QBManager

Vector Algebra

87708 Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ . Let $- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $| A P | = 12$ . Then, the position vector of $A$ can be

1

\hat{i} + \hat{j} + 3 \hat{k}

2

15 \hat{i} + 9 \hat{j} - 19 \hat{k}

3

- \hat{i} - \hat{j} + 3 \hat{k}

4

- 15 \hat{i} - 9 \hat{j} + 19 \hat{k}

Explanation:

(D) : As vector AP is parallel to the vector $(2 \hat{i} + \hat{j} - 2 \hat{k})$
Then,
$A P = | \overset{―}{A P} | \frac{2 \hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{(2)^{2} + (1)^{2} + (- 2)^{2}}} = \frac{12 (2 \hat{i} + \hat{j} - 2 \hat{k})}{3}$
$= 8 \hat{i} + 4 \hat{j} - 8 \hat{k}$
Given, $\overset{―}{OP} = - 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$
$∴ \overset{―}{AP} = \overset{―}{OP} - \overset{―}{OA}$
$Or, \overset{―}{OA} = \overset{―}{OP} - \vec{OA}$
$= (- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}) - (8 \hat{i} + 4 \hat{j} - 8 \hat{k})$
$= - 15 \hat{i} - 9 \hat{j} + 19 \hat{k}$

TS EAMCET-2022-19.07.2022

Vector Algebra

87709 Let the vectors $A B = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $A C = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ be two sides of a $△ A B C$ . If $G$ is the centroid of $△ A B C$ , then $\frac{27}{7} | AG |^{2} + 5 =$

1 25

2 38

3 47

4 52

Explanation:

(B) : Let, vertex $A$ of $△ A B C$ lies at origin. The,
Position vector of $\vec{B} = \vec{AB} = 2 \hat{i} + 2 \hat{j} + \hat{k}$
And, Position vector of $\vec{C} = \vec{AC} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$
original image
$∴$ Mid point $\overset{―}{BC} = D = (\frac{2 + 2}{2}, \frac{2 + 4}{2}, \frac{1 + 4}{2}) \equiv (2, 3, \frac{5}{2})$
Centroid of $△ ABC$ is the intersection points of all the median of $△ ABC$ and is intersect from the vertex in the ratio $2 : 1$ .
$∴$ Co-ordinates of $G$ w.r.t, vertex $A (0, 0, 0)$ is-
$= (\frac{{mx}_{2}}{3}, \frac{{my}_{2}}{3}, \frac{{mz}_{2}}{3}) = (\frac{4}{3}, 2, \frac{5}{3})$
Now,
$| AG |^{2} = {(\frac{4}{3} - 0)}^{2} + (2 - 0)^{2} + {(\frac{5}{3} - 0)}^{2}$
$= \frac{16 + 36 + 25}{9} = \frac{77}{9}$
$\frac{27}{7} | AG |^{2} + 5 = \frac{27}{7} \times \frac{77}{9} + 5 = 33 + 5 = 38$

TS EAMCET-2022-19.07.2022

Vector Algebra

87710 $L_{1}$ is a line passing through the points with position vectors $\hat{i} - 2 \hat{j} - \hat{k}$ and $4 \hat{i} - 3 \hat{k}$ . $L_{2}$ is a line passing through the points with position vectors $\hat{i} + 2 \hat{j} - \hat{k}$ and $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ . Then the distance between $L_{1}$ and $L_{2}$ is

1 0

2

\frac{3}{4}

3

\frac{4}{3}

4

\frac{2}{3}

Explanation:

(C) : Equation of line $L_{1}$ is
$r = (\hat{i} - 2 \hat{j} - \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
Equation of line $L_{2}$ is
$r = (\hat{i} + 2 \hat{j} - \hat{k}) + μ (\hat{i} - 6 \hat{j} - 4 \hat{k})$
Distance between $L_{1}$ and $L_{2}$
$D = \frac{(4 \hat{j}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k})]}{| (3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - \hat{j}) - 4 \hat{k} |}$
$(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k}) = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= - (20 \hat{i} - 10 \hat{j} + 20 \hat{k})$
$D = | \frac{4 \hat{j} \cdot (- 20 \hat{i} + 10 \hat{j} - 20 \hat{k})}{\sqrt{(20)^{2} + (10)^{2} + (20)^{2}}} | = \frac{4}{3}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87711 A vector a has components $2 p$ and 1 with respect to a two dimensional rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise direction. If $a$ has components $p + 1$ and 1 with respect to the new system, then

