QBManager

Three Dimensional Geometry

121163 If $A (3, 4, 5), B (4, 6, 3) C (- 1, 2, 4)$ and $D (1, 0, 5)$ are such that the angle between the lines $D C$ and $A B$ is $θ$ , then $\cos θ$ is equal to

1

\frac{7}{9}

2

\frac{2}{9}

3

\frac{4}{9}

4

\frac{5}{9}

Explanation:

C Given,
$A (3, 4, 5), B (4, 6, 3), C (- 1, 2, 4) and D (1, 0, 5)$
$\vec{DC} = - 2 \hat{i} + 2 \hat{j} - \hat{k}$
$\vec{AB} = \hat{i} + 2 \hat{j} - 2 \hat{k}$
$\Rightarrow \cos θ = \frac{\vec{AB} \cdot \vec{DC}}{| \vec{AB} | \cdot | \vec{DC} |}$
$\Rightarrow \cos θ = \frac{- 2 + 4 + 2}{\sqrt{9} \times \sqrt{9}}$
$\Rightarrow \cos θ = \frac{4}{9}$

AP EAMCET-2014

Three Dimensional Geometry

121164 The direction cosines of a line which makes equal angles with the co-ordinate axes 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{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}} ⟩

4

⟨ \frac{12}{13}, \frac{5}{13}, 0 ⟩

Explanation:

C
Given,
If a line makes angle $α$ with $x$ -axis, $β$ with $y$ -axis \& $γ$ with z-axis.
So, direction cosine of the line be $ℓ = \cos α, m = \cos β, n = \cos γ$
Given, $α = β = γ \Rightarrow ℓ = m = n = \cos α$
We know that: $ℓ^{2} + m^{2} + n^{2} = 1$
$\cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\cos^{2} α = \frac{1}{3}$
$\cos α = \pm \frac{1}{\sqrt{3}}$
Similarly, $\cos β = \pm \frac{1}{\sqrt{3}}, \cos γ = \pm \frac{1}{\sqrt{3}}$
So, direction cosine are,
$ℓ = \pm \frac{1}{\sqrt{3}}, m = \pm \frac{1}{\sqrt{3}}, n = \pm \frac{1}{\sqrt{3}}$

AP EAMCET-19.08.2021

Three Dimensional Geometry

121166 The direction ratios of a normal to a plane are $(3, 12, 4)$ . Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?

1

(2, 2, 1)

2

(2, 1, 2)

3

(2, 1, 1)

4

(3, - 1, 2)

Explanation:

B Let plane be $a x + b y + c z + d = 0$
Direction ratio of normal to the plane $(3, 12, 4)$
So, equation (i) is-
$3 x + 12 y + 4 z + d = 0$
Now, distance of (ii) from origin is 2 unit
$\frac{3 \times 0 + 12 \times 0 + 4 \times 0 + d}{\sqrt{3^{2} + 12^{2} + 4^{2}}} = 2$
$d = \pm (2 \times 13) = \pm 26$
$∴$ Equation of plane is
$3 x + 12 y + 4 z \pm 26 = 0$ This plane is passes through $(2, 1, 2)$

J and K CET-2018

Three Dimensional Geometry

121169 The direction cosines of a line are $⟨ \frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11} ⟩$ respectively. Then its direction ratios are

1

(9, 6, - 2)

2

(- 9, - 6, 2)

3

(- 9, 6, - 2)

4

(9, - 6, - 2)

Explanation:

C Given,
DC's $(\frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11})$
Direction cosine formula
$l = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 9}{11}$
$m = \frac{b}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{6}{11}$
$n = \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{11}$ Here, $(a, b, c)$ is the direction ratio i.e. $(- 9, 6, - 2)$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121163 If $A (3, 4, 5), B (4, 6, 3) C (- 1, 2, 4)$ and $D (1, 0, 5)$ are such that the angle between the lines $D C$ and $A B$ is $θ$ , then $\cos θ$ is equal to

1

\frac{7}{9}

2

\frac{2}{9}

3

\frac{4}{9}

4

\frac{5}{9}

Explanation:

C Given,
$A (3, 4, 5), B (4, 6, 3), C (- 1, 2, 4) and D (1, 0, 5)$
$\vec{DC} = - 2 \hat{i} + 2 \hat{j} - \hat{k}$
$\vec{AB} = \hat{i} + 2 \hat{j} - 2 \hat{k}$
$\Rightarrow \cos θ = \frac{\vec{AB} \cdot \vec{DC}}{| \vec{AB} | \cdot | \vec{DC} |}$
$\Rightarrow \cos θ = \frac{- 2 + 4 + 2}{\sqrt{9} \times \sqrt{9}}$
$\Rightarrow \cos θ = \frac{4}{9}$

AP EAMCET-2014

Three Dimensional Geometry

121164 The direction cosines of a line which makes equal angles with the co-ordinate axes 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{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}} ⟩

4

⟨ \frac{12}{13}, \frac{5}{13}, 0 ⟩

Explanation:

