QBManager

Vector Algebra

87888 If $a, b, c$ are any thr $(\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}})$
ee mutually
perpendicular vectors of equal magnitude a, then $| a + b + c |$

1 a

2

\sqrt{2} a

3

\sqrt{3} a

4

2 a

Explanation:

(C) : We know that,
$| a + b + c |^{2} = | a |^{2} + | b |^{2} + | c |^{2} + 2 (a . b + b . c + c . a)$
Let, $| \vec{a} | = | \vec{b} | = | \vec{c} | = a$
$| a + b + c |^{2} = a^{2} + a^{2} + a^{2} + 0 + 0 + 0$
$| a + b + c |^{2} = 3 a^{2}$
$a + b + c = \sqrt{3 a^{2}}$
$a + b + c = \sqrt{3} a$
Hence, $| \vec{a} + \vec{b} + \vec{c} | = \sqrt{3} a$

Manipal-2017

Vector Algebra

87872 The direction cosines of the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is equal to

1

⟨ \frac{2}{3}, \frac{1}{3}, - \frac{2}{3} ⟩

2

⟨ \frac{2}{3}, \frac{1}{3}, \frac{2}{3} ⟩

3

⟨ \frac{1}{3}, \frac{2}{3}, - \frac{2}{3} ⟩

4

⟨ \frac{2}{3}, \frac{2}{3}, \frac{1}{3} ⟩

Explanation:

(A) : If , $\hat{i}, \hat{j}, \hat{k}$ are the direction cosines and a, b, c are the direction ratio of line.
$2 \hat{i} + \hat{j} - 2 \hat{k}$
$a = 2, b = 1, c = - 2$
$\hat{i} = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{2}{\sqrt{4 + 1 + 4}} = \frac{2}{3}$
$\hat{j} = \frac{\underset{―}{b}}{a^{2} + b^{2} + c^{2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{3}$
$\hat{k} = \frac{\underset{―}{c}}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{\sqrt{4 + 1 + 4}} = \frac{- 2}{3}$
Hence all direction are cosine
$(a, b, c) = (\frac{2}{3}, \frac{1}{3}, - \frac{2}{3})$

COMEDK-2012

Vector Algebra

87873 The projections of a directed line segment on the coordinate axis are $12, 4, 3$ . The direction cosines of the line are

1

\frac{12}{13}, \frac{4}{13}, \frac{3}{13}

2

\frac{12}{13}, \frac{4}{13}, - \frac{3}{13}

3

- \frac{12}{13}, \frac{4}{13}, \frac{3}{13}

4

\frac{12}{13}, - \frac{4}{13}, - \frac{3}{13}

Explanation:

(A) : Assuming position vector of line segment ends points are $\vec{A}$ and $\vec{B}$
Now, given projection of line segment on the axis are $(12, 4, 3)$
So, $A B = 12 \hat{i} + 4 \hat{j} + 3 \hat{k}$
Now the direction cosine of line segment,
$= \frac{12}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{4}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{3}{\sqrt{12^{2} + 4^{2} + 3^{2}}}$
$= \frac{12}{13}, \frac{4}{13}, \frac{3}{13}$

AMU-2011

Vector Algebra

87880 Angle made by the position vector of the point $(5, - 4, - 3)$ with the positive direction of $X$ - axis is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{4}

4

\frac{π}{3}

Explanation:

(C): Let,
A $(5, - 4, - 3)$
Direction ratio of $OA = 5, - 4, - 3$
Direction ratio of $x -$ axis $= 1, 0, 0$
Angle, $θ = \frac{5 \cdot 1 + (- 4) \cdot 0 + (- 3) \cdot 0}{\sqrt{(5)^{2} + (- 4)^{2} + (- 3)^{2}} \sqrt{(1)^{2} + 0^{2} + 0^{2}}}$
$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$
$∴ θ = \frac{π}{4}$

AP EAMCET-2021-19.08.2021

Vector Algebra

87887 The direction cosines of the vector
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$ are

1

\frac{- 2}{\sqrt{8}}, \frac{1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