1

p = 1

or

p = \frac{- 1}{3}

2

p = - 1

or

p = \frac{1}{3}

3

p = 1

or

p = - 1

4

p = 0

or

p = \frac{1}{2}

Explanation:

(A) : Let, point A(2p, 1)
A is rotated about origin in the counter-clock wise direction then new coordinate is $(p + 1, 1)$
Since, $2 p = (p + 1) \cos θ - \sin θ$
And, $1 = (p + 1) \sin θ + \cos θ$
On squaring and adding equation (i) and (ii), we get-
$\begin{array}{r} 4 p^{2} + 1 = (p + 1)^{2} + 1 \Rightarrow 4 p^{2} = p^{2} + 2 p + 1 \\ \Rightarrow 3 p^{2} - 2 p - 1 = 0 \Rightarrow (p - 1) (3 p + 1) = 0 \Rightarrow p = 1, \frac{- 1}{3} \end{array}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87708 Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ . Let $- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $| A P | = 12$ . Then, the position vector of $A$ can be

1

\hat{i} + \hat{j} + 3 \hat{k}

2

15 \hat{i} + 9 \hat{j} - 19 \hat{k}

3

- \hat{i} - \hat{j} + 3 \hat{k}

4

- 15 \hat{i} - 9 \hat{j} + 19 \hat{k}

Explanation:

(D) : As vector AP is parallel to the vector $(2 \hat{i} + \hat{j} - 2 \hat{k})$
Then,
$A P = | \overset{―}{A P} | \frac{2 \hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{(2)^{2} + (1)^{2} + (- 2)^{2}}} = \frac{12 (2 \hat{i} + \hat{j} - 2 \hat{k})}{3}$
$= 8 \hat{i} + 4 \hat{j} - 8 \hat{k}$
Given, $\overset{―}{OP} = - 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$
$∴ \overset{―}{AP} = \overset{―}{OP} - \overset{―}{OA}$
$Or, \overset{―}{OA} = \overset{―}{OP} - \vec{OA}$
$= (- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}) - (8 \hat{i} + 4 \hat{j} - 8 \hat{k})$
$= - 15 \hat{i} - 9 \hat{j} + 19 \hat{k}$

TS EAMCET-2022-19.07.2022

Vector Algebra

87709 Let the vectors $A B = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $A C = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ be two sides of a $△ A B C$ . If $G$ is the centroid of $△ A B C$ , then $\frac{27}{7} | AG |^{2} + 5 =$

1 25

2 38

3 47

4 52

Explanation:

(B) : Let, vertex $A$ of $△ A B C$ lies at origin. The,
Position vector of $\vec{B} = \vec{AB} = 2 \hat{i} + 2 \hat{j} + \hat{k}$
And, Position vector of $\vec{C} = \vec{AC} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$
original image
$∴$ Mid point $\overset{―}{BC} = D = (\frac{2 + 2}{2}, \frac{2 + 4}{2}, \frac{1 + 4}{2}) \equiv (2, 3, \frac{5}{2})$
Centroid of $△ ABC$ is the intersection points of all the median of $△ ABC$ and is intersect from the vertex in the ratio $2 : 1$ .
$∴$ Co-ordinates of $G$ w.r.t, vertex $A (0, 0, 0)$ is-
$= (\frac{{mx}_{2}}{3}, \frac{{my}_{2}}{3}, \frac{{mz}_{2}}{3}) = (\frac{4}{3}, 2, \frac{5}{3})$
Now,
$| AG |^{2} = {(\frac{4}{3} - 0)}^{2} + (2 - 0)^{2} + {(\frac{5}{3} - 0)}^{2}$
$= \frac{16 + 36 + 25}{9} = \frac{77}{9}$
$\frac{27}{7} | AG |^{2} + 5 = \frac{27}{7} \times \frac{77}{9} + 5 = 33 + 5 = 38$

TS EAMCET-2022-19.07.2022

Vector Algebra

87710 $L_{1}$ is a line passing through the points with position vectors $\hat{i} - 2 \hat{j} - \hat{k}$ and $4 \hat{i} - 3 \hat{k}$ . $L_{2}$ is a line passing through the points with position vectors $\hat{i} + 2 \hat{j} - \hat{k}$ and $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ . Then the distance between $L_{1}$ and $L_{2}$ is

1 0

2

\frac{3}{4}

3

\frac{4}{3}

4

\frac{2}{3}

Explanation:

