QBManager

Three Dimensional Geometry

121170 If $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$ , then the projection of $R S$ on $P Q$ is

1

- \frac{2}{3}

2

- \frac{4}{3}

3

\frac{1}{2}

4 2

Explanation:

B $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$
$Direction Ratio of PQ = [(4 - 3), (6 - 4), (3 - 5)]$
$= [1, 2, - 2]$
$Direction Ratio of RS = (2, - 2, 1)$
$We know that, projection of \vec{a} on \vec{b} is given by \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$
$∴$ The projection of RS on PQ is
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$

Manipal UGET-2019

Three Dimensional Geometry

121172 If $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then the value of $n$ is

1

\frac{\sqrt{23}}{6}

2

\frac{23}{36}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

A Since, $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then ${(\frac{1}{2})}^{2} + {(\frac{1}{3})}^{2} + n^{2} = 1$
$\frac{1}{4} + \frac{1}{9} + n^{2} = 1$
$n^{2} = 1 - \frac{1}{4} - \frac{1}{9}$
$n^{2} = \frac{23}{36}$
$n = \sqrt{\frac{23}{36}} = \frac{\sqrt{23}}{6}$ Hence, option (a) is correct

Manipal UGET-2018

Three Dimensional Geometry

121173 A ray makes angles $\frac{π}{3}$ and $\frac{π}{4}$ with $Y$ and $Z$ axes respectively. Then the value of the sine of the angle made by the ray with $X$ -axis is

1

\sqrt{3} / 2

2

1 / 2

3

1 / \sqrt{2}

4 1

Explanation:

A Given that,
A ray make on angle $\frac{π}{3}$ and $\frac{π}{4}$ with $y$ and $z$ axis respectively.
Then we have,
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\cos^{2} α + \cos^{2} (π / 3) + \cos^{2} (π / 4) = 1$
$\cos^{2} α + {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$\cos^{2} α + \frac{1}{4} + \frac{1}{2} = 1$
$\cos^{2} α = 1 - \frac{1}{4} - \frac{1}{2}$
$= \frac{4 - 1 - 2}{4} = \frac{1}{4}$
$∵ \sin^{2} α + \cos^{2} α = 1$
$∴ \sin^{2} α = 1 - \frac{1}{4} = \frac{3}{4}$
$\sin α = \frac{\sqrt{3}}{2}$

AP EAMCET-07.07.2022

Three Dimensional Geometry

121174 If a line makes angles $90^{\circ}, 135^{\circ}$ and $45^{\circ}$ with the positive directions of $x, y, z$ -axes respectively, then its direction cosines are......

1

(0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

2

(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(0, \frac{- 1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

A Dc's are
$l = \cos^{2} α = \cos 90^{\circ} = 0$
$m = \cos^{2} β = \cos 135^{\circ} = \frac{- 1}{\sqrt{2}} and$
$n = \cos^{2} γ = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
So, $(l, m, n) = (0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

AP EAMCET-17.09.2020

Three Dimensional Geometry

121175 If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $60^{\circ}$ , then

1

K^{2} - 56 K - 208 = 0

2

5 K^{2} - 110 K + 112 = 0

3

7 K^{2} - 112 K - 110 = 0

4

7 K^{2} - 112 K + 110 = 0

Explanation:

C Applying the formula for angle between $2$ lines based on direction ratio:
$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
$= \frac{2 K + 3 + (- 10)}{\sqrt{4 + 1 + 4} \sqrt{K^{2} + 9 + 25}}$
$\cos 60^{\circ} = \frac{2 K - 7}{\sqrt{9} \sqrt{K^{2} + 34}}$
$\frac{1}{2} = \frac{2 K - 7}{3 \sqrt{K^{2} + 34}}$
$\Rightarrow 3 \sqrt{K^{2} + 34} = 2 (2 K - 7)$
$\Rightarrow Square on both {side,}^{2}$
$9 (∵ \cos 60^{\circ} = \frac{1}{2})$
$\Rightarrow 9 K^{2} + 306 = 4 (4 K^{2} + 49 - 28 K)$
$\Rightarrow 7 K^{2} - 112 K - 110 = 0$
$\Rightarrow$ Square on both side,

TS EAMCET-05.08.2021

Three Dimensional Geometry

121170 If $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$ , then the projection of $R S$ on $P Q$ is

1

- \frac{2}{3}

2

- \frac{4}{3}

3

\frac{1}{2}

4 2

Explanation:

