QBManager

Vector Algebra

87966 If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} =$

1

\frac{3}{2}

2

- \frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

(B) : Given,
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = ?$
We know that
$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0$
$| \vec{a} |^{2} + 2 \vec{a} \cdot \vec{b} + 2 c \cdot a + | \vec{b} |^{2} + 2 \vec{b} \cdot \vec{c} + | \vec{c} |^{2} = 0$
$∵ Given, | \vec{a} | = | \vec{b} | = | \vec{c} | = 1$
$1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 3$
$(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = \frac{- 3}{2}$

Karnataka CET-2013

Vector Algebra

87967 If $(\vec{a} \times \vec{b})^{2} + (a . \vec{b})^{2} = 144$ and $| \vec{a} | = 4$ , then $| \vec{b} | =$

1 12

2 16

3 8

4 3

Explanation:

(D) : Given,
$| \vec{a} | = 4, | \vec{b} | = ?$
$(\vec{a} \times \vec{b})^{2} + (\vec{a} \cdot \vec{b})^{2} = 144$
We know that
$\vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin θ$
$\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ$
$| \vec{a} |^{2} | \vec{b} |^{2} \sin^{2} θ + | \vec{a} |^{2} | \vec{b} |^{2} \cos^{2} θ = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} (\sin^{2} θ + \cos^{2} θ) = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} = 144$
$| \vec{a} | | \vec{b} | = 12$
$4 | \vec{b} | = 12$
$| \vec{b} | = 3$

Karnataka CET-2012

Vector Algebra

87969 The projection of $\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

1

\frac{8}{\sqrt{14}}

2

\sqrt{14}

3

\frac{8}{\sqrt{35}}

4

\frac{8}{\sqrt{39}}

Explanation:

(A) : Given,
$\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
We know that
The projection of $\vec{a}$ and $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$\vec{a} \cdot \vec{b} = (3 \hat{i} - \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k})$
$= 6 \hat{i} \cdot \hat{i} - 3 \hat{j} \cdot \hat{j} + 5 \hat{k} \cdot \hat{k} (∵ \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1)$
$∴ \vec{a} \cdot \vec{b} = 8$
$\vec{b} = 2 \vec{i} + 3 \vec{j} + \vec{k}$
$| \vec{b} | = \sqrt{2^{2} + 3^{2} + 1} = \sqrt{14}$
Projection of $\vec{a}$ and $\vec{b} = \frac{8}{\sqrt{14}}$

Karnataka CET-2008

Vector Algebra

87970 If $\vec{a}$ is vector perpendicular to both $\vec{b}$ and $\vec{c}$ , then

1

\vec{a} + (\vec{b} + \vec{c}) = \vec{0}

2

\vec{a} \times (\vec{b} + \vec{c}) = \vec{0}

3

\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}

4

\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{0}

Explanation:

(C) : Given,
$\vec{a} is perpendicular to both \vec{b} and \vec{c}$
$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \cdot \vec{b} - (\vec{a} \cdot \vec{b}) \cdot \vec{c}$
$∵ (\vec{a} \cdot \vec{b} = \vec{0}, \vec{a} \cdot \vec{c} = \vec{0})$
$= \vec{0} - \vec{0}$
$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}$

Karnataka CET-2006

Vector Algebra

87966 If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} =$

1

\frac{3}{2}

2

- \frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

(B) : Given,
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = ?$
We know that
$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0$
$| \vec{a} |^{2} + 2 \vec{a} \cdot \vec{b} + 2 c \cdot a + | \vec{b} |^{2} + 2 \vec{b} \cdot \vec{c} + | \vec{c} |^{2} = 0$
$∵ Given, | \vec{a} | = | \vec{b} | = | \vec{c} | = 1$
$1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 3$
$(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = \frac{- 3}{2}$

Karnataka CET-2013

Vector Algebra

87967 If $(\vec{a} \times \vec{b})^{2} + (a . \vec{b})^{2} = 144$ and $| \vec{a} | = 4$ , then $| \vec{b} | =$

1 12

2 16

3 8

4 3

Explanation:

(D) : Given,
$| \vec{a} | = 4, | \vec{b} | = ?$
$(\vec{a} \times \vec{b})^{2} + (\vec{a} \cdot \vec{b})^{2} = 144$
We know that
$\vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin θ$
$\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ$
$| \vec{a} |^{2} | \vec{b} |^{2} \sin^{2} θ + | \vec{a} |^{2} | \vec{b} |^{2} \cos^{2} θ = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} (\sin^{2} θ + \cos^{2} θ) = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} = 144$
$| \vec{a} | | \vec{b} | = 12$
$4 | \vec{b} | = 12$
$| \vec{b} | = 3$