(C) : Equation of line $L_{1}$ is
$r = (\hat{i} - 2 \hat{j} - \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
Equation of line $L_{2}$ is
$r = (\hat{i} + 2 \hat{j} - \hat{k}) + μ (\hat{i} - 6 \hat{j} - 4 \hat{k})$
Distance between $L_{1}$ and $L_{2}$
$D = \frac{(4 \hat{j}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k})]}{| (3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - \hat{j}) - 4 \hat{k} |}$
$(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k}) = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= - (20 \hat{i} - 10 \hat{j} + 20 \hat{k})$
$D = | \frac{4 \hat{j} \cdot (- 20 \hat{i} + 10 \hat{j} - 20 \hat{k})}{\sqrt{(20)^{2} + (10)^{2} + (20)^{2}}} | = \frac{4}{3}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87711 A vector a has components $2 p$ and 1 with respect to a two dimensional rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise direction. If $a$ has components $p + 1$ and 1 with respect to the new system, then

1

p = 1

or

p = \frac{- 1}{3}

2

p = - 1

or

p = \frac{1}{3}

3

p = 1

or

p = - 1

4

p = 0

or

p = \frac{1}{2}

Explanation:

(A) : Let, point A(2p, 1)
A is rotated about origin in the counter-clock wise direction then new coordinate is $(p + 1, 1)$
Since, $2 p = (p + 1) \cos θ - \sin θ$
And, $1 = (p + 1) \sin θ + \cos θ$
On squaring and adding equation (i) and (ii), we get-
$\begin{array}{r} 4 p^{2} + 1 = (p + 1)^{2} + 1 \Rightarrow 4 p^{2} = p^{2} + 2 p + 1 \\ \Rightarrow 3 p^{2} - 2 p - 1 = 0 \Rightarrow (p - 1) (3 p + 1) = 0 \Rightarrow p = 1, \frac{- 1}{3} \end{array}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87708 Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ . Let $- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $| A P | = 12$ . Then, the position vector of $A$ can be

1

\hat{i} + \hat{j} + 3 \hat{k}

2

15 \hat{i} + 9 \hat{j} - 19 \hat{k}

3

- \hat{i} - \hat{j} + 3 \hat{k}

4

- 15 \hat{i} - 9 \hat{j} + 19 \hat{k}

Explanation:

(D) : As vector AP is parallel to the vector $(2 \hat{i} + \hat{j} - 2 \hat{k})$
Then,
$A P = | \overset{―}{A P} | \frac{2 \hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{(2)^{2} + (1)^{2} + (- 2)^{2}}} = \frac{12 (2 \hat{i} + \hat{j} - 2 \hat{k})}{3}$
$= 8 \hat{i} + 4 \hat{j} - 8 \hat{k}$
Given, $\overset{―}{OP} = - 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$
$∴ \overset{―}{AP} = \overset{―}{OP} - \overset{―}{OA}$
$Or, \overset{―}{OA} = \overset{―}{OP} - \vec{OA}$
$= (- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}) - (8 \hat{i} + 4 \hat{j} - 8 \hat{k})$
$= - 15 \hat{i} - 9 \hat{j} + 19 \hat{k}$

TS EAMCET-2022-19.07.2022

Vector Algebra

87709 Let the vectors $A B = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $A C = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ be two sides of a $△ A B C$ . If $G$ is the centroid of $△ A B C$ , then $\frac{27}{7} | AG |^{2} + 5 =$

1 25

2 38

3 47

4 52

Explanation:

(B) : Let, vertex $A$ of $△ A B C$ lies at origin. The,
Position vector of $\vec{B} = \vec{AB} = 2 \hat{i} + 2 \hat{j} + \hat{k}$
And, Position vector of $\vec{C} = \vec{AC} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$
original image
$∴$ Mid point $\overset{―}{BC} = D = (\frac{2 + 2}{2}, \frac{2 + 4}{2}, \frac{1 + 4}{2}) \equiv (2, 3, \frac{5}{2})$
Centroid of $△ ABC$ is the intersection points of all the median of $△ ABC$ and is intersect from the vertex in the ratio $2 : 1$ .
$∴$ Co-ordinates of $G$ w.r.t, vertex $A (0, 0, 0)$ is-
$= (\frac{{mx}_{2}}{3}, \frac{{my}_{2}}{3}, \frac{{mz}_{2}}{3}) = (\frac{4}{3}, 2, \frac{5}{3})$
Now,
$| AG |^{2} = {(\frac{4}{3} - 0)}^{2} + (2 - 0)^{2} + {(\frac{5}{3} - 0)}^{2}$
$= \frac{16 + 36 + 25}{9} = \frac{77}{9}$
$\frac{27}{7} | AG |^{2} + 5 = \frac{27}{7} \times \frac{77}{9} + 5 = 33 + 5 = 38$