B $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$
$Direction Ratio of PQ = [(4 - 3), (6 - 4), (3 - 5)]$
$= [1, 2, - 2]$
$Direction Ratio of RS = (2, - 2, 1)$
$We know that, projection of \vec{a} on \vec{b} is given by \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$
$∴$ The projection of RS on PQ is
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$

Manipal UGET-2019

Three Dimensional Geometry

121172 If $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then the value of $n$ is

1

\frac{\sqrt{23}}{6}

2

\frac{23}{36}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

A Since, $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then ${(\frac{1}{2})}^{2} + {(\frac{1}{3})}^{2} + n^{2} = 1$
$\frac{1}{4} + \frac{1}{9} + n^{2} = 1$
$n^{2} = 1 - \frac{1}{4} - \frac{1}{9}$
$n^{2} = \frac{23}{36}$
$n = \sqrt{\frac{23}{36}} = \frac{\sqrt{23}}{6}$ Hence, option (a) is correct

Manipal UGET-2018

Three Dimensional Geometry

121173 A ray makes angles $\frac{π}{3}$ and $\frac{π}{4}$ with $Y$ and $Z$ axes respectively. Then the value of the sine of the angle made by the ray with $X$ -axis is

1

\sqrt{3} / 2

2

1 / 2

3

1 / \sqrt{2}

4 1

Explanation:

A Given that,
A ray make on angle $\frac{π}{3}$ and $\frac{π}{4}$ with $y$ and $z$ axis respectively.
Then we have,
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\cos^{2} α + \cos^{2} (π / 3) + \cos^{2} (π / 4) = 1$
$\cos^{2} α + {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$\cos^{2} α + \frac{1}{4} + \frac{1}{2} = 1$
$\cos^{2} α = 1 - \frac{1}{4} - \frac{1}{2}$
$= \frac{4 - 1 - 2}{4} = \frac{1}{4}$
$∵ \sin^{2} α + \cos^{2} α = 1$
$∴ \sin^{2} α = 1 - \frac{1}{4} = \frac{3}{4}$
$\sin α = \frac{\sqrt{3}}{2}$

AP EAMCET-07.07.2022

Three Dimensional Geometry

121174 If a line makes angles $90^{\circ}, 135^{\circ}$ and $45^{\circ}$ with the positive directions of $x, y, z$ -axes respectively, then its direction cosines are......

1

(0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

2

(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(0, \frac{- 1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

A Dc's are
$l = \cos^{2} α = \cos 90^{\circ} = 0$
$m = \cos^{2} β = \cos 135^{\circ} = \frac{- 1}{\sqrt{2}} and$
$n = \cos^{2} γ = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
So, $(l, m, n) = (0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

AP EAMCET-17.09.2020

Three Dimensional Geometry

121175 If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $60^{\circ}$ , then

1

K^{2} - 56 K - 208 = 0

2

5 K^{2} - 110 K + 112 = 0

3

7 K^{2} - 112 K - 110 = 0

4

7 K^{2} - 112 K + 110 = 0

Explanation:

C Applying the formula for angle between $2$ lines based on direction ratio:
$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
$= \frac{2 K + 3 + (- 10)}{\sqrt{4 + 1 + 4} \sqrt{K^{2} + 9 + 25}}$
$\cos 60^{\circ} = \frac{2 K - 7}{\sqrt{9} \sqrt{K^{2} + 34}}$
$\frac{1}{2} = \frac{2 K - 7}{3 \sqrt{K^{2} + 34}}$
$\Rightarrow 3 \sqrt{K^{2} + 34} = 2 (2 K - 7)$
$\Rightarrow Square on both {side,}^{2}$
$9 (∵ \cos 60^{\circ} = \frac{1}{2})$
$\Rightarrow 9 K^{2} + 306 = 4 (4 K^{2} + 49 - 28 K)$
$\Rightarrow 7 K^{2} - 112 K - 110 = 0$
$\Rightarrow$ Square on both side,

TS EAMCET-05.08.2021

Three Dimensional Geometry

121170 If $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$ , then the projection of $R S$ on $P Q$ is

1

- \frac{2}{3}

2

- \frac{4}{3}

3

\frac{1}{2}

4 2

Explanation:

B $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$
$Direction Ratio of PQ = [(4 - 3), (6 - 4), (3 - 5)]$
$= [1, 2, - 2]$
$Direction Ratio of RS = (2, - 2, 1)$
$We know that, projection of \vec{a} on \vec{b} is given by \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$
$∴$ The projection of RS on PQ is
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$

