = (y - 2) / 3 = (z - 3) / 4$ and $(x -$ 4) $/ 5 = (y - 1) / 2 = z$ . Their point of intersection is

1

(1, 2, 3)

2

(- 1, - 1, - 1)

3

(4, 1, 0)

4

(1, 1, 1)

Explanation:

C Given line are -
$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$
And $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1}$
Let the point on the line (i) is $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ According to question,
line (i) and (ii) intersect each other-
So, the above point satisfies the line (ii),
$\frac{2 λ + 1 - 4}{5} = \frac{3 λ + 2 - 1}{2} = \frac{4 λ + 3 - 0}{1}$
$\Rightarrow \frac{2 λ - 3}{5} = \frac{3 λ + 1}{2} = \frac{4 λ + 3}{1}$
$\Rightarrow 4 λ - 6 = 15 λ + 5$
$\Rightarrow 11 λ = - 11$
$\Rightarrow λ = - 1$
So, point of intersection is,
$(2 (- 1) + 1), (3 (- 1) + 2), (4 (- 1) + 3)$
$= (- 1, - 1, - 1)$

J and K CET-2018

Three Dimensional Geometry

121100 The line segment joining the points $A (2, 3, 4)$ and $B (- 3, 5, - 4)$ intersects $y z$ plane at the point.

1

(0, \frac{19}{5}, \frac{4}{5})

2

(0, 4, 5)

3

(9, \frac{14}{5}, 4)

4

(0, 0, 0)

Explanation:

A Given,
$A = (2, 3, 4)$
$B = (- 3, 5, - 4)$
Let $P$ divides $AB$ in ratio $λ : 1$ .
Coordinates of $P$ will be -
$P = (\frac{- 3 λ + 2}{λ + 1}, \frac{5 λ + 3}{λ + 1}, \frac{- 4 λ + 4}{λ + 1})$
$∵$ P lies on $yz =$ plane so $x$ -coordinate will be zero.
$\frac{- 3 λ + 2}{λ + 1} = 0$
$λ = \frac{2}{3}$
Put $t = 2 / 3$ in equation (i)
$P = (0, \frac{19}{5}, \frac{4}{5})$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121101 The line $x - 2 y + 4 z + 4 = 0, x + y + z - 8 = 0$ intersect the plane $x - y + 2 z + 1 = 0$ at the point

1

(- 2, 5, 1)

2

(2, - 5, 1)

3

(2, 5, - 1)

4

(2, 5, 1)

Explanation:

D We have equation of line,
$x - 2 y + 4 z + 4 = 0$
$x + y + z - 8 = 0$
Subtracting equation (i) and (ii) we get -
$y - z = 4$
Given, equation of plane -
$x - y + 2 z + 1 = 0$
According to question on solving equation (iii) and (iv) we get $x + z = 3$
Hence, $y = 5, x = 2$ and $z = 1$ ,

WB JEE-2022

Three Dimensional Geometry

121102 What will be the coordinates of foot of perpendicular from the point $(1, 1.5, 2)$ to the plane $2 x - 2 y + 4 z + 5 = 0$ ?

1

(1, 1 / 2, 1)

2

(0, 5 / 2, 0)

3

(0, 0, - 5 / 4) /

4

(6, 1, - 15 / 4)

Explanation:

B Let A $(1, 1.5, 2)$ be the given point and let $B$ be the foot of perpendicular drawn from $A$ to plane
$2 x - 2 y + 4 z + 5 = 0$
Direction ratio of normal to the plane (i) is $(2, - 2, 4)$ equation of line $A B$ is
$\frac{x - 1}{2} = \frac{y - 1.5}{- 2} = \frac{z - 2}{4} = λ$
$B (1, 1.5, 2)$
$x, y, 2)$
original image
Any point on line $AB$ is $B (2 λ + 1, 1.5 - 2 λ, 4 λ + 2)$ This point lies in plane (i)
$2 (2 λ + 1) - 2 (1.5 - 2 λ) + 4 (4 λ + 2) + 5 = 0$
$24 λ = - 12$
$λ = \frac{- 1}{2}$
So, coordinate of $B = (0, 2.5, 0)$
$= (0, 5 / 2, 0)$

J and K CET-2018

Three Dimensional Geometry

121103 Two lines in three-dimensional system are given as $(x - 1) / 2 = (y - 2) / 3 = (z - 3) / 4$ and $(x -$ 4) $/ 5 = (y - 1) / 2 = z$ . Their point of intersection is

1

(1, 2, 3)

2

(- 1, - 1, - 1)

3

(4, 1, 0)

4

(1, 1, 1)

Explanation:

C Given line are -
$\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$
And $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z - 0}{1}$
Let the point on the line (i) is $(2 λ + 1, 3 λ + 2, 4 λ + 3)$ According to question,
line (i) and (ii) intersect each other-
So, the above point satisfies the line (ii),
$\frac{2 λ + 1 - 4}{5} = \frac{3 λ + 2 - 1}{2} = \frac{4 λ + 3 - 0}{1}$
$\Rightarrow \frac{2 λ - 3}{5} = \frac{3 λ + 1}{2} = \frac{4 λ + 3}{1}$
$\Rightarrow 4 λ - 6 = 15 λ + 5$
$\Rightarrow 11 λ = - 11$
$\Rightarrow λ = - 1$
So, point of intersection is,
$(2 (- 1) + 1), (3 (- 1) + 2), (4 (- 1) + 3)$
$= (- 1, - 1, - 1)$

J and K CET-2018