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VECTORS

268984 Three forces ${\vec{F}}_{1} = a (\hat{i} - \hat{j} + \hat{k}), {\vec{F}}_{2} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and ${\vec{F}}_{3} = 8 \hat{i} - 7 \hat{j} + 6 \hat{k}$ act simultaneously on a particle. If the particle is in equilibrium, the value of $a$ is

1 10

2 -10

3 8

4 2

Explanation:

${\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0$

VECTORS

268985 If a particle is displaced from $(0, 0, 0)$ to a point in XY - plane which is at a distance of 4 units in a direction making an angle clock wise $60^{\circ}$ with the negative $x$ -axis. What is the final position vector of the particle?

1

- 2 \hat{i} + 2 \sqrt{3} \hat{j}

2

2 \hat{i} + 2 \sqrt{3} \hat{j}

3

2 \hat{j} + 2 \sqrt{3} \hat{k}

4

2 \sqrt{3} \hat{j} + 2 \hat{k}

Explanation:

$(x, y, z) = (- 4 \cos 60^{\circ}, 4 \sin 60^{\circ}, 0)$

VECTORS

268986 Cosines of angles made by a vector with $X, Y$ axes are $3 / 5 \sqrt{2}, 4 / 5 \sqrt{2}$ respectively. If the magnitude of the vector is $10 \sqrt{2}$ then that vector is

1

8 \hat{i} + 6 \hat{j} - 10 \hat{k}

2

6 \hat{i} - 8 \hat{j} - 10 \hat{k}

3

- 6 \hat{i} - 8 \hat{j} + 10 \hat{k}

4

6 \hat{i} + 8 \hat{j} + 10 \hat{k}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
where $\cos α = \frac{A_{x}}{A}, \cos β = \frac{A_{y}}{A}, \cos γ = \frac{A_{z}}{A}$
$∴ \vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$

VECTORS

268987 If a vector $\vec{A}$ makes angles $45^{\circ}$ and $60^{\circ}$ with $x$ and $y$ axis respectively then the angle made by it with $z$ - axis is

1

30^{0}

2

60^{\circ}

3

90^{0}

4

120^{0}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

VECTORS

268988 A vector $\vec{Q}$ which has a magnitude of 8 is added to the vector $\vec{P}$ which lies along the $X$ -axis. The resultant of these two vectors is a third vector $\vec{R}$ which lies along the $Y$ -axis and has a magnitude twice that of $\vec{P}$ . The magnitude of $\vec{P}$ is

1

\frac{6}{\sqrt{5}}

2

\frac{8}{\sqrt{5}}

3

\frac{12}{\sqrt{5}}

4

\frac{16}{\sqrt{5}}

Explanation:

$| \vec{Q} | = 8, \vec{P} + \vec{Q} = \vec{R}$
$\begin{aligned} R_{y} = 2 P_{x}; (P_{x} + Q_{x}) + Q_{y} = R_{y} \\ Q_{x}^{2} + Q_{y}^{2} = 64 \Rightarrow P_{x}^{2} + 4 P_{x}^{2} = 64 \end{aligned}$

VECTORS

268984 Three forces ${\vec{F}}_{1} = a (\hat{i} - \hat{j} + \hat{k}), {\vec{F}}_{2} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and ${\vec{F}}_{3} = 8 \hat{i} - 7 \hat{j} + 6 \hat{k}$ act simultaneously on a particle. If the particle is in equilibrium, the value of $a$ is

1 10

2 -10

3 8

4 2

Explanation:

${\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0$

VECTORS

268985 If a particle is displaced from $(0, 0, 0)$ to a point in XY - plane which is at a distance of 4 units in a direction making an angle clock wise $60^{\circ}$ with the negative $x$ -axis. What is the final position vector of the particle?