TS EAMCET-2022-19.07.2022

Vector Algebra

87710 $L_{1}$ is a line passing through the points with position vectors $\hat{i} - 2 \hat{j} - \hat{k}$ and $4 \hat{i} - 3 \hat{k}$ . $L_{2}$ is a line passing through the points with position vectors $\hat{i} + 2 \hat{j} - \hat{k}$ and $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ . Then the distance between $L_{1}$ and $L_{2}$ is

1 0

2

\frac{3}{4}

3

\frac{4}{3}

4

\frac{2}{3}

Explanation:

(C) : Equation of line $L_{1}$ is
$r = (\hat{i} - 2 \hat{j} - \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
Equation of line $L_{2}$ is
$r = (\hat{i} + 2 \hat{j} - \hat{k}) + μ (\hat{i} - 6 \hat{j} - 4 \hat{k})$
Distance between $L_{1}$ and $L_{2}$
$D = \frac{(4 \hat{j}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k})]}{| (3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - \hat{j}) - 4 \hat{k} |}$
$(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k}) = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= - (20 \hat{i} - 10 \hat{j} + 20 \hat{k})$
$D = | \frac{4 \hat{j} \cdot (- 20 \hat{i} + 10 \hat{j} - 20 \hat{k})}{\sqrt{(20)^{2} + (10)^{2} + (20)^{2}}} | = \frac{4}{3}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87711 A vector a has components $2 p$ and 1 with respect to a two dimensional rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise direction. If $a$ has components $p + 1$ and 1 with respect to the new system, then

1

p = 1

or

p = \frac{- 1}{3}

2

p = - 1

or

p = \frac{1}{3}

3

p = 1

or

p = - 1

4

p = 0

or

p = \frac{1}{2}

Explanation:

(A) : Let, point A(2p, 1)
A is rotated about origin in the counter-clock wise direction then new coordinate is $(p + 1, 1)$
Since, $2 p = (p + 1) \cos θ - \sin θ$
And, $1 = (p + 1) \sin θ + \cos θ$
On squaring and adding equation (i) and (ii), we get-
$\begin{array}{r} 4 p^{2} + 1 = (p + 1)^{2} + 1 \Rightarrow 4 p^{2} = p^{2} + 2 p + 1 \\ \Rightarrow 3 p^{2} - 2 p - 1 = 0 \Rightarrow (p - 1) (3 p + 1) = 0 \Rightarrow p = 1, \frac{- 1}{3} \end{array}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87708 Let $L$ be a line passing through a point $A$ and parallel to the vector $2 \hat{i} + \hat{j} - 2 \hat{k}$ . Let $- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$ be the position vector of a point $P$ on $L$ such that $| A P | = 12$ . Then, the position vector of $A$ can be

1

\hat{i} + \hat{j} + 3 \hat{k}

2

15 \hat{i} + 9 \hat{j} - 19 \hat{k}

3

- \hat{i} - \hat{j} + 3 \hat{k}

4

- 15 \hat{i} - 9 \hat{j} + 19 \hat{k}

Explanation:

(D) : As vector AP is parallel to the vector $(2 \hat{i} + \hat{j} - 2 \hat{k})$
Then,
$A P = | \overset{―}{A P} | \frac{2 \hat{i} + \hat{j} - 2 \hat{k}}{\sqrt{(2)^{2} + (1)^{2} + (- 2)^{2}}} = \frac{12 (2 \hat{i} + \hat{j} - 2 \hat{k})}{3}$
$= 8 \hat{i} + 4 \hat{j} - 8 \hat{k}$
Given, $\overset{―}{OP} = - 7 \hat{i} - 5 \hat{j} + 11 \hat{k}$
$∴ \overset{―}{AP} = \overset{―}{OP} - \overset{―}{OA}$
$Or, \overset{―}{OA} = \overset{―}{OP} - \vec{OA}$
$= (- 7 \hat{i} - 5 \hat{j} + 11 \hat{k}) - (8 \hat{i} + 4 \hat{j} - 8 \hat{k})$
$= - 15 \hat{i} - 9 \hat{j} + 19 \hat{k}$

TS EAMCET-2022-19.07.2022

Vector Algebra

87709 Let the vectors $A B = 2 \hat{i} + 2 \hat{j} + \hat{k}$ and $A C = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$ be two sides of a $△ A B C$ . If $G$ is the centroid of $△ A B C$ , then $\frac{27}{7} | AG |^{2} + 5 =$