C
Given,
If a line makes angle $α$ with $x$ -axis, $β$ with $y$ -axis \& $γ$ with z-axis.
So, direction cosine of the line be $ℓ = \cos α, m = \cos β, n = \cos γ$
Given, $α = β = γ \Rightarrow ℓ = m = n = \cos α$
We know that: $ℓ^{2} + m^{2} + n^{2} = 1$
$\cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\cos^{2} α = \frac{1}{3}$
$\cos α = \pm \frac{1}{\sqrt{3}}$
Similarly, $\cos β = \pm \frac{1}{\sqrt{3}}, \cos γ = \pm \frac{1}{\sqrt{3}}$
So, direction cosine are,
$ℓ = \pm \frac{1}{\sqrt{3}}, m = \pm \frac{1}{\sqrt{3}}, n = \pm \frac{1}{\sqrt{3}}$

AP EAMCET-19.08.2021

Three Dimensional Geometry

121166 The direction ratios of a normal to a plane are $(3, 12, 4)$ . Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?

1

(2, 2, 1)

2

(2, 1, 2)

3

(2, 1, 1)

4

(3, - 1, 2)

Explanation:

B Let plane be $a x + b y + c z + d = 0$
Direction ratio of normal to the plane $(3, 12, 4)$
So, equation (i) is-
$3 x + 12 y + 4 z + d = 0$
Now, distance of (ii) from origin is 2 unit
$\frac{3 \times 0 + 12 \times 0 + 4 \times 0 + d}{\sqrt{3^{2} + 12^{2} + 4^{2}}} = 2$
$d = \pm (2 \times 13) = \pm 26$
$∴$ Equation of plane is
$3 x + 12 y + 4 z \pm 26 = 0$ This plane is passes through $(2, 1, 2)$

J and K CET-2018

Three Dimensional Geometry

121169 The direction cosines of a line are $⟨ \frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11} ⟩$ respectively. Then its direction ratios are

1

(9, 6, - 2)

2

(- 9, - 6, 2)

3

(- 9, 6, - 2)

4

(9, - 6, - 2)

Explanation:

C Given,
DC's $(\frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11})$
Direction cosine formula
$l = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 9}{11}$
$m = \frac{b}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{6}{11}$
$n = \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{11}$ Here, $(a, b, c)$ is the direction ratio i.e. $(- 9, 6, - 2)$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121163 If $A (3, 4, 5), B (4, 6, 3) C (- 1, 2, 4)$ and $D (1, 0, 5)$ are such that the angle between the lines $D C$ and $A B$ is $θ$ , then $\cos θ$ is equal to

1

\frac{7}{9}

2

\frac{2}{9}

3

\frac{4}{9}

4

\frac{5}{9}

Explanation:

C Given,
$A (3, 4, 5), B (4, 6, 3), C (- 1, 2, 4) and D (1, 0, 5)$
$\vec{DC} = - 2 \hat{i} + 2 \hat{j} - \hat{k}$
$\vec{AB} = \hat{i} + 2 \hat{j} - 2 \hat{k}$
$\Rightarrow \cos θ = \frac{\vec{AB} \cdot \vec{DC}}{| \vec{AB} | \cdot | \vec{DC} |}$
$\Rightarrow \cos θ = \frac{- 2 + 4 + 2}{\sqrt{9} \times \sqrt{9}}$
$\Rightarrow \cos θ = \frac{4}{9}$

AP EAMCET-2014

Three Dimensional Geometry

121164 The direction cosines of a line which makes equal angles with the co-ordinate axes 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{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}} ⟩

4

⟨ \frac{12}{13}, \frac{5}{13}, 0 ⟩

Explanation:

C
Given,
If a line makes angle $α$ with $x$ -axis, $β$ with $y$ -axis \& $γ$ with z-axis.
So, direction cosine of the line be $ℓ = \cos α, m = \cos β, n = \cos γ$
Given, $α = β = γ \Rightarrow ℓ = m = n = \cos α$
We know that: $ℓ^{2} + m^{2} + n^{2} = 1$
$\cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\cos^{2} α = \frac{1}{3}$
$\cos α = \pm \frac{1}{\sqrt{3}}$
Similarly, $\cos β = \pm \frac{1}{\sqrt{3}}, \cos γ = \pm \frac{1}{\sqrt{3}}$
So, direction cosine are,
$ℓ = \pm \frac{1}{\sqrt{3}}, m = \pm \frac{1}{\sqrt{3}}, n = \pm \frac{1}{\sqrt{3}}$

AP EAMCET-19.08.2021

Three Dimensional Geometry

121166 The direction ratios of a normal to a plane are $(3, 12, 4)$ . Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?

