QBManager

Motion in Plane

143640 Let $| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5$ and
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$ . The value of
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$ is

1 -106.5

2 -112.5

3 -99.5

4 -118.5

Explanation:

D Given that
$| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5, | {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = \sqrt{{| {\vec{A}}_{1} |}^{2} + {| {\vec{A}}_{2} |}^{2} + 2 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ}$ $5 = \sqrt{(3)^{2} + (5)^{2} + 2 \times 3 \times 5 \cos θ}$
$5 = \sqrt{9 + 25 + 2 \times 3 \times 5 \cos θ}$
Squaring both side
$25 = 9 + 25 + 2 \times 3 \times 5 \cos θ$
$\cos θ = \frac{- 9}{2 \times 3 \times 5}$
$\cos θ = \frac{- 3}{10}$
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$
$= 6 {| {\vec{A}}_{1} |}^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 {| {\vec{A}}_{2} |}^{2}$
$= 6 (3)^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ - 150$
$= 54 + 5 \times 3 \times 5 (\frac{- 3}{10}) - 150$
$= 54 - 150 - \frac{45}{2}$
$= - 118.5$

JEE Main-08.04.2019

Motion in Plane

143641 In the cube of side ' $a$ ' shown in the figure, the vector from the central point of the face $A B O D$ to the central point of the face BEFO will be

1

\frac{1}{2} a (\hat{i} - \hat{k})

2

\frac{1}{2} a (\hat{j} - \hat{i})

3

\frac{1}{2} a (\hat{j} - \hat{k})

4

\frac{1}{2} a (\hat{k} - \hat{i})

Explanation:

B
Position vector of $G$ is, $G (\frac{a}{2}, 0, \frac{a}{2})$
$\vec{OG} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{k}$
Position vector of $H$ is, $H (0, \frac{a}{2}, \frac{a}{2})$
$\vec{OH} = \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}$
$\vec{GH} = \vec{OH} - \vec{OG}$
$\vec{GH} = (\frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}) - (\frac{a}{2} \hat{i} + \frac{a}{2} \hat{k})$
$\vec{GH} = \frac{a}{2} (\hat{j} - \hat{i})$

JEE Main-10.01.2019

Motion in Plane

143642 If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$ , then the angle between $A$ and $B$ is

1

π

2

π / 3

3

π / 2

4

π / 4

Explanation:

A $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$
$\vec{A} \times \vec{B} = - (\vec{A} \times \vec{B})$
$AB \sin θ = - AB \sin θ$
$2 AB \sin θ = 0$
$\sin θ = 0$
$θ = 0, π, 2 π$

AIEEE-2004

Motion in Plane

143643 $\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$ Find a vector parallel to $\bar{A}$ whose magnitude equal to that of $\overset{―}{B}$

1

\sqrt{\frac{46}{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

2

\sqrt{\frac{46}{29}} (6 \hat{i} - \hat{j} + 3 \hat{k})

3

\sqrt{\frac{29}{46}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

4

\sqrt{\frac{29}{46}} (6 \hat{i} - \hat{j} + 3 \hat{k})

Explanation:

A Givne that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}$
$\vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$
Let $X$ be the vector parallel to $\vec{A}$ whose magnitude is equal to that of $\vec{B}$
$\vec{X} = \frac{\vec{A} | \vec{B} |}{| \vec{A} |}$
$= \frac{3 \hat{i} + 4 \hat{j} + 2 \hat{k}}{\sqrt{9 + 16 + 4}} \sqrt{36 + 1 + 9}$
$\vec{X} = \frac{\sqrt{46}}{\sqrt{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})$

Assam CEE-2020

Motion in Plane

143640 Let $| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5$ and
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$ . The value of
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$ is

1 -106.5

2 -112.5

3 -99.5

4 -118.5

Explanation:

D Given that
$| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5, | {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = \sqrt{{| {\vec{A}}_{1} |}^{2} + {| {\vec{A}}_{2} |}^{2} + 2 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ}$ $5 = \sqrt{(3)^{2} + (5)^{2} + 2 \times 3 \times 5 \cos θ}$
$5 = \sqrt{9 + 25 + 2 \times 3 \times 5 \cos θ}$
Squaring both side
$25 = 9 + 25 + 2 \times 3 \times 5 \cos θ$
$\cos θ = \frac{- 9}{2 \times 3 \times 5}$
$\cos θ = \frac{- 3}{10}$
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$
$= 6 {| {\vec{A}}_{1} |}^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 {| {\vec{A}}_{2} |}^{2}$
$= 6 (3)^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ - 150$
$= 54 + 5 \times 3 \times 5 (\frac{- 3}{10}) - 150$
$= 54 - 150 - \frac{45}{2}$
$= - 118.5$

JEE Main-08.04.2019

Motion in Plane

143641 In the cube of side ' $a$ ' shown in the figure, the vector from the central point of the face $A B O D$ to the central point of the face BEFO will be

1

\frac{1}{2} a (\hat{i} - \hat{k})

2

\frac{1}{2} a (\hat{j} - \hat{i})

3

\frac{1}{2} a (\hat{j} - \hat{k})

4

\frac{1}{2} a (\hat{k} - \hat{i})

Explanation:

B
Position vector of $G$ is, $G (\frac{a}{2}, 0, \frac{a}{2})$
$\vec{OG} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{k}$
Position vector of $H$ is, $H (0, \frac{a}{2}, \frac{a}{2})$
$\vec{OH} = \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}$
$\vec{GH} = \vec{OH} - \vec{OG}$
$\vec{GH} = (\frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}) - (\frac{a}{2} \hat{i} + \frac{a}{2} \hat{k})$
$\vec{GH} = \frac{a}{2} (\hat{j} - \hat{i})$

JEE Main-10.01.2019

Motion in Plane

143642 If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$ , then the angle between $A$ and $B$ is

1

π

2

π / 3

3

π / 2

4

π / 4

Explanation:

A $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$
$\vec{A} \times \vec{B} = - (\vec{A} \times \vec{B})$
$AB \sin θ = - AB \sin θ$
$2 AB \sin θ = 0$
$\sin θ = 0$
$θ = 0, π, 2 π$

AIEEE-2004

Motion in Plane

143643 $\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$ Find a vector parallel to $\bar{A}$ whose magnitude equal to that of $\overset{―}{B}$

1

\sqrt{\frac{46}{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

2

\sqrt{\frac{46}{29}} (6 \hat{i} - \hat{j} + 3 \hat{k})

3

\sqrt{\frac{29}{46}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

4

\sqrt{\frac{29}{46}} (6 \hat{i} - \hat{j} + 3 \hat{k})

Explanation:

A Givne that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}$
$\vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$
Let $X$ be the vector parallel to $\vec{A}$ whose magnitude is equal to that of $\vec{B}$
$\vec{X} = \frac{\vec{A} | \vec{B} |}{| \vec{A} |}$
$= \frac{3 \hat{i} + 4 \hat{j} + 2 \hat{k}}{\sqrt{9 + 16 + 4}} \sqrt{36 + 1 + 9}$
$\vec{X} = \frac{\sqrt{46}}{\sqrt{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})$

Assam CEE-2020

Motion in Plane

143640 Let $| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5$ and
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$ . The value of
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$ is

1 -106.5

2 -112.5

3 -99.5

4 -118.5

Explanation:

D Given that
$| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5, | {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = \sqrt{{| {\vec{A}}_{1} |}^{2} + {| {\vec{A}}_{2} |}^{2} + 2 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ}$ $5 = \sqrt{(3)^{2} + (5)^{2} + 2 \times 3 \times 5 \cos θ}$
$5 = \sqrt{9 + 25 + 2 \times 3 \times 5 \cos θ}$
Squaring both side
$25 = 9 + 25 + 2 \times 3 \times 5 \cos θ$
$\cos θ = \frac{- 9}{2 \times 3 \times 5}$
$\cos θ = \frac{- 3}{10}$
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$
$= 6 {| {\vec{A}}_{1} |}^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 {| {\vec{A}}_{2} |}^{2}$
$= 6 (3)^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ - 150$
$= 54 + 5 \times 3 \times 5 (\frac{- 3}{10}) - 150$
$= 54 - 150 - \frac{45}{2}$
$= - 118.5$

JEE Main-08.04.2019

Motion in Plane

143641 In the cube of side ' $a$ ' shown in the figure, the vector from the central point of the face $A B O D$ to the central point of the face BEFO will be

1

\frac{1}{2} a (\hat{i} - \hat{k})

2

\frac{1}{2} a (\hat{j} - \hat{i})

3

\frac{1}{2} a (\hat{j} - \hat{k})

4

\frac{1}{2} a (\hat{k} - \hat{i})

Explanation:

B
Position vector of $G$ is, $G (\frac{a}{2}, 0, \frac{a}{2})$
$\vec{OG} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{k}$
Position vector of $H$ is, $H (0, \frac{a}{2}, \frac{a}{2})$
$\vec{OH} = \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}$
$\vec{GH} = \vec{OH} - \vec{OG}$
$\vec{GH} = (\frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}) - (\frac{a}{2} \hat{i} + \frac{a}{2} \hat{k})$
$\vec{GH} = \frac{a}{2} (\hat{j} - \hat{i})$

JEE Main-10.01.2019

Motion in Plane

143642 If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$ , then the angle between $A$ and $B$ is

1

π

2

π / 3

3

π / 2

4

π / 4

Explanation:

A $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$
$\vec{A} \times \vec{B} = - (\vec{A} \times \vec{B})$
$AB \sin θ = - AB \sin θ$
$2 AB \sin θ = 0$
$\sin θ = 0$
$θ = 0, π, 2 π$

AIEEE-2004

Motion in Plane

143643 $\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$ Find a vector parallel to $\bar{A}$ whose magnitude equal to that of $\overset{―}{B}$

1

\sqrt{\frac{46}{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

2

\sqrt{\frac{46}{29}} (6 \hat{i} - \hat{j} + 3 \hat{k})

3

\sqrt{\frac{29}{46}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

4

\sqrt{\frac{29}{46}} (6 \hat{i} - \hat{j} + 3 \hat{k})

Explanation:

A Givne that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}$
$\vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$
Let $X$ be the vector parallel to $\vec{A}$ whose magnitude is equal to that of $\vec{B}$
$\vec{X} = \frac{\vec{A} | \vec{B} |}{| \vec{A} |}$
$= \frac{3 \hat{i} + 4 \hat{j} + 2 \hat{k}}{\sqrt{9 + 16 + 4}} \sqrt{36 + 1 + 9}$
$\vec{X} = \frac{\sqrt{46}}{\sqrt{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})$

Assam CEE-2020

Motion in Plane

143640 Let $| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5$ and
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$ . The value of
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$ is

1 -106.5

2 -112.5

3 -99.5

4 -118.5

Explanation:

D Given that
$| {\vec{A}}_{1} | = 3, | {\vec{A}}_{2} | = 5, | {\vec{A}}_{1} + {\vec{A}}_{2} | = 5$
$| {\vec{A}}_{1} + {\vec{A}}_{2} | = \sqrt{{| {\vec{A}}_{1} |}^{2} + {| {\vec{A}}_{2} |}^{2} + 2 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ}$ $5 = \sqrt{(3)^{2} + (5)^{2} + 2 \times 3 \times 5 \cos θ}$
$5 = \sqrt{9 + 25 + 2 \times 3 \times 5 \cos θ}$
Squaring both side
$25 = 9 + 25 + 2 \times 3 \times 5 \cos θ$
$\cos θ = \frac{- 9}{2 \times 3 \times 5}$
$\cos θ = \frac{- 3}{10}$
$(2 {\vec{A}}_{1} + 3 {\vec{A}}_{2}) \cdot (3 {\vec{A}}_{1} - 2 {\vec{A}}_{2})$
$= 6 {| {\vec{A}}_{1} |}^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 {| {\vec{A}}_{2} |}^{2}$
$= 6 (3)^{2} + 9 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 4 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 {\vec{A}}_{1} \cdot {\vec{A}}_{2} - 6 \times 25$
$= 54 + 5 | {\vec{A}}_{1} | | {\vec{A}}_{2} | \cos θ - 150$
$= 54 + 5 \times 3 \times 5 (\frac{- 3}{10}) - 150$
$= 54 - 150 - \frac{45}{2}$
$= - 118.5$

JEE Main-08.04.2019

Motion in Plane

143641 In the cube of side ' $a$ ' shown in the figure, the vector from the central point of the face $A B O D$ to the central point of the face BEFO will be

1

\frac{1}{2} a (\hat{i} - \hat{k})

2

\frac{1}{2} a (\hat{j} - \hat{i})

3

\frac{1}{2} a (\hat{j} - \hat{k})

4

\frac{1}{2} a (\hat{k} - \hat{i})

Explanation:

B
Position vector of $G$ is, $G (\frac{a}{2}, 0, \frac{a}{2})$
$\vec{OG} = \frac{a}{2} \hat{i} + \frac{a}{2} \hat{k}$
Position vector of $H$ is, $H (0, \frac{a}{2}, \frac{a}{2})$
$\vec{OH} = \frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}$
$\vec{GH} = \vec{OH} - \vec{OG}$
$\vec{GH} = (\frac{a}{2} \hat{j} + \frac{a}{2} \hat{k}) - (\frac{a}{2} \hat{i} + \frac{a}{2} \hat{k})$
$\vec{GH} = \frac{a}{2} (\hat{j} - \hat{i})$

JEE Main-10.01.2019

Motion in Plane

143642 If $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$ , then the angle between $A$ and $B$ is

1

π

2

π / 3

3

π / 2

4

π / 4

Explanation:

A $\vec{A} \times \vec{B} = \vec{B} \times \vec{A}$
$\vec{A} \times \vec{B} = - (\vec{A} \times \vec{B})$
$AB \sin θ = - AB \sin θ$
$2 AB \sin θ = 0$
$\sin θ = 0$
$θ = 0, π, 2 π$

AIEEE-2004

Motion in Plane

143643 $\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$ Find a vector parallel to $\bar{A}$ whose magnitude equal to that of $\overset{―}{B}$

1

\sqrt{\frac{46}{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

2

\sqrt{\frac{46}{29}} (6 \hat{i} - \hat{j} + 3 \hat{k})

3

\sqrt{\frac{29}{46}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})

4

\sqrt{\frac{29}{46}} (6 \hat{i} - \hat{j} + 3 \hat{k})

Explanation:

A Givne that,
$\vec{A} = 3 \hat{i} + 4 \hat{j} + 2 \hat{k}$
$\vec{B} = 6 \hat{i} - \hat{j} + 3 \hat{k}$
Let $X$ be the vector parallel to $\vec{A}$ whose magnitude is equal to that of $\vec{B}$
$\vec{X} = \frac{\vec{A} | \vec{B} |}{| \vec{A} |}$
$= \frac{3 \hat{i} + 4 \hat{j} + 2 \hat{k}}{\sqrt{9 + 16 + 4}} \sqrt{36 + 1 + 9}$
$\vec{X} = \frac{\sqrt{46}}{\sqrt{29}} (3 \hat{i} + 4 \hat{j} + 2 \hat{k})$

Assam CEE-2020

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Question Bank Series

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