2

\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

3

\frac{2}{\sqrt{8}}, \frac{- 1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

4

\frac{- 2}{\sqrt{30}}, \frac{- 1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

Explanation:

(B) : Given that,
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$
$| \vec{a} | = \sqrt{(- 2)^{2} + (1)^{2} + (- 5)^{2}}$
$= \sqrt{4 + 1 + 25} = \sqrt{30}$
Direction cosine are $(l, m, n)$
$l = \frac{a_{1}}{| \vec{a} |}, m = \frac{a_{2}}{| \vec{a} |}, n = \frac{a_{3}}{| \vec{a} |}$
$∴$ Direction cosine are -

AP EAMCET-2020-21.09.2020

Vector Algebra

87888 If $a, b, c$ are any thr $(\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}})$
ee mutually
perpendicular vectors of equal magnitude a, then $| a + b + c |$

1 a

2

\sqrt{2} a

3

\sqrt{3} a

4

2 a

Explanation:

(C) : We know that,
$| a + b + c |^{2} = | a |^{2} + | b |^{2} + | c |^{2} + 2 (a . b + b . c + c . a)$
Let, $| \vec{a} | = | \vec{b} | = | \vec{c} | = a$
$| a + b + c |^{2} = a^{2} + a^{2} + a^{2} + 0 + 0 + 0$
$| a + b + c |^{2} = 3 a^{2}$
$a + b + c = \sqrt{3 a^{2}}$
$a + b + c = \sqrt{3} a$
Hence, $| \vec{a} + \vec{b} + \vec{c} | = \sqrt{3} a$

Manipal-2017

Vector Algebra

87872 The direction cosines of the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is equal to

1

⟨ \frac{2}{3}, \frac{1}{3}, - \frac{2}{3} ⟩

2

⟨ \frac{2}{3}, \frac{1}{3}, \frac{2}{3} ⟩

3

⟨ \frac{1}{3}, \frac{2}{3}, - \frac{2}{3} ⟩

4

⟨ \frac{2}{3}, \frac{2}{3}, \frac{1}{3} ⟩

Explanation:

(A) : If , $\hat{i}, \hat{j}, \hat{k}$ are the direction cosines and a, b, c are the direction ratio of line.
$2 \hat{i} + \hat{j} - 2 \hat{k}$
$a = 2, b = 1, c = - 2$
$\hat{i} = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{2}{\sqrt{4 + 1 + 4}} = \frac{2}{3}$
$\hat{j} = \frac{\underset{―}{b}}{a^{2} + b^{2} + c^{2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{3}$
$\hat{k} = \frac{\underset{―}{c}}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{\sqrt{4 + 1 + 4}} = \frac{- 2}{3}$
Hence all direction are cosine
$(a, b, c) = (\frac{2}{3}, \frac{1}{3}, - \frac{2}{3})$

COMEDK-2012

Vector Algebra

87873 The projections of a directed line segment on the coordinate axis are $12, 4, 3$ . The direction cosines of the line are

1

\frac{12}{13}, \frac{4}{13}, \frac{3}{13}

2

\frac{12}{13}, \frac{4}{13}, - \frac{3}{13}

3

- \frac{12}{13}, \frac{4}{13}, \frac{3}{13}

4

\frac{12}{13}, - \frac{4}{13}, - \frac{3}{13}

Explanation:

(A) : Assuming position vector of line segment ends points are $\vec{A}$ and $\vec{B}$
Now, given projection of line segment on the axis are $(12, 4, 3)$
So, $A B = 12 \hat{i} + 4 \hat{j} + 3 \hat{k}$
Now the direction cosine of line segment,
$= \frac{12}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{4}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{3}{\sqrt{12^{2} + 4^{2} + 3^{2}}}$
$= \frac{12}{13}, \frac{4}{13}, \frac{3}{13}$

AMU-2011

Vector Algebra

87880 Angle made by the position vector of the point $(5, - 4, - 3)$ with the positive direction of $X$ - axis is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{4}

4

\frac{π}{3}

Explanation:

(C): Let,
A $(5, - 4, - 3)$
Direction ratio of $OA = 5, - 4, - 3$
Direction ratio of $x -$ axis $= 1, 0, 0$
Angle, $θ = \frac{5 \cdot 1 + (- 4) \cdot 0 + (- 3) \cdot 0}{\sqrt{(5)^{2} + (- 4)^{2} + (- 3)^{2}} \sqrt{(1)^{2} + 0^{2} + 0^{2}}}$
$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$
$∴ θ = \frac{π}{4}$

AP EAMCET-2021-19.08.2021

Vector Algebra

87887 The direction cosines of the vector
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$ are

1

\frac{- 2}{\sqrt{8}}, \frac{1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

2

\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

3

\frac{2}{\sqrt{8}}, \frac{- 1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

4

\frac{- 2}{\sqrt{30}}, \frac{- 1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

Explanation:

(B) : Given that,
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$
$| \vec{a} | = \sqrt{(- 2)^{2} + (1)^{2} + (- 5)^{2}}$
$= \sqrt{4 + 1 + 25} = \sqrt{30}$
Direction cosine are $(l, m, n)$
$l = \frac{a_{1}}{| \vec{a} |}, m = \frac{a_{2}}{| \vec{a} |}, n = \frac{a_{3}}{| \vec{a} |}$
$∴$ Direction cosine are -

AP EAMCET-2020-21.09.2020

Vector Algebra

87888 If $a, b, c$ are any thr $(\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}})$
ee mutually
perpendicular vectors of equal magnitude a, then $| a + b + c |$

1 a

2

\sqrt{2} a

3

\sqrt{3} a

4

2 a

Explanation:

(C) : We know that,
$| a + b + c |^{2} = | a |^{2} + | b |^{2} + | c |^{2} + 2 (a . b + b . c + c . a)$
Let, $| \vec{a} | = | \vec{b} | = | \vec{c} | = a$
$| a + b + c |^{2} = a^{2} + a^{2} + a^{2} + 0 + 0 + 0$
$| a + b + c |^{2} = 3 a^{2}$
$a + b + c = \sqrt{3 a^{2}}$
$a + b + c = \sqrt{3} a$
Hence, $| \vec{a} + \vec{b} + \vec{c} | = \sqrt{3} a$

Manipal-2017

Vector Algebra

87872 The direction cosines of the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is equal to

1

⟨ \frac{2}{3}, \frac{1}{3}, - \frac{2}{3} ⟩

2

⟨ \frac{2}{3}, \frac{1}{3}, \frac{2}{3} ⟩

3

⟨ \frac{1}{3}, \frac{2}{3}, - \frac{2}{3} ⟩

4

⟨ \frac{2}{3}, \frac{2}{3}, \frac{1}{3} ⟩

Explanation:

(A) : If , $\hat{i}, \hat{j}, \hat{k}$ are the direction cosines and a, b, c are the direction ratio of line.
$2 \hat{i} + \hat{j} - 2 \hat{k}$
$a = 2, b = 1, c = - 2$
$\hat{i} = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{2}{\sqrt{4 + 1 + 4}} = \frac{2}{3}$
$\hat{j} = \frac{\underset{―}{b}}{a^{2} + b^{2} + c^{2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{3}$
$\hat{k} = \frac{\underset{―}{c}}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{\sqrt{4 + 1 + 4}} = \frac{- 2}{3}$
Hence all direction are cosine
$(a, b, c) = (\frac{2}{3}, \frac{1}{3}, - \frac{2}{3})$

COMEDK-2012

Vector Algebra

87873 The projections of a directed line segment on the coordinate axis are $12, 4, 3$ . The direction cosines of the line are

1

\frac{12}{13}, \frac{4}{13}, \frac{3}{13}

2

\frac{12}{13}, \frac{4}{13}, - \frac{3}{13}

3

- \frac{12}{13}, \frac{4}{13}, \frac{3}{13}

4

\frac{12}{13}, - \frac{4}{13}, - \frac{3}{13}

Explanation:

(A) : Assuming position vector of line segment ends points are $\vec{A}$ and $\vec{B}$
Now, given projection of line segment on the axis are $(12, 4, 3)$
So, $A B = 12 \hat{i} + 4 \hat{j} + 3 \hat{k}$
Now the direction cosine of line segment,
$= \frac{12}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{4}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{3}{\sqrt{12^{2} + 4^{2} + 3^{2}}}$
$= \frac{12}{13}, \frac{4}{13}, \frac{3}{13}$

AMU-2011

Vector Algebra

87880 Angle made by the position vector of the point $(5, - 4, - 3)$ with the positive direction of $X$ - axis is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{4}

4

\frac{π}{3}

Explanation:

(C): Let,
A $(5, - 4, - 3)$
Direction ratio of $OA = 5, - 4, - 3$
Direction ratio of $x -$ axis $= 1, 0, 0$
Angle, $θ = \frac{5 \cdot 1 + (- 4) \cdot 0 + (- 3) \cdot 0}{\sqrt{(5)^{2} + (- 4)^{2} + (- 3)^{2}} \sqrt{(1)^{2} + 0^{2} + 0^{2}}}$
$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$
$∴ θ = \frac{π}{4}$

AP EAMCET-2021-19.08.2021

Vector Algebra

87887 The direction cosines of the vector
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$ are

1

\frac{- 2}{\sqrt{8}}, \frac{1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

2

\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

3

\frac{2}{\sqrt{8}}, \frac{- 1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

4

\frac{- 2}{\sqrt{30}}, \frac{- 1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

Explanation:

(B) : Given that,
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$
$| \vec{a} | = \sqrt{(- 2)^{2} + (1)^{2} + (- 5)^{2}}$
$= \sqrt{4 + 1 + 25} = \sqrt{30}$
Direction cosine are $(l, m, n)$
$l = \frac{a_{1}}{| \vec{a} |}, m = \frac{a_{2}}{| \vec{a} |}, n = \frac{a_{3}}{| \vec{a} |}$
$∴$ Direction cosine are -

AP EAMCET-2020-21.09.2020

Vector Algebra

87888 If $a, b, c$ are any thr $(\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}})$
ee mutually
perpendicular vectors of equal magnitude a, then $| a + b + c |$

1 a

2

\sqrt{2} a

3

\sqrt{3} a

4

2 a

Explanation:

(C) : We know that,
$| a + b + c |^{2} = | a |^{2} + | b |^{2} + | c |^{2} + 2 (a . b + b . c + c . a)$
Let, $| \vec{a} | = | \vec{b} | = | \vec{c} | = a$
$| a + b + c |^{2} = a^{2} + a^{2} + a^{2} + 0 + 0 + 0$
$| a + b + c |^{2} = 3 a^{2}$
$a + b + c = \sqrt{3 a^{2}}$
$a + b + c = \sqrt{3} a$
Hence, $| \vec{a} + \vec{b} + \vec{c} | = \sqrt{3} a$

Manipal-2017

Vector Algebra

87872 The direction cosines of the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is equal to

1

⟨ \frac{2}{3}, \frac{1}{3}, - \frac{2}{3} ⟩

2

⟨ \frac{2}{3}, \frac{1}{3}, \frac{2}{3} ⟩

3

⟨ \frac{1}{3}, \frac{2}{3}, - \frac{2}{3} ⟩

4

⟨ \frac{2}{3}, \frac{2}{3}, \frac{1}{3} ⟩

Explanation:

(A) : If , $\hat{i}, \hat{j}, \hat{k}$ are the direction cosines and a, b, c are the direction ratio of line.
$2 \hat{i} + \hat{j} - 2 \hat{k}$
$a = 2, b = 1, c = - 2$
$\hat{i} = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{2}{\sqrt{4 + 1 + 4}} = \frac{2}{3}$
$\hat{j} = \frac{\underset{―}{b}}{a^{2} + b^{2} + c^{2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{3}$
$\hat{k} = \frac{\underset{―}{c}}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{\sqrt{4 + 1 + 4}} = \frac{- 2}{3}$
Hence all direction are cosine
$(a, b, c) = (\frac{2}{3}, \frac{1}{3}, - \frac{2}{3})$