Manipal UGET-2019

Three Dimensional Geometry

121172 If $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then the value of $n$ is

1

\frac{\sqrt{23}}{6}

2

\frac{23}{36}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

A Since, $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then ${(\frac{1}{2})}^{2} + {(\frac{1}{3})}^{2} + n^{2} = 1$
$\frac{1}{4} + \frac{1}{9} + n^{2} = 1$
$n^{2} = 1 - \frac{1}{4} - \frac{1}{9}$
$n^{2} = \frac{23}{36}$
$n = \sqrt{\frac{23}{36}} = \frac{\sqrt{23}}{6}$ Hence, option (a) is correct

Manipal UGET-2018

Three Dimensional Geometry

121173 A ray makes angles $\frac{π}{3}$ and $\frac{π}{4}$ with $Y$ and $Z$ axes respectively. Then the value of the sine of the angle made by the ray with $X$ -axis is

1

\sqrt{3} / 2

2

1 / 2

3

1 / \sqrt{2}

4 1

Explanation:

A Given that,
A ray make on angle $\frac{π}{3}$ and $\frac{π}{4}$ with $y$ and $z$ axis respectively.
Then we have,
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\cos^{2} α + \cos^{2} (π / 3) + \cos^{2} (π / 4) = 1$
$\cos^{2} α + {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$\cos^{2} α + \frac{1}{4} + \frac{1}{2} = 1$
$\cos^{2} α = 1 - \frac{1}{4} - \frac{1}{2}$
$= \frac{4 - 1 - 2}{4} = \frac{1}{4}$
$∵ \sin^{2} α + \cos^{2} α = 1$
$∴ \sin^{2} α = 1 - \frac{1}{4} = \frac{3}{4}$
$\sin α = \frac{\sqrt{3}}{2}$

AP EAMCET-07.07.2022

Three Dimensional Geometry

121174 If a line makes angles $90^{\circ}, 135^{\circ}$ and $45^{\circ}$ with the positive directions of $x, y, z$ -axes respectively, then its direction cosines are......

1

(0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

2

(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(0, \frac{- 1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

A Dc's are
$l = \cos^{2} α = \cos 90^{\circ} = 0$
$m = \cos^{2} β = \cos 135^{\circ} = \frac{- 1}{\sqrt{2}} and$
$n = \cos^{2} γ = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
So, $(l, m, n) = (0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

AP EAMCET-17.09.2020

Three Dimensional Geometry

121175 If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $60^{\circ}$ , then

1

K^{2} - 56 K - 208 = 0

2

5 K^{2} - 110 K + 112 = 0

3

7 K^{2} - 112 K - 110 = 0

4

7 K^{2} - 112 K + 110 = 0

Explanation:

C Applying the formula for angle between $2$ lines based on direction ratio:
$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
$= \frac{2 K + 3 + (- 10)}{\sqrt{4 + 1 + 4} \sqrt{K^{2} + 9 + 25}}$
$\cos 60^{\circ} = \frac{2 K - 7}{\sqrt{9} \sqrt{K^{2} + 34}}$
$\frac{1}{2} = \frac{2 K - 7}{3 \sqrt{K^{2} + 34}}$
$\Rightarrow 3 \sqrt{K^{2} + 34} = 2 (2 K - 7)$
$\Rightarrow Square on both {side,}^{2}$
$9 (∵ \cos 60^{\circ} = \frac{1}{2})$
$\Rightarrow 9 K^{2} + 306 = 4 (4 K^{2} + 49 - 28 K)$
$\Rightarrow 7 K^{2} - 112 K - 110 = 0$
$\Rightarrow$ Square on both side,

TS EAMCET-05.08.2021

Three Dimensional Geometry

121170 If $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$ , then the projection of $R S$ on $P Q$ is

1

- \frac{2}{3}

2

- \frac{4}{3}

3

\frac{1}{2}

4 2

Explanation:

B $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$
$Direction Ratio of PQ = [(4 - 3), (6 - 4), (3 - 5)]$
$= [1, 2, - 2]$
$Direction Ratio of RS = (2, - 2, 1)$
$We know that, projection of \vec{a} on \vec{b} is given by \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$
$∴$ The projection of RS on PQ is
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$

