QBManager

Vector Algebra

87813 If the position vectors of $A, B, C$ are respectively $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$ , then the projection of $A B$ on $B C$ is equal to

1

\frac{- 14}{\sqrt{10}}

2

\sqrt{5}

3

\sqrt{7}

4 2

Explanation:

(A) : given, $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$
$∵ A B = \hat{i} + 5 \hat{j} - \hat{k}$
And, $B C = - 3 \hat{j} - \hat{k}$
Therefore,
$Projection of A B on B C = \frac{A B \cdot B C}{| B C |}$
$= \frac{(\hat{i} + 5 \hat{j} - \hat{k}) (- 3 \hat{j} - \hat{k})}{\sqrt{(- 3)^{2} + (- 1)^{2}}} = \frac{- 15 + 1}{\sqrt{10}} = - \frac{14}{\sqrt{10}}$

Manipal-2017

Vector Algebra

87814 If the projection of $PQ$ on $OX, OY, OZ$ are respectively 12,3 and 4 , then the magnitude of $P Q$ is

1 169

2 19

3 13

4 144

Explanation:

(C) : As given that the projection of $\overset{―}{PQ}$ on $OX$ , $OY, OZ$ are respectively 12,3 and 4,30 ,
$\overset{―}{PQ} = 12 \hat{i} + 3 \hat{j} + 4 \hat{k}$
So, $| \overset{―}{PQ} | = \sqrt{(12)^{2} + (3)^{2} + (4)^{2}}$
$= \sqrt{144 + 9 + 16} = \sqrt{169} = 13$

Manipal-2017

Vector Algebra

87815 Consider a tetrahedron with $f$ aces $F_{1}, F_{2}, F_{3}, F_{4}$ Let ${\vec{V}}_{1}, {\vec{V}}_{2}, {\vec{V}}_{3}, {\vec{V}}_{4}$ be the vectors whose magnitudes are respectively equal to areas of $F_{1}, F_{2}, F_{3}, F_{4}$ and whose directions are perpendicular to these faces in outward direction, then $| {\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} |$ equals

1 1

2 4

3 0

4 None of these

Explanation:

(C) :
According to the question, problem is represented in the figure,
original image
Form figure,
Area of ${\vec{V}}_{1}$ ,
${\vec{V}}_{1} = \frac{1}{2} (\vec{a} \times \vec{b})$
Area of ${\vec{V}}_{2}$ ,
${\vec{V}}_{2} = \frac{1}{2} (\vec{b} \times \vec{c})$
Area of ${\vec{V}}_{3}$ ,
${\vec{V}}_{3} = \frac{1}{2} (\vec{c} \times \vec{a})$
Area of ${\vec{V}}_{4}$ ,
${\vec{V}}_{4} = \frac{1}{2} [(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})]$
$= \frac{1}{2} [\vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = \frac{1}{2} [\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$
$+ \vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} + \vec{a} \times \vec{c}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} - \vec{c} \times \vec{a}]$
$= \frac{1}{2} [0]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = 0$

Manipal UGET-2010

Vector Algebra

87816 In a trapezoid of the vector $\vec{BC} = λ \vec{AD}$ . We will, then find that $\vec{P} = \vec{AC} + \vec{BD}$ is collinear with $\vec{AD}$ .If $\vec{P} = μ \vec{AD}$ , then

1

μ = λ + 1

2

λ = μ + 1

3

λ + μ = 1

4

μ = 2 + λ

Explanation:

(A) : Given,
$\vec{P} = \vec{AC} + \vec{BD}$
$= \vec{AC} + \vec{BC} + \vec{CD}$
$= \vec{AC} + λ \vec{AD} + \vec{CD}$
$= λ \vec{AD} + (\vec{AC} + \vec{CD})$
$= (λ + 1) \overset{―}{AD}$
But $\vec{P} = μ \vec{AD}$
$μ = λ + 1$