1

- 2 \hat{i} + 2 \sqrt{3} \hat{j}

2

2 \hat{i} + 2 \sqrt{3} \hat{j}

3

2 \hat{j} + 2 \sqrt{3} \hat{k}

4

2 \sqrt{3} \hat{j} + 2 \hat{k}

Explanation:

$(x, y, z) = (- 4 \cos 60^{\circ}, 4 \sin 60^{\circ}, 0)$

VECTORS

268986 Cosines of angles made by a vector with $X, Y$ axes are $3 / 5 \sqrt{2}, 4 / 5 \sqrt{2}$ respectively. If the magnitude of the vector is $10 \sqrt{2}$ then that vector is

1

8 \hat{i} + 6 \hat{j} - 10 \hat{k}

2

6 \hat{i} - 8 \hat{j} - 10 \hat{k}

3

- 6 \hat{i} - 8 \hat{j} + 10 \hat{k}

4

6 \hat{i} + 8 \hat{j} + 10 \hat{k}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
where $\cos α = \frac{A_{x}}{A}, \cos β = \frac{A_{y}}{A}, \cos γ = \frac{A_{z}}{A}$
$∴ \vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$

VECTORS

268987 If a vector $\vec{A}$ makes angles $45^{\circ}$ and $60^{\circ}$ with $x$ and $y$ axis respectively then the angle made by it with $z$ - axis is

1

30^{0}

2

60^{\circ}

3

90^{0}

4

120^{0}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

VECTORS

268988 A vector $\vec{Q}$ which has a magnitude of 8 is added to the vector $\vec{P}$ which lies along the $X$ -axis. The resultant of these two vectors is a third vector $\vec{R}$ which lies along the $Y$ -axis and has a magnitude twice that of $\vec{P}$ . The magnitude of $\vec{P}$ is

1

\frac{6}{\sqrt{5}}

2

\frac{8}{\sqrt{5}}

3

\frac{12}{\sqrt{5}}

4

\frac{16}{\sqrt{5}}

Explanation:

$| \vec{Q} | = 8, \vec{P} + \vec{Q} = \vec{R}$
$\begin{aligned} R_{y} = 2 P_{x}; (P_{x} + Q_{x}) + Q_{y} = R_{y} \\ Q_{x}^{2} + Q_{y}^{2} = 64 \Rightarrow P_{x}^{2} + 4 P_{x}^{2} = 64 \end{aligned}$

VECTORS

268984 Three forces ${\vec{F}}_{1} = a (\hat{i} - \hat{j} + \hat{k}), {\vec{F}}_{2} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and ${\vec{F}}_{3} = 8 \hat{i} - 7 \hat{j} + 6 \hat{k}$ act simultaneously on a particle. If the particle is in equilibrium, the value of $a$ is

1 10

2 -10

3 8

4 2

Explanation:

${\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0$

VECTORS

268985 If a particle is displaced from $(0, 0, 0)$ to a point in XY - plane which is at a distance of 4 units in a direction making an angle clock wise $60^{\circ}$ with the negative $x$ -axis. What is the final position vector of the particle?

1

- 2 \hat{i} + 2 \sqrt{3} \hat{j}

2

2 \hat{i} + 2 \sqrt{3} \hat{j}

3

2 \hat{j} + 2 \sqrt{3} \hat{k}

4

2 \sqrt{3} \hat{j} + 2 \hat{k}

Explanation:

$(x, y, z) = (- 4 \cos 60^{\circ}, 4 \sin 60^{\circ}, 0)$

VECTORS

268986 Cosines of angles made by a vector with $X, Y$ axes are $3 / 5 \sqrt{2}, 4 / 5 \sqrt{2}$ respectively. If the magnitude of the vector is $10 \sqrt{2}$ then that vector is

1

8 \hat{i} + 6 \hat{j} - 10 \hat{k}

2

6 \hat{i} - 8 \hat{j} - 10 \hat{k}

3

- 6 \hat{i} - 8 \hat{j} + 10 \hat{k}

4

6 \hat{i} + 8 \hat{j} + 10 \hat{k}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
where $\cos α = \frac{A_{x}}{A}, \cos β = \frac{A_{y}}{A}, \cos γ = \frac{A_{z}}{A}$
$∴ \vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$