Manipal UGET-2019

Three Dimensional Geometry

121172 If $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then the value of $n$ is

1

\frac{\sqrt{23}}{6}

2

\frac{23}{36}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

A Since, $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then ${(\frac{1}{2})}^{2} + {(\frac{1}{3})}^{2} + n^{2} = 1$
$\frac{1}{4} + \frac{1}{9} + n^{2} = 1$
$n^{2} = 1 - \frac{1}{4} - \frac{1}{9}$
$n^{2} = \frac{23}{36}$
$n = \sqrt{\frac{23}{36}} = \frac{\sqrt{23}}{6}$ Hence, option (a) is correct

Manipal UGET-2018

Three Dimensional Geometry

121173 A ray makes angles $\frac{π}{3}$ and $\frac{π}{4}$ with $Y$ and $Z$ axes respectively. Then the value of the sine of the angle made by the ray with $X$ -axis is

1

\sqrt{3} / 2

2

1 / 2

3

1 / \sqrt{2}

4 1

Explanation:

A Given that,
A ray make on angle $\frac{π}{3}$ and $\frac{π}{4}$ with $y$ and $z$ axis respectively.
Then we have,
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\cos^{2} α + \cos^{2} (π / 3) + \cos^{2} (π / 4) = 1$
$\cos^{2} α + {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$\cos^{2} α + \frac{1}{4} + \frac{1}{2} = 1$
$\cos^{2} α = 1 - \frac{1}{4} - \frac{1}{2}$
$= \frac{4 - 1 - 2}{4} = \frac{1}{4}$
$∵ \sin^{2} α + \cos^{2} α = 1$
$∴ \sin^{2} α = 1 - \frac{1}{4} = \frac{3}{4}$
$\sin α = \frac{\sqrt{3}}{2}$

AP EAMCET-07.07.2022

Three Dimensional Geometry

121174 If a line makes angles $90^{\circ}, 135^{\circ}$ and $45^{\circ}$ with the positive directions of $x, y, z$ -axes respectively, then its direction cosines are......

1

(0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

2

(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(0, \frac{- 1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

A Dc's are
$l = \cos^{2} α = \cos 90^{\circ} = 0$
$m = \cos^{2} β = \cos 135^{\circ} = \frac{- 1}{\sqrt{2}} and$
$n = \cos^{2} γ = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
So, $(l, m, n) = (0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

AP EAMCET-17.09.2020

Three Dimensional Geometry

121175 If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $60^{\circ}$ , then

1

K^{2} - 56 K - 208 = 0

2

5 K^{2} - 110 K + 112 = 0

3

7 K^{2} - 112 K - 110 = 0

4

7 K^{2} - 112 K + 110 = 0

Explanation:

C Applying the formula for angle between $2$ lines based on direction ratio:
$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
$= \frac{2 K + 3 + (- 10)}{\sqrt{4 + 1 + 4} \sqrt{K^{2} + 9 + 25}}$
$\cos 60^{\circ} = \frac{2 K - 7}{\sqrt{9} \sqrt{K^{2} + 34}}$
$\frac{1}{2} = \frac{2 K - 7}{3 \sqrt{K^{2} + 34}}$
$\Rightarrow 3 \sqrt{K^{2} + 34} = 2 (2 K - 7)$
$\Rightarrow Square on both {side,}^{2}$
$9 (∵ \cos 60^{\circ} = \frac{1}{2})$
$\Rightarrow 9 K^{2} + 306 = 4 (4 K^{2} + 49 - 28 K)$
$\Rightarrow 7 K^{2} - 112 K - 110 = 0$
$\Rightarrow$ Square on both side,

TS EAMCET-05.08.2021

Three Dimensional Geometry

121170 If $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$ , then the projection of $R S$ on $P Q$ is

1

- \frac{2}{3}

2

- \frac{4}{3}

3

\frac{1}{2}

4 2

Explanation:

B $P (3, 4, 5), Q (4, 6, 3), R (- 1, 2, 4), S (1, 0, 5)$
$Direction Ratio of PQ = [(4 - 3), (6 - 4), (3 - 5)]$
$= [1, 2, - 2]$
$Direction Ratio of RS = (2, - 2, 1)$
$We know that, projection of \vec{a} on \vec{b} is given by \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$
$∴$ The projection of RS on PQ is
$= \frac{(RS) \cdot (PQ)}{| PQ |}$
$= \frac{(2 \hat{i} - 2 \hat{j} + \hat{k}) \cdot (\hat{i} + 2 \hat{j} - 2 \hat{k})}{\sqrt{(1)^{2} + (2)^{2} + (- 2)^{2}}}$
$= \frac{2 - 4 - 2}{\sqrt{9}} = - \frac{4}{3}$