Manipal-2010

Vector Algebra

87817 The area of a parallelogram with diagonals as $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$ is

1

10 \sqrt{3}

2

\frac{10}{\sqrt{3}}

3

5 \sqrt{3}

4

\frac{5}{\sqrt{3}}

Explanation:

(C) : Given,
$\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$
$\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$
Now,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{array} |$
$= \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1)$
$= - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$
Therefore, Area of required parallelogram,
$= \frac{1}{2} | \vec{a} \times \vec{b} |$
$= \frac{1}{2} \sqrt{4 + 196 + 100}$
$= \frac{10 \sqrt{3}}{2} = 5 \sqrt{3}$

Manipal-2018

Vector Algebra

87813 If the position vectors of $A, B, C$ are respectively $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$ , then the projection of $A B$ on $B C$ is equal to

1

\frac{- 14}{\sqrt{10}}

2

\sqrt{5}

3

\sqrt{7}

4 2

Explanation:

(A) : given, $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$
$∵ A B = \hat{i} + 5 \hat{j} - \hat{k}$
And, $B C = - 3 \hat{j} - \hat{k}$
Therefore,
$Projection of A B on B C = \frac{A B \cdot B C}{| B C |}$
$= \frac{(\hat{i} + 5 \hat{j} - \hat{k}) (- 3 \hat{j} - \hat{k})}{\sqrt{(- 3)^{2} + (- 1)^{2}}} = \frac{- 15 + 1}{\sqrt{10}} = - \frac{14}{\sqrt{10}}$

Manipal-2017

Vector Algebra

87814 If the projection of $PQ$ on $OX, OY, OZ$ are respectively 12,3 and 4 , then the magnitude of $P Q$ is

1 169

2 19

3 13

4 144

Explanation:

(C) : As given that the projection of $\overset{―}{PQ}$ on $OX$ , $OY, OZ$ are respectively 12,3 and 4,30 ,
$\overset{―}{PQ} = 12 \hat{i} + 3 \hat{j} + 4 \hat{k}$
So, $| \overset{―}{PQ} | = \sqrt{(12)^{2} + (3)^{2} + (4)^{2}}$
$= \sqrt{144 + 9 + 16} = \sqrt{169} = 13$

Manipal-2017

Vector Algebra

87815 Consider a tetrahedron with $f$ aces $F_{1}, F_{2}, F_{3}, F_{4}$ Let ${\vec{V}}_{1}, {\vec{V}}_{2}, {\vec{V}}_{3}, {\vec{V}}_{4}$ be the vectors whose magnitudes are respectively equal to areas of $F_{1}, F_{2}, F_{3}, F_{4}$ and whose directions are perpendicular to these faces in outward direction, then $| {\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} |$ equals

1 1

2 4

3 0

4 None of these

Explanation:

(C) :
According to the question, problem is represented in the figure,
original image
Form figure,
Area of ${\vec{V}}_{1}$ ,
${\vec{V}}_{1} = \frac{1}{2} (\vec{a} \times \vec{b})$
Area of ${\vec{V}}_{2}$ ,
${\vec{V}}_{2} = \frac{1}{2} (\vec{b} \times \vec{c})$
Area of ${\vec{V}}_{3}$ ,
${\vec{V}}_{3} = \frac{1}{2} (\vec{c} \times \vec{a})$
Area of ${\vec{V}}_{4}$ ,
${\vec{V}}_{4} = \frac{1}{2} [(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})]$
$= \frac{1}{2} [\vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = \frac{1}{2} [\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$
$+ \vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} + \vec{a} \times \vec{c}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} - \vec{c} \times \vec{a}]$
$= \frac{1}{2} [0]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = 0$

Manipal UGET-2010

Vector Algebra

87816 In a trapezoid of the vector $\vec{BC} = λ \vec{AD}$ . We will, then find that $\vec{P} = \vec{AC} + \vec{BD}$ is collinear with $\vec{AD}$ .If $\vec{P} = μ \vec{AD}$ , then