VECTORS

268987 If a vector $\vec{A}$ makes angles $45^{\circ}$ and $60^{\circ}$ with $x$ and $y$ axis respectively then the angle made by it with $z$ - axis is

1

30^{0}

2

60^{\circ}

3

90^{0}

4

120^{0}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

VECTORS

268988 A vector $\vec{Q}$ which has a magnitude of 8 is added to the vector $\vec{P}$ which lies along the $X$ -axis. The resultant of these two vectors is a third vector $\vec{R}$ which lies along the $Y$ -axis and has a magnitude twice that of $\vec{P}$ . The magnitude of $\vec{P}$ is

1

\frac{6}{\sqrt{5}}

2

\frac{8}{\sqrt{5}}

3

\frac{12}{\sqrt{5}}

4

\frac{16}{\sqrt{5}}

Explanation:

$| \vec{Q} | = 8, \vec{P} + \vec{Q} = \vec{R}$
$\begin{aligned} R_{y} = 2 P_{x}; (P_{x} + Q_{x}) + Q_{y} = R_{y} \\ Q_{x}^{2} + Q_{y}^{2} = 64 \Rightarrow P_{x}^{2} + 4 P_{x}^{2} = 64 \end{aligned}$

VECTORS

268984 Three forces ${\vec{F}}_{1} = a (\hat{i} - \hat{j} + \hat{k}), {\vec{F}}_{2} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and ${\vec{F}}_{3} = 8 \hat{i} - 7 \hat{j} + 6 \hat{k}$ act simultaneously on a particle. If the particle is in equilibrium, the value of $a$ is

1 10

2 -10

3 8

4 2

Explanation:

${\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0$

VECTORS

268985 If a particle is displaced from $(0, 0, 0)$ to a point in XY - plane which is at a distance of 4 units in a direction making an angle clock wise $60^{\circ}$ with the negative $x$ -axis. What is the final position vector of the particle?

1

- 2 \hat{i} + 2 \sqrt{3} \hat{j}

2

2 \hat{i} + 2 \sqrt{3} \hat{j}

3

2 \hat{j} + 2 \sqrt{3} \hat{k}

4

2 \sqrt{3} \hat{j} + 2 \hat{k}

Explanation:

$(x, y, z) = (- 4 \cos 60^{\circ}, 4 \sin 60^{\circ}, 0)$

VECTORS

268986 Cosines of angles made by a vector with $X, Y$ axes are $3 / 5 \sqrt{2}, 4 / 5 \sqrt{2}$ respectively. If the magnitude of the vector is $10 \sqrt{2}$ then that vector is

1

8 \hat{i} + 6 \hat{j} - 10 \hat{k}

2

6 \hat{i} - 8 \hat{j} - 10 \hat{k}

3

- 6 \hat{i} - 8 \hat{j} + 10 \hat{k}

4

6 \hat{i} + 8 \hat{j} + 10 \hat{k}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
where $\cos α = \frac{A_{x}}{A}, \cos β = \frac{A_{y}}{A}, \cos γ = \frac{A_{z}}{A}$
$∴ \vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$

VECTORS

268987 If a vector $\vec{A}$ makes angles $45^{\circ}$ and $60^{\circ}$ with $x$ and $y$ axis respectively then the angle made by it with $z$ - axis is

1

30^{0}

2

60^{\circ}

3

90^{0}

4

120^{0}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

VECTORS

268988 A vector $\vec{Q}$ which has a magnitude of 8 is added to the vector $\vec{P}$ which lies along the $X$ -axis. The resultant of these two vectors is a third vector $\vec{R}$ which lies along the $Y$ -axis and has a magnitude twice that of $\vec{P}$ . The magnitude of $\vec{P}$ is

1

\frac{6}{\sqrt{5}}

2

\frac{8}{\sqrt{5}}

3

\frac{12}{\sqrt{5}}

4

\frac{16}{\sqrt{5}}

Explanation:

$| \vec{Q} | = 8, \vec{P} + \vec{Q} = \vec{R}$
$\begin{aligned} R_{y} = 2 P_{x}; (P_{x} + Q_{x}) + Q_{y} = R_{y} \\ Q_{x}^{2} + Q_{y}^{2} = 64 \Rightarrow P_{x}^{2} + 4 P_{x}^{2} = 64 \end{aligned}$

VECTORS

268984 Three forces ${\vec{F}}_{1} = a (\hat{i} - \hat{j} + \hat{k}), {\vec{F}}_{2} = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and ${\vec{F}}_{3} = 8 \hat{i} - 7 \hat{j} + 6 \hat{k}$ act simultaneously on a particle. If the particle is in equilibrium, the value of $a$ is

1 10

2 -10

3 8

4 2

Explanation:

${\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = 0$

VECTORS

268985 If a particle is displaced from $(0, 0, 0)$ to a point in XY - plane which is at a distance of 4 units in a direction making an angle clock wise $60^{\circ}$ with the negative $x$ -axis. What is the final position vector of the particle?

1

- 2 \hat{i} + 2 \sqrt{3} \hat{j}

2

2 \hat{i} + 2 \sqrt{3} \hat{j}

3

2 \hat{j} + 2 \sqrt{3} \hat{k}

4

2 \sqrt{3} \hat{j} + 2 \hat{k}

Explanation:

$(x, y, z) = (- 4 \cos 60^{\circ}, 4 \sin 60^{\circ}, 0)$

VECTORS

268986 Cosines of angles made by a vector with $X, Y$ axes are $3 / 5 \sqrt{2}, 4 / 5 \sqrt{2}$ respectively. If the magnitude of the vector is $10 \sqrt{2}$ then that vector is

1

8 \hat{i} + 6 \hat{j} - 10 \hat{k}

2

6 \hat{i} - 8 \hat{j} - 10 \hat{k}

3

- 6 \hat{i} - 8 \hat{j} + 10 \hat{k}

4

6 \hat{i} + 8 \hat{j} + 10 \hat{k}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$
where $\cos α = \frac{A_{x}}{A}, \cos β = \frac{A_{y}}{A}, \cos γ = \frac{A_{z}}{A}$
$∴ \vec{A} = A_{x} \hat{i} + A_{y} \hat{j} + A_{z} \hat{k}$

VECTORS

268987 If a vector $\vec{A}$ makes angles $45^{\circ}$ and $60^{\circ}$ with $x$ and $y$ axis respectively then the angle made by it with $z$ - axis is

1

30^{0}

2

60^{\circ}

3

90^{0}

4

120^{0}

Explanation:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = 1$

VECTORS

268988 A vector $\vec{Q}$ which has a magnitude of 8 is added to the vector $\vec{P}$ which lies along the $X$ -axis. The resultant of these two vectors is a third vector $\vec{R}$ which lies along the $Y$ -axis and has a magnitude twice that of $\vec{P}$ . The magnitude of $\vec{P}$ is

1

\frac{6}{\sqrt{5}}

2

\frac{8}{\sqrt{5}}

3

\frac{12}{\sqrt{5}}

4

\frac{16}{\sqrt{5}}

Explanation:

$| \vec{Q} | = 8, \vec{P} + \vec{Q} = \vec{R}$
$\begin{aligned} R_{y} = 2 P_{x}; (P_{x} + Q_{x}) + Q_{y} = R_{y} \\ Q_{x}^{2} + Q_{y}^{2} = 64 \Rightarrow P_{x}^{2} + 4 P_{x}^{2} = 64 \end{aligned}$

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