COMEDK-2012

Vector Algebra

87873 The projections of a directed line segment on the coordinate axis are $12, 4, 3$ . The direction cosines of the line are

1

\frac{12}{13}, \frac{4}{13}, \frac{3}{13}

2

\frac{12}{13}, \frac{4}{13}, - \frac{3}{13}

3

- \frac{12}{13}, \frac{4}{13}, \frac{3}{13}

4

\frac{12}{13}, - \frac{4}{13}, - \frac{3}{13}

Explanation:

(A) : Assuming position vector of line segment ends points are $\vec{A}$ and $\vec{B}$
Now, given projection of line segment on the axis are $(12, 4, 3)$
So, $A B = 12 \hat{i} + 4 \hat{j} + 3 \hat{k}$
Now the direction cosine of line segment,
$= \frac{12}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{4}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{3}{\sqrt{12^{2} + 4^{2} + 3^{2}}}$
$= \frac{12}{13}, \frac{4}{13}, \frac{3}{13}$

AMU-2011

Vector Algebra

87880 Angle made by the position vector of the point $(5, - 4, - 3)$ with the positive direction of $X$ - axis is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{4}

4

\frac{π}{3}

Explanation:

(C): Let,
A $(5, - 4, - 3)$
Direction ratio of $OA = 5, - 4, - 3$
Direction ratio of $x -$ axis $= 1, 0, 0$
Angle, $θ = \frac{5 \cdot 1 + (- 4) \cdot 0 + (- 3) \cdot 0}{\sqrt{(5)^{2} + (- 4)^{2} + (- 3)^{2}} \sqrt{(1)^{2} + 0^{2} + 0^{2}}}$
$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$
$∴ θ = \frac{π}{4}$

AP EAMCET-2021-19.08.2021

Vector Algebra

87887 The direction cosines of the vector
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$ are

1

\frac{- 2}{\sqrt{8}}, \frac{1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

2

\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

3

\frac{2}{\sqrt{8}}, \frac{- 1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

4

\frac{- 2}{\sqrt{30}}, \frac{- 1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

Explanation:

(B) : Given that,
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$
$| \vec{a} | = \sqrt{(- 2)^{2} + (1)^{2} + (- 5)^{2}}$
$= \sqrt{4 + 1 + 25} = \sqrt{30}$
Direction cosine are $(l, m, n)$
$l = \frac{a_{1}}{| \vec{a} |}, m = \frac{a_{2}}{| \vec{a} |}, n = \frac{a_{3}}{| \vec{a} |}$
$∴$ Direction cosine are -

AP EAMCET-2020-21.09.2020

Vector Algebra

87888 If $a, b, c$ are any thr $(\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}})$
ee mutually
perpendicular vectors of equal magnitude a, then $| a + b + c |$

1 a

2

\sqrt{2} a

3

\sqrt{3} a

4

2 a

Explanation:

(C) : We know that,
$| a + b + c |^{2} = | a |^{2} + | b |^{2} + | c |^{2} + 2 (a . b + b . c + c . a)$
Let, $| \vec{a} | = | \vec{b} | = | \vec{c} | = a$
$| a + b + c |^{2} = a^{2} + a^{2} + a^{2} + 0 + 0 + 0$
$| a + b + c |^{2} = 3 a^{2}$
$a + b + c = \sqrt{3 a^{2}}$
$a + b + c = \sqrt{3} a$
Hence, $| \vec{a} + \vec{b} + \vec{c} | = \sqrt{3} a$

Manipal-2017

Vector Algebra

87872 The direction cosines of the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ is equal to

1

⟨ \frac{2}{3}, \frac{1}{3}, - \frac{2}{3} ⟩

2

⟨ \frac{2}{3}, \frac{1}{3}, \frac{2}{3} ⟩

3

⟨ \frac{1}{3}, \frac{2}{3}, - \frac{2}{3} ⟩

4

⟨ \frac{2}{3}, \frac{2}{3}, \frac{1}{3} ⟩

Explanation:

(A) : If , $\hat{i}, \hat{j}, \hat{k}$ are the direction cosines and a, b, c are the direction ratio of line.
$2 \hat{i} + \hat{j} - 2 \hat{k}$
$a = 2, b = 1, c = - 2$
$\hat{i} = \frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{2}{\sqrt{4 + 1 + 4}} = \frac{2}{3}$
$\hat{j} = \frac{\underset{―}{b}}{a^{2} + b^{2} + c^{2}} = \frac{1}{\sqrt{4 + 1 + 4}} = \frac{1}{3}$
$\hat{k} = \frac{\underset{―}{c}}{\sqrt{a^{2} + b^{2} + c^{2}}} = \frac{- 2}{\sqrt{4 + 1 + 4}} = \frac{- 2}{3}$
Hence all direction are cosine
$(a, b, c) = (\frac{2}{3}, \frac{1}{3}, - \frac{2}{3})$

COMEDK-2012

Vector Algebra

87873 The projections of a directed line segment on the coordinate axis are $12, 4, 3$ . The direction cosines of the line are

1

\frac{12}{13}, \frac{4}{13}, \frac{3}{13}

2

\frac{12}{13}, \frac{4}{13}, - \frac{3}{13}

3

- \frac{12}{13}, \frac{4}{13}, \frac{3}{13}

4

\frac{12}{13}, - \frac{4}{13}, - \frac{3}{13}

Explanation:

(A) : Assuming position vector of line segment ends points are $\vec{A}$ and $\vec{B}$
Now, given projection of line segment on the axis are $(12, 4, 3)$
So, $A B = 12 \hat{i} + 4 \hat{j} + 3 \hat{k}$
Now the direction cosine of line segment,
$= \frac{12}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{4}{\sqrt{12^{2} + 4^{2} + 3^{2}}}, \frac{3}{\sqrt{12^{2} + 4^{2} + 3^{2}}}$
$= \frac{12}{13}, \frac{4}{13}, \frac{3}{13}$

AMU-2011

Vector Algebra

87880 Angle made by the position vector of the point $(5, - 4, - 3)$ with the positive direction of $X$ - axis is

1

\frac{π}{2}

2

\frac{π}{6}

3

\frac{π}{4}

4

\frac{π}{3}

Explanation:

(C): Let,
A $(5, - 4, - 3)$
Direction ratio of $OA = 5, - 4, - 3$
Direction ratio of $x -$ axis $= 1, 0, 0$
Angle, $θ = \frac{5 \cdot 1 + (- 4) \cdot 0 + (- 3) \cdot 0}{\sqrt{(5)^{2} + (- 4)^{2} + (- 3)^{2}} \sqrt{(1)^{2} + 0^{2} + 0^{2}}}$
$= \frac{5}{\sqrt{50}} = \frac{1}{\sqrt{2}}$
$∴ θ = \frac{π}{4}$

AP EAMCET-2021-19.08.2021

Vector Algebra

87887 The direction cosines of the vector
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$ are

1

\frac{- 2}{\sqrt{8}}, \frac{1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

2

\frac{- 2}{\sqrt{30}}, \frac{1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

3

\frac{2}{\sqrt{8}}, \frac{- 1}{\sqrt{8}}, \frac{- 5}{\sqrt{8}}

4

\frac{- 2}{\sqrt{30}}, \frac{- 1}{\sqrt{30}}, \frac{- 5}{\sqrt{30}}

Explanation:

(B) : Given that,
$\vec{a} = - 2 \hat{i} + \hat{j} - 5 \hat{k}$
$| \vec{a} | = \sqrt{(- 2)^{2} + (1)^{2} + (- 5)^{2}}$
$= \sqrt{4 + 1 + 25} = \sqrt{30}$
Direction cosine are $(l, m, n)$
$l = \frac{a_{1}}{| \vec{a} |}, m = \frac{a_{2}}{| \vec{a} |}, n = \frac{a_{3}}{| \vec{a} |}$
$∴$ Direction cosine are -

AP EAMCET-2020-21.09.2020