1

μ = λ + 1

2

λ = μ + 1

3

λ + μ = 1

4

μ = 2 + λ

Explanation:

(A) : Given,
$\vec{P} = \vec{AC} + \vec{BD}$
$= \vec{AC} + \vec{BC} + \vec{CD}$
$= \vec{AC} + λ \vec{AD} + \vec{CD}$
$= λ \vec{AD} + (\vec{AC} + \vec{CD})$
$= (λ + 1) \overset{―}{AD}$
But $\vec{P} = μ \vec{AD}$
$μ = λ + 1$

Manipal-2010

Vector Algebra

87817 The area of a parallelogram with diagonals as $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$ is

1

10 \sqrt{3}

2

\frac{10}{\sqrt{3}}

3

5 \sqrt{3}

4

\frac{5}{\sqrt{3}}

Explanation:

(C) : Given,
$\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$
$\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$
Now,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{array} |$
$= \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1)$
$= - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$
Therefore, Area of required parallelogram,
$= \frac{1}{2} | \vec{a} \times \vec{b} |$
$= \frac{1}{2} \sqrt{4 + 196 + 100}$
$= \frac{10 \sqrt{3}}{2} = 5 \sqrt{3}$

Manipal-2018

Vector Algebra

87813 If the position vectors of $A, B, C$ are respectively $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$ , then the projection of $A B$ on $B C$ is equal to

1

\frac{- 14}{\sqrt{10}}

2

\sqrt{5}

3

\sqrt{7}

4 2

Explanation:

(A) : given, $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$
$∵ A B = \hat{i} + 5 \hat{j} - \hat{k}$
And, $B C = - 3 \hat{j} - \hat{k}$
Therefore,
$Projection of A B on B C = \frac{A B \cdot B C}{| B C |}$
$= \frac{(\hat{i} + 5 \hat{j} - \hat{k}) (- 3 \hat{j} - \hat{k})}{\sqrt{(- 3)^{2} + (- 1)^{2}}} = \frac{- 15 + 1}{\sqrt{10}} = - \frac{14}{\sqrt{10}}$

Manipal-2017

Vector Algebra

87814 If the projection of $PQ$ on $OX, OY, OZ$ are respectively 12,3 and 4 , then the magnitude of $P Q$ is

1 169

2 19

3 13

4 144

Explanation:

(C) : As given that the projection of $\overset{―}{PQ}$ on $OX$ , $OY, OZ$ are respectively 12,3 and 4,30 ,
$\overset{―}{PQ} = 12 \hat{i} + 3 \hat{j} + 4 \hat{k}$
So, $| \overset{―}{PQ} | = \sqrt{(12)^{2} + (3)^{2} + (4)^{2}}$
$= \sqrt{144 + 9 + 16} = \sqrt{169} = 13$

Manipal-2017

Vector Algebra

87815 Consider a tetrahedron with $f$ aces $F_{1}, F_{2}, F_{3}, F_{4}$ Let ${\vec{V}}_{1}, {\vec{V}}_{2}, {\vec{V}}_{3}, {\vec{V}}_{4}$ be the vectors whose magnitudes are respectively equal to areas of $F_{1}, F_{2}, F_{3}, F_{4}$ and whose directions are perpendicular to these faces in outward direction, then $| {\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} |$ equals

1 1

2 4

3 0

4 None of these

Explanation:

(C) :
According to the question, problem is represented in the figure,
original image
Form figure,
Area of ${\vec{V}}_{1}$ ,
${\vec{V}}_{1} = \frac{1}{2} (\vec{a} \times \vec{b})$
Area of ${\vec{V}}_{2}$ ,
${\vec{V}}_{2} = \frac{1}{2} (\vec{b} \times \vec{c})$
Area of ${\vec{V}}_{3}$ ,
${\vec{V}}_{3} = \frac{1}{2} (\vec{c} \times \vec{a})$
Area of ${\vec{V}}_{4}$ ,
${\vec{V}}_{4} = \frac{1}{2} [(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})]$
$= \frac{1}{2} [\vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = \frac{1}{2} [\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$
$+ \vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} + \vec{a} \times \vec{c}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} - \vec{c} \times \vec{a}]$
$= \frac{1}{2} [0]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = 0$

Manipal UGET-2010

Vector Algebra

87816 In a trapezoid of the vector $\vec{BC} = λ \vec{AD}$ . We will, then find that $\vec{P} = \vec{AC} + \vec{BD}$ is collinear with $\vec{AD}$ .If $\vec{P} = μ \vec{AD}$ , then

1

μ = λ + 1

2

λ = μ + 1

3

λ + μ = 1

4

μ = 2 + λ

Explanation:

(A) : Given,
$\vec{P} = \vec{AC} + \vec{BD}$
$= \vec{AC} + \vec{BC} + \vec{CD}$
$= \vec{AC} + λ \vec{AD} + \vec{CD}$
$= λ \vec{AD} + (\vec{AC} + \vec{CD})$
$= (λ + 1) \overset{―}{AD}$
But $\vec{P} = μ \vec{AD}$
$μ = λ + 1$

Manipal-2010

Vector Algebra

87817 The area of a parallelogram with diagonals as $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$ is

1

10 \sqrt{3}

2

\frac{10}{\sqrt{3}}

3

5 \sqrt{3}

4

\frac{5}{\sqrt{3}}

Explanation:

(C) : Given,
$\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$
$\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$
Now,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{array} |$
$= \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1)$
$= - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$
Therefore, Area of required parallelogram,
$= \frac{1}{2} | \vec{a} \times \vec{b} |$
$= \frac{1}{2} \sqrt{4 + 196 + 100}$
$= \frac{10 \sqrt{3}}{2} = 5 \sqrt{3}$

Manipal-2018

Vector Algebra

87813 If the position vectors of $A, B, C$ are respectively $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$ , then the projection of $A B$ on $B C$ is equal to

1

\frac{- 14}{\sqrt{10}}

2

\sqrt{5}

3

\sqrt{7}

4 2

Explanation:

(A) : given, $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$
$∵ A B = \hat{i} + 5 \hat{j} - \hat{k}$
And, $B C = - 3 \hat{j} - \hat{k}$
Therefore,
$Projection of A B on B C = \frac{A B \cdot B C}{| B C |}$
$= \frac{(\hat{i} + 5 \hat{j} - \hat{k}) (- 3 \hat{j} - \hat{k})}{\sqrt{(- 3)^{2} + (- 1)^{2}}} = \frac{- 15 + 1}{\sqrt{10}} = - \frac{14}{\sqrt{10}}$

Manipal-2017

Vector Algebra

87814 If the projection of $PQ$ on $OX, OY, OZ$ are respectively 12,3 and 4 , then the magnitude of $P Q$ is

1 169

2 19

3 13

4 144

Explanation:

(C) : As given that the projection of $\overset{―}{PQ}$ on $OX$ , $OY, OZ$ are respectively 12,3 and 4,30 ,
$\overset{―}{PQ} = 12 \hat{i} + 3 \hat{j} + 4 \hat{k}$
So, $| \overset{―}{PQ} | = \sqrt{(12)^{2} + (3)^{2} + (4)^{2}}$
$= \sqrt{144 + 9 + 16} = \sqrt{169} = 13$

Manipal-2017

Vector Algebra

87815 Consider a tetrahedron with $f$ aces $F_{1}, F_{2}, F_{3}, F_{4}$ Let ${\vec{V}}_{1}, {\vec{V}}_{2}, {\vec{V}}_{3}, {\vec{V}}_{4}$ be the vectors whose magnitudes are respectively equal to areas of $F_{1}, F_{2}, F_{3}, F_{4}$ and whose directions are perpendicular to these faces in outward direction, then $| {\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} |$ equals

1 1

2 4

3 0

4 None of these

Explanation:

(C) :
According to the question, problem is represented in the figure,
original image
Form figure,
Area of ${\vec{V}}_{1}$ ,
${\vec{V}}_{1} = \frac{1}{2} (\vec{a} \times \vec{b})$
Area of ${\vec{V}}_{2}$ ,
${\vec{V}}_{2} = \frac{1}{2} (\vec{b} \times \vec{c})$
Area of ${\vec{V}}_{3}$ ,
${\vec{V}}_{3} = \frac{1}{2} (\vec{c} \times \vec{a})$
Area of ${\vec{V}}_{4}$ ,
${\vec{V}}_{4} = \frac{1}{2} [(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})]$
$= \frac{1}{2} [\vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = \frac{1}{2} [\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$
$+ \vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} + \vec{a} \times \vec{c}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} - \vec{c} \times \vec{a}]$
$= \frac{1}{2} [0]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = 0$

Manipal UGET-2010

Vector Algebra

87816 In a trapezoid of the vector $\vec{BC} = λ \vec{AD}$ . We will, then find that $\vec{P} = \vec{AC} + \vec{BD}$ is collinear with $\vec{AD}$ .If $\vec{P} = μ \vec{AD}$ , then

1

μ = λ + 1

2

λ = μ + 1

3

λ + μ = 1

4

μ = 2 + λ

Explanation:

(A) : Given,
$\vec{P} = \vec{AC} + \vec{BD}$
$= \vec{AC} + \vec{BC} + \vec{CD}$
$= \vec{AC} + λ \vec{AD} + \vec{CD}$
$= λ \vec{AD} + (\vec{AC} + \vec{CD})$
$= (λ + 1) \overset{―}{AD}$
But $\vec{P} = μ \vec{AD}$
$μ = λ + 1$

Manipal-2010

Vector Algebra

87817 The area of a parallelogram with diagonals as $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$ is

1

10 \sqrt{3}

2

\frac{10}{\sqrt{3}}

3

5 \sqrt{3}

4

\frac{5}{\sqrt{3}}

Explanation:

(C) : Given,
$\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$
$\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$
Now,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{array} |$
$= \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1)$
$= - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$
Therefore, Area of required parallelogram,
$= \frac{1}{2} | \vec{a} \times \vec{b} |$
$= \frac{1}{2} \sqrt{4 + 196 + 100}$
$= \frac{10 \sqrt{3}}{2} = 5 \sqrt{3}$

Manipal-2018

Vector Algebra

87813 If the position vectors of $A, B, C$ are respectively $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$ , then the projection of $A B$ on $B C$ is equal to

1

\frac{- 14}{\sqrt{10}}

2

\sqrt{5}

3

\sqrt{7}

4 2

Explanation:

(A) : given, $\hat{i} - 2 \hat{j} + \hat{k}, 2 \hat{i} + 3 \hat{j}$ and $2 \hat{i} - \hat{k}$
$∵ A B = \hat{i} + 5 \hat{j} - \hat{k}$
And, $B C = - 3 \hat{j} - \hat{k}$
Therefore,
$Projection of A B on B C = \frac{A B \cdot B C}{| B C |}$
$= \frac{(\hat{i} + 5 \hat{j} - \hat{k}) (- 3 \hat{j} - \hat{k})}{\sqrt{(- 3)^{2} + (- 1)^{2}}} = \frac{- 15 + 1}{\sqrt{10}} = - \frac{14}{\sqrt{10}}$

Manipal-2017

Vector Algebra

87814 If the projection of $PQ$ on $OX, OY, OZ$ are respectively 12,3 and 4 , then the magnitude of $P Q$ is

1 169

2 19

3 13

4 144

Explanation:

(C) : As given that the projection of $\overset{―}{PQ}$ on $OX$ , $OY, OZ$ are respectively 12,3 and 4,30 ,
$\overset{―}{PQ} = 12 \hat{i} + 3 \hat{j} + 4 \hat{k}$
So, $| \overset{―}{PQ} | = \sqrt{(12)^{2} + (3)^{2} + (4)^{2}}$
$= \sqrt{144 + 9 + 16} = \sqrt{169} = 13$

Manipal-2017

Vector Algebra

87815 Consider a tetrahedron with $f$ aces $F_{1}, F_{2}, F_{3}, F_{4}$ Let ${\vec{V}}_{1}, {\vec{V}}_{2}, {\vec{V}}_{3}, {\vec{V}}_{4}$ be the vectors whose magnitudes are respectively equal to areas of $F_{1}, F_{2}, F_{3}, F_{4}$ and whose directions are perpendicular to these faces in outward direction, then $| {\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} |$ equals

1 1

2 4

3 0

4 None of these

Explanation:

(C) :
According to the question, problem is represented in the figure,
original image
Form figure,
Area of ${\vec{V}}_{1}$ ,
${\vec{V}}_{1} = \frac{1}{2} (\vec{a} \times \vec{b})$
Area of ${\vec{V}}_{2}$ ,
${\vec{V}}_{2} = \frac{1}{2} (\vec{b} \times \vec{c})$
Area of ${\vec{V}}_{3}$ ,
${\vec{V}}_{3} = \frac{1}{2} (\vec{c} \times \vec{a})$
Area of ${\vec{V}}_{4}$ ,
${\vec{V}}_{4} = \frac{1}{2} [(\vec{a} - \vec{b}) \times (\vec{c} - \vec{b})]$
$= \frac{1}{2} [\vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = \frac{1}{2} [\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}$
$+ \vec{a} \times \vec{c} - \vec{b} \times \vec{c} - \vec{a} \times \vec{b}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} + \vec{a} \times \vec{c}]$
$= \frac{1}{2} [\vec{c} \times \vec{a} - \vec{c} \times \vec{a}]$
$= \frac{1}{2} [0]$
Therefore, ${\vec{V}}_{1} + {\vec{V}}_{2} + {\vec{V}}_{3} + {\vec{V}}_{4} = 0$

Manipal UGET-2010

Vector Algebra

87816 In a trapezoid of the vector $\vec{BC} = λ \vec{AD}$ . We will, then find that $\vec{P} = \vec{AC} + \vec{BD}$ is collinear with $\vec{AD}$ .If $\vec{P} = μ \vec{AD}$ , then

1

μ = λ + 1

2

λ = μ + 1

3

λ + μ = 1

4

μ = 2 + λ

Explanation:

(A) : Given,
$\vec{P} = \vec{AC} + \vec{BD}$
$= \vec{AC} + \vec{BC} + \vec{CD}$
$= \vec{AC} + λ \vec{AD} + \vec{CD}$
$= λ \vec{AD} + (\vec{AC} + \vec{CD})$
$= (λ + 1) \overset{―}{AD}$
But $\vec{P} = μ \vec{AD}$
$μ = λ + 1$

Manipal-2010

Vector Algebra

87817 The area of a parallelogram with diagonals as $\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$ is

1

10 \sqrt{3}

2

\frac{10}{\sqrt{3}}

3

5 \sqrt{3}

4

\frac{5}{\sqrt{3}}

Explanation:

(C) : Given,
$\vec{a} = 3 \hat{i} + \hat{j} - 2 \hat{k}$
$\vec{b} = \hat{i} - 3 \hat{j} + 4 \hat{k}$
Now,
$\vec{a} \times \vec{b} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & - 2 \\ 1 & - 3 & 4 \end{array} |$
$= \hat{i} (4 - 6) - \hat{j} (12 + 2) + \hat{k} (- 9 - 1)$
$= - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$
Therefore, Area of required parallelogram,
$= \frac{1}{2} | \vec{a} \times \vec{b} |$
$= \frac{1}{2} \sqrt{4 + 196 + 100}$
$= \frac{10 \sqrt{3}}{2} = 5 \sqrt{3}$

Manipal-2018

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