1 25

2 38

3 47

4 52

Explanation:

(B) : Let, vertex $A$ of $△ A B C$ lies at origin. The,
Position vector of $\vec{B} = \vec{AB} = 2 \hat{i} + 2 \hat{j} + \hat{k}$
And, Position vector of $\vec{C} = \vec{AC} = 2 \hat{i} + 4 \hat{j} + 4 \hat{k}$
original image
$∴$ Mid point $\overset{―}{BC} = D = (\frac{2 + 2}{2}, \frac{2 + 4}{2}, \frac{1 + 4}{2}) \equiv (2, 3, \frac{5}{2})$
Centroid of $△ ABC$ is the intersection points of all the median of $△ ABC$ and is intersect from the vertex in the ratio $2 : 1$ .
$∴$ Co-ordinates of $G$ w.r.t, vertex $A (0, 0, 0)$ is-
$= (\frac{{mx}_{2}}{3}, \frac{{my}_{2}}{3}, \frac{{mz}_{2}}{3}) = (\frac{4}{3}, 2, \frac{5}{3})$
Now,
$| AG |^{2} = {(\frac{4}{3} - 0)}^{2} + (2 - 0)^{2} + {(\frac{5}{3} - 0)}^{2}$
$= \frac{16 + 36 + 25}{9} = \frac{77}{9}$
$\frac{27}{7} | AG |^{2} + 5 = \frac{27}{7} \times \frac{77}{9} + 5 = 33 + 5 = 38$

TS EAMCET-2022-19.07.2022

Vector Algebra

87710 $L_{1}$ is a line passing through the points with position vectors $\hat{i} - 2 \hat{j} - \hat{k}$ and $4 \hat{i} - 3 \hat{k}$ . $L_{2}$ is a line passing through the points with position vectors $\hat{i} + 2 \hat{j} - \hat{k}$ and $2 \hat{i} - 4 \hat{j} - 5 \hat{k}$ . Then the distance between $L_{1}$ and $L_{2}$ is

1 0

2

\frac{3}{4}

3

\frac{4}{3}

4

\frac{2}{3}

Explanation:

(C) : Equation of line $L_{1}$ is
$r = (\hat{i} - 2 \hat{j} - \hat{k}) + λ (3 \hat{i} + 2 \hat{j} - 2 \hat{k})$
Equation of line $L_{2}$ is
$r = (\hat{i} + 2 \hat{j} - \hat{k}) + μ (\hat{i} - 6 \hat{j} - 4 \hat{k})$
Distance between $L_{1}$ and $L_{2}$
$D = \frac{(4 \hat{j}) \cdot [(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k})]}{| (3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - \hat{j}) - 4 \hat{k} |}$
$(3 \hat{i} + 2 \hat{j} - 2 \hat{k}) \times (\hat{i} - 6 \hat{j} - 4 \hat{k}) = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & - 2 \\ 1 & - 6 & - 4 \end{array} |$
$= - (20 \hat{i} - 10 \hat{j} + 20 \hat{k})$
$D = | \frac{4 \hat{j} \cdot (- 20 \hat{i} + 10 \hat{j} - 20 \hat{k})}{\sqrt{(20)^{2} + (10)^{2} + (20)^{2}}} | = \frac{4}{3}$

TS EAMCET-2020-14.09.2020

Vector Algebra

87711 A vector a has components $2 p$ and 1 with respect to a two dimensional rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter-clockwise direction. If $a$ has components $p + 1$ and 1 with respect to the new system, then

1

p = 1

or

p = \frac{- 1}{3}

2

p = - 1

or

p = \frac{1}{3}

3

p = 1

or

p = - 1

4

p = 0

or

p = \frac{1}{2}

Explanation:

(A) : Let, point A(2p, 1)
A is rotated about origin in the counter-clock wise direction then new coordinate is $(p + 1, 1)$
Since, $2 p = (p + 1) \cos θ - \sin θ$
And, $1 = (p + 1) \sin θ + \cos θ$
On squaring and adding equation (i) and (ii), we get-
$\begin{array}{r} 4 p^{2} + 1 = (p + 1)^{2} + 1 \Rightarrow 4 p^{2} = p^{2} + 2 p + 1 \\ \Rightarrow 3 p^{2} - 2 p - 1 = 0 \Rightarrow (p - 1) (3 p + 1) = 0 \Rightarrow p = 1, \frac{- 1}{3} \end{array}$

TS EAMCET-2020-14.09.2020

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Explanation:

Question Bank Series

Explanation:

Explanation:

Explanation:

Explanation: