QBManager

Three Dimensional Geometry

121113 XY-plane divides the line joining the points $A (2, 3, - 5)$ and $B (- 1, - 2, - 3)$ in the ratio

1

2 : 1

internally

2

5 : 3

internally

3

3 : 2

externally

4

5 : 3

externally

Explanation:

D Let, the XY plane divides the line segment joining the given, points in the ratio $λ : 1$ a the point $P (x$ , y, z)
$∴ x = \frac{λ a + 2}{λ + 1}, y = \frac{- 2 λ + 3}{λ + 1}, z = \frac{- 3 λ - 5}{λ + 1}$
Since, $P (x, y, z)$ lie on the $XY$ plane, its $Z$ co-ordinate will be zero.
$∴ 0 = \frac{- 3 λ - 5}{λ + 1}$
$- 3 λ - 5 = 0$
$λ = \frac{- 5}{3}$ $∴$ Ratio, $5 : 3$ externally

Karnataka CET-2019

Three Dimensional Geometry

121114 A vector $\overset{―}{v}$ is equally inclined to the $X$ -axis, $Y$ -axis and $Z$ -axis respectively, its direction cosines are

1

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

2

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

3

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

or

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

4 None of these

Explanation:

C Given,
$α = β = γ$ i.e. $\cos α = \cos β = \cos γ$
Then. $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$∴ \cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\Rightarrow \cos^{2} α = \frac{1}{3}$
$\Rightarrow \cos α = \pm \frac{1}{\sqrt{3}}$
Hence, the $x$ -axis, $y$ -axis and $z$ -axis are respectively of the direction of cosines are -
$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$

MHT CET-2011

Three Dimensional Geometry

121115 What is the value of $n$ so that the angle between the lines having direction ratios $(1, 1, 1)$ and $(1$ , $- 1, n)$ is $60^{\circ}$ ?

1

\sqrt{3}

2

\sqrt{6}

3 3

4 None of these

Explanation:

B Given, $l_{1} = 1, m_{1} = 1, n_{1} = 1$ and $l_{2} = 1, m_{2} = - 1, n_{2} = n$ and $θ = 60^{\circ}$
We know that, If $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2})$ are the direction ratios then angle between the lines is
$\cos θ = \frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$
$∴ \cos 60^{0} = \frac{1 \times 1 + 1 \times (- 1) + 1 \times n}{\sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + n^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{n}{\sqrt{3} \sqrt{2 + n^{2}}}$
$\Rightarrow 3 (2 + n^{2}) = 4 n^{2}$
$\Rightarrow n^{2} = 6$
$\Rightarrow n = \pm \sqrt{6}$

BITSAT-2012

Three Dimensional Geometry

121116 The projection of line joining $(3, 4, 5)$ and $(4, 6$ , $3)$ on the line joining $(- 1, 2, 4)$ and $(1, 0, 5)$ is

1

\frac{4}{3}

2

\frac{2}{3}

3

\frac{8}{3}

4

\frac{1}{3}

Explanation:

A Given, $A = (3, 4, 5), P = (- 1, 2, 4), B = (4, 6, 3)$ and $Q = (1, 0, 5)$ $∴$ Direction ratio of line $AB$ are $(4 - 3), (6 - 4), (3 - 5)$ $= (1, 2, - 2)$
And direction ratio of line PQ are $(1 + 1), (0 - 2)$ , $(5 -$ $4) = (2, - 2, 1)$
$∴$ Direction cosine of line -
$PQ = \frac{2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{- 2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{1}{\sqrt{2^{2} + (- 2)^{2} + 1}}$
$= \frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}$
$∴$ Projection of line segment $AB$ on the line $PQ$ is
$| \frac{2}{3} (1) + (\frac{- 2}{3}) (2) + (\frac{1}{3}) (- 2) | = \frac{4}{3}$

BITSAT-2018

Three Dimensional Geometry

121117 The ratio in which the line joining $(2, 1, 5)$ and $(3, 4, 3)$ is divided by the plane $(x + y - z) = \frac{1}{2}$ is :

1

3 : 5

2

5 : 7

3

1 : 3

4

4 : 5

Explanation:

B Given, $x + y - z = \frac{1}{2}$ divides the line joining the points $A (2, 1, 5)$ and $B (3, 4, 3)$ at a point $C$ in the
Applying section formula at a point $C$ ,
$(\frac{3 k + 2}{k + 1}, \frac{4 k + 1}{k + 1}, \frac{3 k + 5}{k + 1})$
Point $C$ lies on the plane, Coordinates of $C$ must satisfy the equation of plane
So, $(\frac{3 k + 2}{k + 1}) + (\frac{4 k + 1}{k + 1}) - (\frac{3 k + 5}{k + 1}) = \frac{1}{2}$
$\Rightarrow 3 k + 2 + 4 k + 1 - 3 k - 5 = \frac{1}{2} (k + 1)$
$\Rightarrow 4 k - 2 = \frac{1}{2} (k + 1)$
$8 k - 4 = k + 1$
$\Rightarrow 7 k = 5 \Rightarrow k = \frac{5}{7}$ Hence, ratio is $5 : 7$

BITSAT-2017

Three Dimensional Geometry

121113 XY-plane divides the line joining the points $A (2, 3, - 5)$ and $B (- 1, - 2, - 3)$ in the ratio

1

2 : 1

internally

2

5 : 3

internally

3

3 : 2

externally

4

5 : 3

externally

Explanation:

D Let, the XY plane divides the line segment joining the given, points in the ratio $λ : 1$ a the point $P (x$ , y, z)
$∴ x = \frac{λ a + 2}{λ + 1}, y = \frac{- 2 λ + 3}{λ + 1}, z = \frac{- 3 λ - 5}{λ + 1}$
Since, $P (x, y, z)$ lie on the $XY$ plane, its $Z$ co-ordinate will be zero.
$∴ 0 = \frac{- 3 λ - 5}{λ + 1}$
$- 3 λ - 5 = 0$
$λ = \frac{- 5}{3}$ $∴$ Ratio, $5 : 3$ externally

Karnataka CET-2019

Three Dimensional Geometry

121114 A vector $\overset{―}{v}$ is equally inclined to the $X$ -axis, $Y$ -axis and $Z$ -axis respectively, its direction cosines are

1

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

2

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

3

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

or

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

4 None of these

Explanation:

C Given,
$α = β = γ$ i.e. $\cos α = \cos β = \cos γ$
Then. $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$∴ \cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\Rightarrow \cos^{2} α = \frac{1}{3}$
$\Rightarrow \cos α = \pm \frac{1}{\sqrt{3}}$
Hence, the $x$ -axis, $y$ -axis and $z$ -axis are respectively of the direction of cosines are -
$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$

MHT CET-2011

Three Dimensional Geometry

121115 What is the value of $n$ so that the angle between the lines having direction ratios $(1, 1, 1)$ and $(1$ , $- 1, n)$ is $60^{\circ}$ ?

1

\sqrt{3}

2

\sqrt{6}

3 3

4 None of these

Explanation:

B Given, $l_{1} = 1, m_{1} = 1, n_{1} = 1$ and $l_{2} = 1, m_{2} = - 1, n_{2} = n$ and $θ = 60^{\circ}$
We know that, If $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2})$ are the direction ratios then angle between the lines is
$\cos θ = \frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$
$∴ \cos 60^{0} = \frac{1 \times 1 + 1 \times (- 1) + 1 \times n}{\sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + n^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{n}{\sqrt{3} \sqrt{2 + n^{2}}}$
$\Rightarrow 3 (2 + n^{2}) = 4 n^{2}$
$\Rightarrow n^{2} = 6$
$\Rightarrow n = \pm \sqrt{6}$

BITSAT-2012

Three Dimensional Geometry

121116 The projection of line joining $(3, 4, 5)$ and $(4, 6$ , $3)$ on the line joining $(- 1, 2, 4)$ and $(1, 0, 5)$ is

1

\frac{4}{3}

2

\frac{2}{3}

3

\frac{8}{3}

4

\frac{1}{3}

Explanation:

A Given, $A = (3, 4, 5), P = (- 1, 2, 4), B = (4, 6, 3)$ and $Q = (1, 0, 5)$ $∴$ Direction ratio of line $AB$ are $(4 - 3), (6 - 4), (3 - 5)$ $= (1, 2, - 2)$
And direction ratio of line PQ are $(1 + 1), (0 - 2)$ , $(5 -$ $4) = (2, - 2, 1)$
$∴$ Direction cosine of line -
$PQ = \frac{2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{- 2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{1}{\sqrt{2^{2} + (- 2)^{2} + 1}}$
$= \frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}$
$∴$ Projection of line segment $AB$ on the line $PQ$ is
$| \frac{2}{3} (1) + (\frac{- 2}{3}) (2) + (\frac{1}{3}) (- 2) | = \frac{4}{3}$

BITSAT-2018

Three Dimensional Geometry

121117 The ratio in which the line joining $(2, 1, 5)$ and $(3, 4, 3)$ is divided by the plane $(x + y - z) = \frac{1}{2}$ is :

1

3 : 5

2

5 : 7

3

1 : 3

4

4 : 5

Explanation:

B Given, $x + y - z = \frac{1}{2}$ divides the line joining the points $A (2, 1, 5)$ and $B (3, 4, 3)$ at a point $C$ in the
Applying section formula at a point $C$ ,
$(\frac{3 k + 2}{k + 1}, \frac{4 k + 1}{k + 1}, \frac{3 k + 5}{k + 1})$
Point $C$ lies on the plane, Coordinates of $C$ must satisfy the equation of plane
So, $(\frac{3 k + 2}{k + 1}) + (\frac{4 k + 1}{k + 1}) - (\frac{3 k + 5}{k + 1}) = \frac{1}{2}$
$\Rightarrow 3 k + 2 + 4 k + 1 - 3 k - 5 = \frac{1}{2} (k + 1)$
$\Rightarrow 4 k - 2 = \frac{1}{2} (k + 1)$
$8 k - 4 = k + 1$
$\Rightarrow 7 k = 5 \Rightarrow k = \frac{5}{7}$ Hence, ratio is $5 : 7$

BITSAT-2017

Three Dimensional Geometry

121113 XY-plane divides the line joining the points $A (2, 3, - 5)$ and $B (- 1, - 2, - 3)$ in the ratio

1

2 : 1

internally

2

5 : 3

internally

3

3 : 2

externally

4

5 : 3

externally

Explanation:

D Let, the XY plane divides the line segment joining the given, points in the ratio $λ : 1$ a the point $P (x$ , y, z)
$∴ x = \frac{λ a + 2}{λ + 1}, y = \frac{- 2 λ + 3}{λ + 1}, z = \frac{- 3 λ - 5}{λ + 1}$
Since, $P (x, y, z)$ lie on the $XY$ plane, its $Z$ co-ordinate will be zero.
$∴ 0 = \frac{- 3 λ - 5}{λ + 1}$
$- 3 λ - 5 = 0$
$λ = \frac{- 5}{3}$ $∴$ Ratio, $5 : 3$ externally

Karnataka CET-2019

Three Dimensional Geometry

121114 A vector $\overset{―}{v}$ is equally inclined to the $X$ -axis, $Y$ -axis and $Z$ -axis respectively, its direction cosines are

1

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

2

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

3

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

or

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

4 None of these

Explanation:

C Given,
$α = β = γ$ i.e. $\cos α = \cos β = \cos γ$
Then. $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$∴ \cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\Rightarrow \cos^{2} α = \frac{1}{3}$
$\Rightarrow \cos α = \pm \frac{1}{\sqrt{3}}$
Hence, the $x$ -axis, $y$ -axis and $z$ -axis are respectively of the direction of cosines are -
$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$

MHT CET-2011

Three Dimensional Geometry

121115 What is the value of $n$ so that the angle between the lines having direction ratios $(1, 1, 1)$ and $(1$ , $- 1, n)$ is $60^{\circ}$ ?

1

\sqrt{3}

2

\sqrt{6}

3 3

4 None of these

Explanation:

B Given, $l_{1} = 1, m_{1} = 1, n_{1} = 1$ and $l_{2} = 1, m_{2} = - 1, n_{2} = n$ and $θ = 60^{\circ}$
We know that, If $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2})$ are the direction ratios then angle between the lines is
$\cos θ = \frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$
$∴ \cos 60^{0} = \frac{1 \times 1 + 1 \times (- 1) + 1 \times n}{\sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + n^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{n}{\sqrt{3} \sqrt{2 + n^{2}}}$
$\Rightarrow 3 (2 + n^{2}) = 4 n^{2}$
$\Rightarrow n^{2} = 6$
$\Rightarrow n = \pm \sqrt{6}$

BITSAT-2012

Three Dimensional Geometry

121116 The projection of line joining $(3, 4, 5)$ and $(4, 6$ , $3)$ on the line joining $(- 1, 2, 4)$ and $(1, 0, 5)$ is

1

\frac{4}{3}

2

\frac{2}{3}

3

\frac{8}{3}

4

\frac{1}{3}

Explanation:

A Given, $A = (3, 4, 5), P = (- 1, 2, 4), B = (4, 6, 3)$ and $Q = (1, 0, 5)$ $∴$ Direction ratio of line $AB$ are $(4 - 3), (6 - 4), (3 - 5)$ $= (1, 2, - 2)$
And direction ratio of line PQ are $(1 + 1), (0 - 2)$ , $(5 -$ $4) = (2, - 2, 1)$
$∴$ Direction cosine of line -
$PQ = \frac{2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{- 2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{1}{\sqrt{2^{2} + (- 2)^{2} + 1}}$
$= \frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}$
$∴$ Projection of line segment $AB$ on the line $PQ$ is
$| \frac{2}{3} (1) + (\frac{- 2}{3}) (2) + (\frac{1}{3}) (- 2) | = \frac{4}{3}$

BITSAT-2018

Three Dimensional Geometry

121117 The ratio in which the line joining $(2, 1, 5)$ and $(3, 4, 3)$ is divided by the plane $(x + y - z) = \frac{1}{2}$ is :

1

3 : 5

2

5 : 7

3

1 : 3

4

4 : 5

Explanation:

B Given, $x + y - z = \frac{1}{2}$ divides the line joining the points $A (2, 1, 5)$ and $B (3, 4, 3)$ at a point $C$ in the
Applying section formula at a point $C$ ,
$(\frac{3 k + 2}{k + 1}, \frac{4 k + 1}{k + 1}, \frac{3 k + 5}{k + 1})$
Point $C$ lies on the plane, Coordinates of $C$ must satisfy the equation of plane
So, $(\frac{3 k + 2}{k + 1}) + (\frac{4 k + 1}{k + 1}) - (\frac{3 k + 5}{k + 1}) = \frac{1}{2}$
$\Rightarrow 3 k + 2 + 4 k + 1 - 3 k - 5 = \frac{1}{2} (k + 1)$
$\Rightarrow 4 k - 2 = \frac{1}{2} (k + 1)$
$8 k - 4 = k + 1$
$\Rightarrow 7 k = 5 \Rightarrow k = \frac{5}{7}$ Hence, ratio is $5 : 7$

BITSAT-2017

Three Dimensional Geometry

121113 XY-plane divides the line joining the points $A (2, 3, - 5)$ and $B (- 1, - 2, - 3)$ in the ratio

1

2 : 1

internally

2

5 : 3

internally

3

3 : 2

externally

4

5 : 3

externally

Explanation:

D Let, the XY plane divides the line segment joining the given, points in the ratio $λ : 1$ a the point $P (x$ , y, z)
$∴ x = \frac{λ a + 2}{λ + 1}, y = \frac{- 2 λ + 3}{λ + 1}, z = \frac{- 3 λ - 5}{λ + 1}$
Since, $P (x, y, z)$ lie on the $XY$ plane, its $Z$ co-ordinate will be zero.
$∴ 0 = \frac{- 3 λ - 5}{λ + 1}$
$- 3 λ - 5 = 0$
$λ = \frac{- 5}{3}$ $∴$ Ratio, $5 : 3$ externally

Karnataka CET-2019

Three Dimensional Geometry

121114 A vector $\overset{―}{v}$ is equally inclined to the $X$ -axis, $Y$ -axis and $Z$ -axis respectively, its direction cosines are

1

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

2

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

3

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

or

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

4 None of these

Explanation:

C Given,
$α = β = γ$ i.e. $\cos α = \cos β = \cos γ$
Then. $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$∴ \cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\Rightarrow \cos^{2} α = \frac{1}{3}$
$\Rightarrow \cos α = \pm \frac{1}{\sqrt{3}}$
Hence, the $x$ -axis, $y$ -axis and $z$ -axis are respectively of the direction of cosines are -
$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$

MHT CET-2011

Three Dimensional Geometry

121115 What is the value of $n$ so that the angle between the lines having direction ratios $(1, 1, 1)$ and $(1$ , $- 1, n)$ is $60^{\circ}$ ?

1

\sqrt{3}

2

\sqrt{6}

3 3

4 None of these

Explanation:

B Given, $l_{1} = 1, m_{1} = 1, n_{1} = 1$ and $l_{2} = 1, m_{2} = - 1, n_{2} = n$ and $θ = 60^{\circ}$
We know that, If $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2})$ are the direction ratios then angle between the lines is
$\cos θ = \frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$
$∴ \cos 60^{0} = \frac{1 \times 1 + 1 \times (- 1) + 1 \times n}{\sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + n^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{n}{\sqrt{3} \sqrt{2 + n^{2}}}$
$\Rightarrow 3 (2 + n^{2}) = 4 n^{2}$
$\Rightarrow n^{2} = 6$
$\Rightarrow n = \pm \sqrt{6}$

BITSAT-2012

Three Dimensional Geometry

121116 The projection of line joining $(3, 4, 5)$ and $(4, 6$ , $3)$ on the line joining $(- 1, 2, 4)$ and $(1, 0, 5)$ is

1

\frac{4}{3}

2

\frac{2}{3}

3

\frac{8}{3}

4

\frac{1}{3}

Explanation:

A Given, $A = (3, 4, 5), P = (- 1, 2, 4), B = (4, 6, 3)$ and $Q = (1, 0, 5)$ $∴$ Direction ratio of line $AB$ are $(4 - 3), (6 - 4), (3 - 5)$ $= (1, 2, - 2)$
And direction ratio of line PQ are $(1 + 1), (0 - 2)$ , $(5 -$ $4) = (2, - 2, 1)$
$∴$ Direction cosine of line -
$PQ = \frac{2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{- 2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{1}{\sqrt{2^{2} + (- 2)^{2} + 1}}$
$= \frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}$
$∴$ Projection of line segment $AB$ on the line $PQ$ is
$| \frac{2}{3} (1) + (\frac{- 2}{3}) (2) + (\frac{1}{3}) (- 2) | = \frac{4}{3}$

BITSAT-2018

Three Dimensional Geometry

121117 The ratio in which the line joining $(2, 1, 5)$ and $(3, 4, 3)$ is divided by the plane $(x + y - z) = \frac{1}{2}$ is :

1

3 : 5

2

5 : 7

3

1 : 3

4

4 : 5

Explanation:

B Given, $x + y - z = \frac{1}{2}$ divides the line joining the points $A (2, 1, 5)$ and $B (3, 4, 3)$ at a point $C$ in the
Applying section formula at a point $C$ ,
$(\frac{3 k + 2}{k + 1}, \frac{4 k + 1}{k + 1}, \frac{3 k + 5}{k + 1})$
Point $C$ lies on the plane, Coordinates of $C$ must satisfy the equation of plane
So, $(\frac{3 k + 2}{k + 1}) + (\frac{4 k + 1}{k + 1}) - (\frac{3 k + 5}{k + 1}) = \frac{1}{2}$
$\Rightarrow 3 k + 2 + 4 k + 1 - 3 k - 5 = \frac{1}{2} (k + 1)$
$\Rightarrow 4 k - 2 = \frac{1}{2} (k + 1)$
$8 k - 4 = k + 1$
$\Rightarrow 7 k = 5 \Rightarrow k = \frac{5}{7}$ Hence, ratio is $5 : 7$

BITSAT-2017

Three Dimensional Geometry

121113 XY-plane divides the line joining the points $A (2, 3, - 5)$ and $B (- 1, - 2, - 3)$ in the ratio

1

2 : 1

internally

2

5 : 3

internally

3

3 : 2

externally

4

5 : 3

externally

Explanation:

D Let, the XY plane divides the line segment joining the given, points in the ratio $λ : 1$ a the point $P (x$ , y, z)
$∴ x = \frac{λ a + 2}{λ + 1}, y = \frac{- 2 λ + 3}{λ + 1}, z = \frac{- 3 λ - 5}{λ + 1}$
Since, $P (x, y, z)$ lie on the $XY$ plane, its $Z$ co-ordinate will be zero.
$∴ 0 = \frac{- 3 λ - 5}{λ + 1}$
$- 3 λ - 5 = 0$
$λ = \frac{- 5}{3}$ $∴$ Ratio, $5 : 3$ externally

Karnataka CET-2019

Three Dimensional Geometry

121114 A vector $\overset{―}{v}$ is equally inclined to the $X$ -axis, $Y$ -axis and $Z$ -axis respectively, its direction cosines are

1

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

2

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

3

\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

or

- \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}

4 None of these

Explanation:

C Given,
$α = β = γ$ i.e. $\cos α = \cos β = \cos γ$
Then. $\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$∴ \cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\Rightarrow \cos^{2} α = \frac{1}{3}$
$\Rightarrow \cos α = \pm \frac{1}{\sqrt{3}}$
Hence, the $x$ -axis, $y$ -axis and $z$ -axis are respectively of the direction of cosines are -
$\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} or - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, - \frac{1}{\sqrt{3}}$

MHT CET-2011

Three Dimensional Geometry

121115 What is the value of $n$ so that the angle between the lines having direction ratios $(1, 1, 1)$ and $(1$ , $- 1, n)$ is $60^{\circ}$ ?

1

\sqrt{3}

2

\sqrt{6}

3 3

4 None of these

Explanation:

B Given, $l_{1} = 1, m_{1} = 1, n_{1} = 1$ and $l_{2} = 1, m_{2} = - 1, n_{2} = n$ and $θ = 60^{\circ}$
We know that, If $(l_{1}, m_{1}, n_{1})$ and $(l_{2}, m_{2}, n_{2})$ are the direction ratios then angle between the lines is
$\cos θ = \frac{l_{1} l_{2} + m_{1} m_{2} + n_{1} n_{2}}{\sqrt{l_{1}^{2} + m_{1}^{2} + n_{1}^{2}} \sqrt{l_{2}^{2} + m_{2}^{2} + n_{2}^{2}}}$
$∴ \cos 60^{0} = \frac{1 \times 1 + 1 \times (- 1) + 1 \times n}{\sqrt{1^{2} + 1^{2} + 1^{2}} \sqrt{1^{2} + 1^{2} + n^{2}}}$
$\Rightarrow \frac{1}{2} = \frac{n}{\sqrt{3} \sqrt{2 + n^{2}}}$
$\Rightarrow 3 (2 + n^{2}) = 4 n^{2}$
$\Rightarrow n^{2} = 6$
$\Rightarrow n = \pm \sqrt{6}$

BITSAT-2012

Three Dimensional Geometry

121116 The projection of line joining $(3, 4, 5)$ and $(4, 6$ , $3)$ on the line joining $(- 1, 2, 4)$ and $(1, 0, 5)$ is

1

\frac{4}{3}

2

\frac{2}{3}

3

\frac{8}{3}

4

\frac{1}{3}

Explanation:

A Given, $A = (3, 4, 5), P = (- 1, 2, 4), B = (4, 6, 3)$ and $Q = (1, 0, 5)$ $∴$ Direction ratio of line $AB$ are $(4 - 3), (6 - 4), (3 - 5)$ $= (1, 2, - 2)$
And direction ratio of line PQ are $(1 + 1), (0 - 2)$ , $(5 -$ $4) = (2, - 2, 1)$
$∴$ Direction cosine of line -
$PQ = \frac{2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{- 2}{\sqrt{2^{2} + (- 2)^{2} + 1}}, \frac{1}{\sqrt{2^{2} + (- 2)^{2} + 1}}$
$= \frac{2}{3}, \frac{- 2}{3}, \frac{1}{3}$
$∴$ Projection of line segment $AB$ on the line $PQ$ is
$| \frac{2}{3} (1) + (\frac{- 2}{3}) (2) + (\frac{1}{3}) (- 2) | = \frac{4}{3}$

BITSAT-2018

Three Dimensional Geometry

121117 The ratio in which the line joining $(2, 1, 5)$ and $(3, 4, 3)$ is divided by the plane $(x + y - z) = \frac{1}{2}$ is :

1

3 : 5

2

5 : 7

3

1 : 3

4

4 : 5

Explanation:

B Given, $x + y - z = \frac{1}{2}$ divides the line joining the points $A (2, 1, 5)$ and $B (3, 4, 3)$ at a point $C$ in the
Applying section formula at a point $C$ ,
$(\frac{3 k + 2}{k + 1}, \frac{4 k + 1}{k + 1}, \frac{3 k + 5}{k + 1})$
Point $C$ lies on the plane, Coordinates of $C$ must satisfy the equation of plane
So, $(\frac{3 k + 2}{k + 1}) + (\frac{4 k + 1}{k + 1}) - (\frac{3 k + 5}{k + 1}) = \frac{1}{2}$
$\Rightarrow 3 k + 2 + 4 k + 1 - 3 k - 5 = \frac{1}{2} (k + 1)$
$\Rightarrow 4 k - 2 = \frac{1}{2} (k + 1)$
$8 k - 4 = k + 1$
$\Rightarrow 7 k = 5 \Rightarrow k = \frac{5}{7}$ Hence, ratio is $5 : 7$

BITSAT-2017