Manipal UGET-2019

Three Dimensional Geometry

121172 If $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then the value of $n$ is

1

\frac{\sqrt{23}}{6}

2

\frac{23}{36}

3

\frac{2}{3}

4

\frac{3}{2}

Explanation:

A Since, $(\frac{1}{2}, \frac{1}{3}, n)$ are the direction cosines of a line, then ${(\frac{1}{2})}^{2} + {(\frac{1}{3})}^{2} + n^{2} = 1$
$\frac{1}{4} + \frac{1}{9} + n^{2} = 1$
$n^{2} = 1 - \frac{1}{4} - \frac{1}{9}$
$n^{2} = \frac{23}{36}$
$n = \sqrt{\frac{23}{36}} = \frac{\sqrt{23}}{6}$ Hence, option (a) is correct

Manipal UGET-2018

Three Dimensional Geometry

121173 A ray makes angles $\frac{π}{3}$ and $\frac{π}{4}$ with $Y$ and $Z$ axes respectively. Then the value of the sine of the angle made by the ray with $X$ -axis is

1

\sqrt{3} / 2

2

1 / 2

3

1 / \sqrt{2}

4 1

Explanation:

A Given that,
A ray make on angle $\frac{π}{3}$ and $\frac{π}{4}$ with $y$ and $z$ axis respectively.
Then we have,
$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
$\cos^{2} α + \cos^{2} (π / 3) + \cos^{2} (π / 4) = 1$
$\cos^{2} α + {(\frac{1}{2})}^{2} + {(\frac{1}{\sqrt{2}})}^{2} = 1$
$\cos^{2} α + \frac{1}{4} + \frac{1}{2} = 1$
$\cos^{2} α = 1 - \frac{1}{4} - \frac{1}{2}$
$= \frac{4 - 1 - 2}{4} = \frac{1}{4}$
$∵ \sin^{2} α + \cos^{2} α = 1$
$∴ \sin^{2} α = 1 - \frac{1}{4} = \frac{3}{4}$
$\sin α = \frac{\sqrt{3}}{2}$

AP EAMCET-07.07.2022

Three Dimensional Geometry

121174 If a line makes angles $90^{\circ}, 135^{\circ}$ and $45^{\circ}$ with the positive directions of $x, y, z$ -axes respectively, then its direction cosines are......

1

(0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

2

(0, \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})

3

(0, \frac{- 1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

4

(0, \frac{1}{\sqrt{2}}, \frac{- 1}{\sqrt{2}})

Explanation:

A Dc's are
$l = \cos^{2} α = \cos 90^{\circ} = 0$
$m = \cos^{2} β = \cos 135^{\circ} = \frac{- 1}{\sqrt{2}} and$
$n = \cos^{2} γ = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$
So, $(l, m, n) = (0, \frac{- 1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$

AP EAMCET-17.09.2020

Three Dimensional Geometry

121175 If $(2, - 1, 2)$ and $(K, - 3, - 5)$ are the triads of direction ratios of two lines and the angle between the lines is $60^{\circ}$ , then

1

K^{2} - 56 K - 208 = 0

2

5 K^{2} - 110 K + 112 = 0

3

7 K^{2} - 112 K - 110 = 0

4

7 K^{2} - 112 K + 110 = 0

Explanation:

C Applying the formula for angle between $2$ lines based on direction ratio:
$\cos θ = \frac{a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2}}{\sqrt{a_{1}^{2} + b_{1}^{2} + c_{1}^{2}} \sqrt{a_{2}^{2} + b_{2}^{2} + c_{2}^{2}}}$
$= \frac{2 K + 3 + (- 10)}{\sqrt{4 + 1 + 4} \sqrt{K^{2} + 9 + 25}}$
$\cos 60^{\circ} = \frac{2 K - 7}{\sqrt{9} \sqrt{K^{2} + 34}}$
$\frac{1}{2} = \frac{2 K - 7}{3 \sqrt{K^{2} + 34}}$
$\Rightarrow 3 \sqrt{K^{2} + 34} = 2 (2 K - 7)$
$\Rightarrow Square on both {side,}^{2}$
$9 (∵ \cos 60^{\circ} = \frac{1}{2})$
$\Rightarrow 9 K^{2} + 306 = 4 (4 K^{2} + 49 - 28 K)$
$\Rightarrow 7 K^{2} - 112 K - 110 = 0$
$\Rightarrow$ Square on both side,

TS EAMCET-05.08.2021