QBManager

Vector Algebra

87791 In quadrilateral $ABCD, \vec{AB} = \vec{a}, \vec{BC} = \vec{b}$ . $\vec{A D} = \vec{b} - \vec{a}$ if $M$ is the midpoint of $B C$ and $N$ is $\vec{a}$ point on DM such that $\vec{DN} = (\frac{4}{5}) \vec{DM}$ , then $5 \vec{AN} =$

1

\vec{AC}

2

2 \vec{AC}

3

3 \vec{AC}

4

4 \vec{AC}

Explanation:

(C) : Given quadrilateral $A B C D$ ,
$\vec{AB} = a, \vec{BC} = b, \vec{AD} = \vec{b} - \vec{a}$
Therefore, According to the question, $\overset{―}{BM} = \frac{\vec{b}}{2}$
And,
$\frac{\vec{DN}}{\overset{―}{NM}} = \frac{4}{1} {since, \vec{DN} = (\frac{4}{5}) \vec{DM}}$
Therefore,
$\vec{AN} - \vec{AD} = \frac{4}{1} (\vec{AM} - \vec{AN})$
$5 \vec{AN} = 4 \vec{AM} + \vec{AD} = 4 (\vec{a} + \frac{\vec{b}}{2}) + (\vec{b} - \vec{a})$
$= 4 \vec{a} + 2 \vec{b} + \vec{b} - \vec{a} = 3 (\vec{a} + \vec{b}) = 3 \vec{AC}$

AP EAMCET-2020-21.09.2020

Vector Algebra

87820 If $P Q R S T$ is a pentagon, then the resultant of forces, PQ, PT, QR, SR, TS and PS is

1 3 PT

2 3 PQ

3

3 PS

4 0

Explanation:

(C) : According to the question the following of the figure is consider.
original image
$= PQ + PT + QR + \overset{P}{SR} + TS + PS$
$= (PQ + QR) + (PT + TS) + (SR + PS)$
$= PR + PS + SR + PS$
$= (PR + SR) + 2 PS$
$= PS + 2 PS = 3 PS$

AP EAMCET-2020-23.09.2020

Vector Algebra

87792 Let ABCDEF be a regular hexagon with the vertices $A, B, C, D, E, F$ counter clock-wise. Then the vector $\vec{A B} + \overset{―}{A F} + \overset{―}{C D} + \overset{―}{E F}$ is equal to

1

\vec{DE} +

2

\overset{―}{CB} + \overset{―}{ED}

3

\overset{―}{BC} + \overset{―}{FA}

4

\overset{―}{BC} + \vec{DE}

Explanation:

(D) : Given,
A regular hexagon $A B C D E F$ ,
Now, as we know that,
$\vec{AB} + \vec{BC} + \overset{―}{CD} = \vec{AD} = 2 \overset{―}{BC}$
original image
Therefore,
$\vec{AB} + \vec{CD} = \vec{BC}$
Similarly,
$\overset{―}{AF} = \overset{―}{CD}$
And we know that,
$\overset{―}{CD} + \overset{―}{DE} + \overset{―}{EF} = \overset{―}{CF} = 2 \overset{―}{DF}$
$\overset{―}{CD} + \overset{―}{EF} = \overset{―}{DE}$
From equation (i) and (iii), we get -
$\vec{AB} + \vec{CD} + \vec{CD} + \vec{EF} = \overset{―}{BC} + \overset{―}{DE}$
$\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF} = \vec{BC} + \overset{―}{DE}$ (from equation (ii))

AP EAMCET-2020-21.09.2020

Vector Algebra

87793 If the vectors
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ are
coplanar, where $(a, b, c \neq l)$ , then the value of
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} =$

1 2

2 0

3 -1

4 1

Explanation:

(D) : Given vectors,
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k} and \hat{i} + \hat{j} + c \hat{k}$
For, coplanar vectors,
$\vec{a} \cdot | \vec{b} \times \vec{c} | = 0$
$| \begin{array}{lll} a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z} \end{array} | = 0$
substituting values we have, $| \begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array} | = 0$ $R_{2} \to R_{2} \to R_{1}$ and $R_{3} \to R_{3} \to R_{1}$
$| \begin{array}{ccc} a & 1 & 1 \\ 1 - a & b - 1 & 0 \\ 1 - a & 0 & c - 1 \end{array} | = 0$
$a (b - 1) (c - 1) - (1 - a) (c - 1) - (1 - a) (b - 1) = 0$ Diving by $(1 - a) (1 - b) (1 - c)$ , we get -
$\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} - \frac{a}{1 - a} - \frac{1}{1 - b} - \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} - \frac{a}{1 - a} = 1$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{1}{1 - a} - \frac{a}{1 - a} = 1$

AP EAMCET-2020-17.09.2020

Vector Algebra

87791 In quadrilateral $ABCD, \vec{AB} = \vec{a}, \vec{BC} = \vec{b}$ . $\vec{A D} = \vec{b} - \vec{a}$ if $M$ is the midpoint of $B C$ and $N$ is $\vec{a}$ point on DM such that $\vec{DN} = (\frac{4}{5}) \vec{DM}$ , then $5 \vec{AN} =$

1

\vec{AC}

2

2 \vec{AC}

3

3 \vec{AC}

4

4 \vec{AC}

Explanation:

(C) : Given quadrilateral $A B C D$ ,
$\vec{AB} = a, \vec{BC} = b, \vec{AD} = \vec{b} - \vec{a}$
Therefore, According to the question, $\overset{―}{BM} = \frac{\vec{b}}{2}$
And,
$\frac{\vec{DN}}{\overset{―}{NM}} = \frac{4}{1} {since, \vec{DN} = (\frac{4}{5}) \vec{DM}}$
Therefore,
$\vec{AN} - \vec{AD} = \frac{4}{1} (\vec{AM} - \vec{AN})$
$5 \vec{AN} = 4 \vec{AM} + \vec{AD} = 4 (\vec{a} + \frac{\vec{b}}{2}) + (\vec{b} - \vec{a})$
$= 4 \vec{a} + 2 \vec{b} + \vec{b} - \vec{a} = 3 (\vec{a} + \vec{b}) = 3 \vec{AC}$

AP EAMCET-2020-21.09.2020

Vector Algebra

87820 If $P Q R S T$ is a pentagon, then the resultant of forces, PQ, PT, QR, SR, TS and PS is

1 3 PT

2 3 PQ

3

3 PS

4 0

Explanation:

(C) : According to the question the following of the figure is consider.
original image
$= PQ + PT + QR + \overset{P}{SR} + TS + PS$
$= (PQ + QR) + (PT + TS) + (SR + PS)$
$= PR + PS + SR + PS$
$= (PR + SR) + 2 PS$
$= PS + 2 PS = 3 PS$

AP EAMCET-2020-23.09.2020

Vector Algebra

87792 Let ABCDEF be a regular hexagon with the vertices $A, B, C, D, E, F$ counter clock-wise. Then the vector $\vec{A B} + \overset{―}{A F} + \overset{―}{C D} + \overset{―}{E F}$ is equal to

1

\vec{DE} +

2

\overset{―}{CB} + \overset{―}{ED}

3

\overset{―}{BC} + \overset{―}{FA}

4

\overset{―}{BC} + \vec{DE}

Explanation:

(D) : Given,
A regular hexagon $A B C D E F$ ,
Now, as we know that,
$\vec{AB} + \vec{BC} + \overset{―}{CD} = \vec{AD} = 2 \overset{―}{BC}$
original image
Therefore,
$\vec{AB} + \vec{CD} = \vec{BC}$
Similarly,
$\overset{―}{AF} = \overset{―}{CD}$
And we know that,
$\overset{―}{CD} + \overset{―}{DE} + \overset{―}{EF} = \overset{―}{CF} = 2 \overset{―}{DF}$
$\overset{―}{CD} + \overset{―}{EF} = \overset{―}{DE}$
From equation (i) and (iii), we get -
$\vec{AB} + \vec{CD} + \vec{CD} + \vec{EF} = \overset{―}{BC} + \overset{―}{DE}$
$\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF} = \vec{BC} + \overset{―}{DE}$ (from equation (ii))

AP EAMCET-2020-21.09.2020

Vector Algebra

87793 If the vectors
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ are
coplanar, where $(a, b, c \neq l)$ , then the value of
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} =$

1 2

2 0

3 -1

4 1

Explanation:

(D) : Given vectors,
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k} and \hat{i} + \hat{j} + c \hat{k}$
For, coplanar vectors,
$\vec{a} \cdot | \vec{b} \times \vec{c} | = 0$
$| \begin{array}{lll} a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z} \end{array} | = 0$
substituting values we have, $| \begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array} | = 0$ $R_{2} \to R_{2} \to R_{1}$ and $R_{3} \to R_{3} \to R_{1}$
$| \begin{array}{ccc} a & 1 & 1 \\ 1 - a & b - 1 & 0 \\ 1 - a & 0 & c - 1 \end{array} | = 0$
$a (b - 1) (c - 1) - (1 - a) (c - 1) - (1 - a) (b - 1) = 0$ Diving by $(1 - a) (1 - b) (1 - c)$ , we get -
$\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} - \frac{a}{1 - a} - \frac{1}{1 - b} - \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} - \frac{a}{1 - a} = 1$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{1}{1 - a} - \frac{a}{1 - a} = 1$

AP EAMCET-2020-17.09.2020

Vector Algebra

87791 In quadrilateral $ABCD, \vec{AB} = \vec{a}, \vec{BC} = \vec{b}$ . $\vec{A D} = \vec{b} - \vec{a}$ if $M$ is the midpoint of $B C$ and $N$ is $\vec{a}$ point on DM such that $\vec{DN} = (\frac{4}{5}) \vec{DM}$ , then $5 \vec{AN} =$

1

\vec{AC}

2

2 \vec{AC}

3

3 \vec{AC}

4

4 \vec{AC}

Explanation:

(C) : Given quadrilateral $A B C D$ ,
$\vec{AB} = a, \vec{BC} = b, \vec{AD} = \vec{b} - \vec{a}$
Therefore, According to the question, $\overset{―}{BM} = \frac{\vec{b}}{2}$
And,
$\frac{\vec{DN}}{\overset{―}{NM}} = \frac{4}{1} {since, \vec{DN} = (\frac{4}{5}) \vec{DM}}$
Therefore,
$\vec{AN} - \vec{AD} = \frac{4}{1} (\vec{AM} - \vec{AN})$
$5 \vec{AN} = 4 \vec{AM} + \vec{AD} = 4 (\vec{a} + \frac{\vec{b}}{2}) + (\vec{b} - \vec{a})$
$= 4 \vec{a} + 2 \vec{b} + \vec{b} - \vec{a} = 3 (\vec{a} + \vec{b}) = 3 \vec{AC}$

AP EAMCET-2020-21.09.2020

Vector Algebra

87820 If $P Q R S T$ is a pentagon, then the resultant of forces, PQ, PT, QR, SR, TS and PS is

1 3 PT

2 3 PQ

3

3 PS

4 0

Explanation:

(C) : According to the question the following of the figure is consider.
original image
$= PQ + PT + QR + \overset{P}{SR} + TS + PS$
$= (PQ + QR) + (PT + TS) + (SR + PS)$
$= PR + PS + SR + PS$
$= (PR + SR) + 2 PS$
$= PS + 2 PS = 3 PS$

AP EAMCET-2020-23.09.2020

Vector Algebra

87792 Let ABCDEF be a regular hexagon with the vertices $A, B, C, D, E, F$ counter clock-wise. Then the vector $\vec{A B} + \overset{―}{A F} + \overset{―}{C D} + \overset{―}{E F}$ is equal to

1

\vec{DE} +

2

\overset{―}{CB} + \overset{―}{ED}

3

\overset{―}{BC} + \overset{―}{FA}

4

\overset{―}{BC} + \vec{DE}

Explanation:

(D) : Given,
A regular hexagon $A B C D E F$ ,
Now, as we know that,
$\vec{AB} + \vec{BC} + \overset{―}{CD} = \vec{AD} = 2 \overset{―}{BC}$
original image
Therefore,
$\vec{AB} + \vec{CD} = \vec{BC}$
Similarly,
$\overset{―}{AF} = \overset{―}{CD}$
And we know that,
$\overset{―}{CD} + \overset{―}{DE} + \overset{―}{EF} = \overset{―}{CF} = 2 \overset{―}{DF}$
$\overset{―}{CD} + \overset{―}{EF} = \overset{―}{DE}$
From equation (i) and (iii), we get -
$\vec{AB} + \vec{CD} + \vec{CD} + \vec{EF} = \overset{―}{BC} + \overset{―}{DE}$
$\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF} = \vec{BC} + \overset{―}{DE}$ (from equation (ii))

AP EAMCET-2020-21.09.2020

Vector Algebra

87793 If the vectors
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ are
coplanar, where $(a, b, c \neq l)$ , then the value of
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} =$

1 2

2 0

3 -1

4 1

Explanation:

(D) : Given vectors,
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k} and \hat{i} + \hat{j} + c \hat{k}$
For, coplanar vectors,
$\vec{a} \cdot | \vec{b} \times \vec{c} | = 0$
$| \begin{array}{lll} a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z} \end{array} | = 0$
substituting values we have, $| \begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array} | = 0$ $R_{2} \to R_{2} \to R_{1}$ and $R_{3} \to R_{3} \to R_{1}$
$| \begin{array}{ccc} a & 1 & 1 \\ 1 - a & b - 1 & 0 \\ 1 - a & 0 & c - 1 \end{array} | = 0$
$a (b - 1) (c - 1) - (1 - a) (c - 1) - (1 - a) (b - 1) = 0$ Diving by $(1 - a) (1 - b) (1 - c)$ , we get -
$\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} - \frac{a}{1 - a} - \frac{1}{1 - b} - \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} - \frac{a}{1 - a} = 1$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{1}{1 - a} - \frac{a}{1 - a} = 1$

AP EAMCET-2020-17.09.2020

Vector Algebra

87791 In quadrilateral $ABCD, \vec{AB} = \vec{a}, \vec{BC} = \vec{b}$ . $\vec{A D} = \vec{b} - \vec{a}$ if $M$ is the midpoint of $B C$ and $N$ is $\vec{a}$ point on DM such that $\vec{DN} = (\frac{4}{5}) \vec{DM}$ , then $5 \vec{AN} =$

1

\vec{AC}

2

2 \vec{AC}

3

3 \vec{AC}

4

4 \vec{AC}

Explanation:

(C) : Given quadrilateral $A B C D$ ,
$\vec{AB} = a, \vec{BC} = b, \vec{AD} = \vec{b} - \vec{a}$
Therefore, According to the question, $\overset{―}{BM} = \frac{\vec{b}}{2}$
And,
$\frac{\vec{DN}}{\overset{―}{NM}} = \frac{4}{1} {since, \vec{DN} = (\frac{4}{5}) \vec{DM}}$
Therefore,
$\vec{AN} - \vec{AD} = \frac{4}{1} (\vec{AM} - \vec{AN})$
$5 \vec{AN} = 4 \vec{AM} + \vec{AD} = 4 (\vec{a} + \frac{\vec{b}}{2}) + (\vec{b} - \vec{a})$
$= 4 \vec{a} + 2 \vec{b} + \vec{b} - \vec{a} = 3 (\vec{a} + \vec{b}) = 3 \vec{AC}$

AP EAMCET-2020-21.09.2020

Vector Algebra

87820 If $P Q R S T$ is a pentagon, then the resultant of forces, PQ, PT, QR, SR, TS and PS is

1 3 PT

2 3 PQ

3

3 PS

4 0

Explanation:

(C) : According to the question the following of the figure is consider.
original image
$= PQ + PT + QR + \overset{P}{SR} + TS + PS$
$= (PQ + QR) + (PT + TS) + (SR + PS)$
$= PR + PS + SR + PS$
$= (PR + SR) + 2 PS$
$= PS + 2 PS = 3 PS$

AP EAMCET-2020-23.09.2020

Vector Algebra

87792 Let ABCDEF be a regular hexagon with the vertices $A, B, C, D, E, F$ counter clock-wise. Then the vector $\vec{A B} + \overset{―}{A F} + \overset{―}{C D} + \overset{―}{E F}$ is equal to

1

\vec{DE} +

2

\overset{―}{CB} + \overset{―}{ED}

3

\overset{―}{BC} + \overset{―}{FA}

4

\overset{―}{BC} + \vec{DE}

Explanation:

(D) : Given,
A regular hexagon $A B C D E F$ ,
Now, as we know that,
$\vec{AB} + \vec{BC} + \overset{―}{CD} = \vec{AD} = 2 \overset{―}{BC}$
original image
Therefore,
$\vec{AB} + \vec{CD} = \vec{BC}$
Similarly,
$\overset{―}{AF} = \overset{―}{CD}$
And we know that,
$\overset{―}{CD} + \overset{―}{DE} + \overset{―}{EF} = \overset{―}{CF} = 2 \overset{―}{DF}$
$\overset{―}{CD} + \overset{―}{EF} = \overset{―}{DE}$
From equation (i) and (iii), we get -
$\vec{AB} + \vec{CD} + \vec{CD} + \vec{EF} = \overset{―}{BC} + \overset{―}{DE}$
$\vec{AB} + \vec{AF} + \vec{CD} + \vec{EF} = \vec{BC} + \overset{―}{DE}$ (from equation (ii))

AP EAMCET-2020-21.09.2020

Vector Algebra

87793 If the vectors
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + c \hat{k}$ are
coplanar, where $(a, b, c \neq l)$ , then the value of
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} =$

1 2

2 0

3 -1

4 1

Explanation:

(D) : Given vectors,
$a \hat{i} + \hat{j} + \hat{k}, \hat{i} + b \hat{j} + \hat{k} and \hat{i} + \hat{j} + c \hat{k}$
For, coplanar vectors,
$\vec{a} \cdot | \vec{b} \times \vec{c} | = 0$
$| \begin{array}{lll} a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \\ c_{x} & c_{y} & c_{z} \end{array} | = 0$
substituting values we have, $| \begin{array}{lll} a & 1 & 1 \\ 1 & b & 1 \\ 1 & 1 & c \end{array} | = 0$ $R_{2} \to R_{2} \to R_{1}$ and $R_{3} \to R_{3} \to R_{1}$
$| \begin{array}{ccc} a & 1 & 1 \\ 1 - a & b - 1 & 0 \\ 1 - a & 0 & c - 1 \end{array} | = 0$
$a (b - 1) (c - 1) - (1 - a) (c - 1) - (1 - a) (b - 1) = 0$ Diving by $(1 - a) (1 - b) (1 - c)$ , we get -
$\frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{a}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} - \frac{a}{1 - a} - \frac{1}{1 - b} - \frac{1}{1 - c} = 0$
$\frac{1}{1 - a} - \frac{a}{1 - a} = 1$
$\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c} = \frac{1}{1 - a} - \frac{a}{1 - a} = 1$

AP EAMCET-2020-17.09.2020