Karnataka CET-2012

Vector Algebra

87969 The projection of $\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

1

\frac{8}{\sqrt{14}}

2

\sqrt{14}

3

\frac{8}{\sqrt{35}}

4

\frac{8}{\sqrt{39}}

Explanation:

(A) : Given,
$\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
We know that
The projection of $\vec{a}$ and $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$\vec{a} \cdot \vec{b} = (3 \hat{i} - \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k})$
$= 6 \hat{i} \cdot \hat{i} - 3 \hat{j} \cdot \hat{j} + 5 \hat{k} \cdot \hat{k} (∵ \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1)$
$∴ \vec{a} \cdot \vec{b} = 8$
$\vec{b} = 2 \vec{i} + 3 \vec{j} + \vec{k}$
$| \vec{b} | = \sqrt{2^{2} + 3^{2} + 1} = \sqrt{14}$
Projection of $\vec{a}$ and $\vec{b} = \frac{8}{\sqrt{14}}$

Karnataka CET-2008

Vector Algebra

87970 If $\vec{a}$ is vector perpendicular to both $\vec{b}$ and $\vec{c}$ , then

1

\vec{a} + (\vec{b} + \vec{c}) = \vec{0}

2

\vec{a} \times (\vec{b} + \vec{c}) = \vec{0}

3

\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}

4

\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{0}

Explanation:

(C) : Given,
$\vec{a} is perpendicular to both \vec{b} and \vec{c}$
$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \cdot \vec{b} - (\vec{a} \cdot \vec{b}) \cdot \vec{c}$
$∵ (\vec{a} \cdot \vec{b} = \vec{0}, \vec{a} \cdot \vec{c} = \vec{0})$
$= \vec{0} - \vec{0}$
$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}$

Karnataka CET-2006

Vector Algebra

87966 If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} =$

1

\frac{3}{2}

2

- \frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

(B) : Given,
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = ?$
We know that
$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0$
$| \vec{a} |^{2} + 2 \vec{a} \cdot \vec{b} + 2 c \cdot a + | \vec{b} |^{2} + 2 \vec{b} \cdot \vec{c} + | \vec{c} |^{2} = 0$
$∵ Given, | \vec{a} | = | \vec{b} | = | \vec{c} | = 1$
$1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 3$
$(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = \frac{- 3}{2}$

Karnataka CET-2013

Vector Algebra

87967 If $(\vec{a} \times \vec{b})^{2} + (a . \vec{b})^{2} = 144$ and $| \vec{a} | = 4$ , then $| \vec{b} | =$

1 12

2 16

3 8

4 3

Explanation:

(D) : Given,
$| \vec{a} | = 4, | \vec{b} | = ?$
$(\vec{a} \times \vec{b})^{2} + (\vec{a} \cdot \vec{b})^{2} = 144$
We know that
$\vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin θ$
$\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ$
$| \vec{a} |^{2} | \vec{b} |^{2} \sin^{2} θ + | \vec{a} |^{2} | \vec{b} |^{2} \cos^{2} θ = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} (\sin^{2} θ + \cos^{2} θ) = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} = 144$
$| \vec{a} | | \vec{b} | = 12$
$4 | \vec{b} | = 12$
$| \vec{b} | = 3$

Karnataka CET-2012

Vector Algebra

87969 The projection of $\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

1

\frac{8}{\sqrt{14}}

2

\sqrt{14}

3

\frac{8}{\sqrt{35}}

4

\frac{8}{\sqrt{39}}

Explanation:

(A) : Given,
$\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
We know that
The projection of $\vec{a}$ and $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$\vec{a} \cdot \vec{b} = (3 \hat{i} - \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k})$
$= 6 \hat{i} \cdot \hat{i} - 3 \hat{j} \cdot \hat{j} + 5 \hat{k} \cdot \hat{k} (∵ \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1)$
$∴ \vec{a} \cdot \vec{b} = 8$
$\vec{b} = 2 \vec{i} + 3 \vec{j} + \vec{k}$
$| \vec{b} | = \sqrt{2^{2} + 3^{2} + 1} = \sqrt{14}$
Projection of $\vec{a}$ and $\vec{b} = \frac{8}{\sqrt{14}}$

Karnataka CET-2008

Vector Algebra

87970 If $\vec{a}$ is vector perpendicular to both $\vec{b}$ and $\vec{c}$ , then

1

\vec{a} + (\vec{b} + \vec{c}) = \vec{0}

2

\vec{a} \times (\vec{b} + \vec{c}) = \vec{0}

3

\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}

4

\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{0}

Explanation:

(C) : Given,
$\vec{a} is perpendicular to both \vec{b} and \vec{c}$
$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \cdot \vec{b} - (\vec{a} \cdot \vec{b}) \cdot \vec{c}$
$∵ (\vec{a} \cdot \vec{b} = \vec{0}, \vec{a} \cdot \vec{c} = \vec{0})$
$= \vec{0} - \vec{0}$
$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}$

Karnataka CET-2006

Vector Algebra

87966 If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ , then $\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} =$

1

\frac{3}{2}

2

- \frac{3}{2}

3

\frac{2}{3}

4

\frac{1}{2}

Explanation:

(B) : Given,
$\vec{a} + \vec{b} + \vec{c} = \vec{0}$
$\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = ?$
We know that
$(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$
$\vec{a} \cdot \vec{a} + \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} + \vec{c} \cdot \vec{b} + \vec{c} \cdot \vec{c} = 0$
$| \vec{a} |^{2} + 2 \vec{a} \cdot \vec{b} + 2 c \cdot a + | \vec{b} |^{2} + 2 \vec{b} \cdot \vec{c} + | \vec{c} |^{2} = 0$
$∵ Given, | \vec{a} | = | \vec{b} | = | \vec{c} | = 1$
$1 + 1 + 1 + 2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
$2 (\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = - 3$
$(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = \frac{- 3}{2}$

Karnataka CET-2013

Vector Algebra

87967 If $(\vec{a} \times \vec{b})^{2} + (a . \vec{b})^{2} = 144$ and $| \vec{a} | = 4$ , then $| \vec{b} | =$

1 12

2 16

3 8

4 3

Explanation:

(D) : Given,
$| \vec{a} | = 4, | \vec{b} | = ?$
$(\vec{a} \times \vec{b})^{2} + (\vec{a} \cdot \vec{b})^{2} = 144$
We know that
$\vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin θ$
$\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ$
$| \vec{a} |^{2} | \vec{b} |^{2} \sin^{2} θ + | \vec{a} |^{2} | \vec{b} |^{2} \cos^{2} θ = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} (\sin^{2} θ + \cos^{2} θ) = 144$
$| \vec{a} |^{2} | \vec{b} |^{2} = 144$
$| \vec{a} | | \vec{b} | = 12$
$4 | \vec{b} | = 12$
$| \vec{b} | = 3$

Karnataka CET-2012

Vector Algebra

87969 The projection of $\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$ on $\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$ is

1

\frac{8}{\sqrt{14}}

2

\sqrt{14}

3

\frac{8}{\sqrt{35}}

4

\frac{8}{\sqrt{39}}

Explanation:

(A) : Given,
$\vec{a} = 3 \hat{i} - \hat{j} + 5 \hat{k}$
$\vec{b} = 2 \hat{i} + 3 \hat{j} + \hat{k}$
We know that
The projection of $\vec{a}$ and $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}$
$\vec{a} \cdot \vec{b} = (3 \hat{i} - \hat{j} + 5 \hat{k}) \cdot (2 \hat{i} + 3 \hat{j} + \hat{k})$
$= 6 \hat{i} \cdot \hat{i} - 3 \hat{j} \cdot \hat{j} + 5 \hat{k} \cdot \hat{k} (∵ \hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1)$
$∴ \vec{a} \cdot \vec{b} = 8$
$\vec{b} = 2 \vec{i} + 3 \vec{j} + \vec{k}$
$| \vec{b} | = \sqrt{2^{2} + 3^{2} + 1} = \sqrt{14}$
Projection of $\vec{a}$ and $\vec{b} = \frac{8}{\sqrt{14}}$

Karnataka CET-2008

Vector Algebra

87970 If $\vec{a}$ is vector perpendicular to both $\vec{b}$ and $\vec{c}$ , then

1

\vec{a} + (\vec{b} + \vec{c}) = \vec{0}

2

\vec{a} \times (\vec{b} + \vec{c}) = \vec{0}

3

\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}

4

\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{0}

Explanation:

(C) : Given,
$\vec{a} is perpendicular to both \vec{b} and \vec{c}$
$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \cdot \vec{b} - (\vec{a} \cdot \vec{b}) \cdot \vec{c}$
$∵ (\vec{a} \cdot \vec{b} = \vec{0}, \vec{a} \cdot \vec{c} = \vec{0})$
$= \vec{0} - \vec{0}$
$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{0}$

Karnataka CET-2006

Question Bank Series

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