1

(2, 2, 1)

2

(2, 1, 2)

3

(2, 1, 1)

4

(3, - 1, 2)

Explanation:

B Let plane be $a x + b y + c z + d = 0$
Direction ratio of normal to the plane $(3, 12, 4)$
So, equation (i) is-
$3 x + 12 y + 4 z + d = 0$
Now, distance of (ii) from origin is 2 unit
$\frac{3 \times 0 + 12 \times 0 + 4 \times 0 + d}{\sqrt{3^{2} + 12^{2} + 4^{2}}} = 2$
$d = \pm (2 \times 13) = \pm 26$
$∴$ Equation of plane is
$3 x + 12 y + 4 z \pm 26 = 0$ This plane is passes through $(2, 1, 2)$

J and K CET-2018

Three Dimensional Geometry

121169 The direction cosines of a line are $⟨ \frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11} ⟩$ respectively. Then its direction ratios are

1

(9, 6, - 2)

2

(- 9, - 6, 2)

3

(- 9, 6, - 2)

4

(9, - 6, - 2)

Explanation:

C Given,
DC's $(\frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11})$
Direction cosine formula
$l = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 9}{11}$
$m = \frac{b}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{6}{11}$
$n = \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{11}$ Here, $(a, b, c)$ is the direction ratio i.e. $(- 9, 6, - 2)$

AP EAMCET-21.09.2020

Three Dimensional Geometry

121163 If $A (3, 4, 5), B (4, 6, 3) C (- 1, 2, 4)$ and $D (1, 0, 5)$ are such that the angle between the lines $D C$ and $A B$ is $θ$ , then $\cos θ$ is equal to

1

\frac{7}{9}

2

\frac{2}{9}

3

\frac{4}{9}

4

\frac{5}{9}

Explanation:

C Given,
$A (3, 4, 5), B (4, 6, 3), C (- 1, 2, 4) and D (1, 0, 5)$
$\vec{DC} = - 2 \hat{i} + 2 \hat{j} - \hat{k}$
$\vec{AB} = \hat{i} + 2 \hat{j} - 2 \hat{k}$
$\Rightarrow \cos θ = \frac{\vec{AB} \cdot \vec{DC}}{| \vec{AB} | \cdot | \vec{DC} |}$
$\Rightarrow \cos θ = \frac{- 2 + 4 + 2}{\sqrt{9} \times \sqrt{9}}$
$\Rightarrow \cos θ = \frac{4}{9}$

AP EAMCET-2014

Three Dimensional Geometry

121164 The direction cosines of a line which makes equal angles with the co-ordinate axes 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{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}}, \frac{\pm 1}{\sqrt{3}} ⟩

4

⟨ \frac{12}{13}, \frac{5}{13}, 0 ⟩

Explanation:

C
Given,
If a line makes angle $α$ with $x$ -axis, $β$ with $y$ -axis \& $γ$ with z-axis.
So, direction cosine of the line be $ℓ = \cos α, m = \cos β, n = \cos γ$
Given, $α = β = γ \Rightarrow ℓ = m = n = \cos α$
We know that: $ℓ^{2} + m^{2} + n^{2} = 1$
$\cos^{2} α + \cos^{2} α + \cos^{2} α = 1$
$3 \cos^{2} α = 1$
$\cos^{2} α = \frac{1}{3}$
$\cos α = \pm \frac{1}{\sqrt{3}}$
Similarly, $\cos β = \pm \frac{1}{\sqrt{3}}, \cos γ = \pm \frac{1}{\sqrt{3}}$
So, direction cosine are,
$ℓ = \pm \frac{1}{\sqrt{3}}, m = \pm \frac{1}{\sqrt{3}}, n = \pm \frac{1}{\sqrt{3}}$

AP EAMCET-19.08.2021

Three Dimensional Geometry

121166 The direction ratios of a normal to a plane are $(3, 12, 4)$ . Through which of the following points the plane passes if origin is at a distance of 2 units from this plane?

1

(2, 2, 1)

2

(2, 1, 2)

3

(2, 1, 1)

4

(3, - 1, 2)

Explanation:

B Let plane be $a x + b y + c z + d = 0$
Direction ratio of normal to the plane $(3, 12, 4)$
So, equation (i) is-
$3 x + 12 y + 4 z + d = 0$
Now, distance of (ii) from origin is 2 unit
$\frac{3 \times 0 + 12 \times 0 + 4 \times 0 + d}{\sqrt{3^{2} + 12^{2} + 4^{2}}} = 2$
$d = \pm (2 \times 13) = \pm 26$
$∴$ Equation of plane is
$3 x + 12 y + 4 z \pm 26 = 0$ This plane is passes through $(2, 1, 2)$

J and K CET-2018

Three Dimensional Geometry

121169 The direction cosines of a line are $⟨ \frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11} ⟩$ respectively. Then its direction ratios are

1

(9, 6, - 2)

2

(- 9, - 6, 2)

3

(- 9, 6, - 2)

4

(9, - 6, - 2)

Explanation:

C Given,
DC's $(\frac{- 9}{11}, \frac{6}{11}, \frac{- 2}{11})$
Direction cosine formula
$l = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 9}{11}$
$m = \frac{b}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{6}{11}$
$n = \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{11}$ Here, $(a, b, c)$ is the direction ratio i.e. $(- 9, 6, - 2)$

AP EAMCET-21.09.2020

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