QBManager

Motion in Plane

143602 Consider the vectors $A = \hat{i} + \hat{j} - \hat{k}, B =$ $2 \hat{i} - \hat{j} + \hat{k}$ and $C = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ . What is the value of $C . (A \times B)$ ?

1 1

2 0

3

3 \sqrt{2}

4

18 \sqrt{5}

Explanation:

B Given, $\vec{A} = \hat{i} + \hat{j} - \hat{k}, \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$
$\vec{C} = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$
Now,
$\vec{A} \times \vec{B} = | \begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 11 - 1 \\ 2 - 11 \end{array} |$
$\vec{A} \times \vec{B} = - 3 \hat{j} - 3 \hat{k}$
According to the question,
$\vec{C} \cdot (\vec{A} \times \vec{B}) = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (0 \hat{i} - 3 \hat{j} - 3 \hat{k})$
$= \frac{1}{\sqrt{5}} (6 - 6)$
$= 0$

WB JEE 2020

Motion in Plane

143603 In a triangle $ABC$ , the sides $AB$ and $AC$ are represented by the vectors $3 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Calculate the angle $∠ ABC$ .

1

\cos^{- 1} \sqrt{\frac{5}{11}}

2

\cos^{- 1} \sqrt{\frac{6}{11}}

3

(90^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

4

(180^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

Explanation:

A
Given that, $\vec{A B} = 3 \hat{i} + \hat{j} + \hat{k}, \vec{A C} = \hat{i} + 2 \hat{j} + \hat{k}$
$∴ \vec{CB} = \vec{AB} - \vec{AC}$
$\vec{CB} = 3 \hat{i} + \hat{j} + \hat{k} - (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{CB} = 2 \hat{i} - \hat{j}$
$∠ ABC$ is angle between $\vec{AB}$ and $\vec{CB}$
$∴ \vec{AB} \cdot \vec{CB} = | \vec{AB} | | \vec{CB} | \cos θ$
$(6 - 1) = \sqrt{3^{2} + 1^{2} + 1^{2}} \times \sqrt{(2)^{2} + (- 1)^{2}} \times \cos θ$
$5 = \sqrt{11} \times \sqrt{5} \times \cos θ$
$\cos θ = \frac{5}{\sqrt{11} \times \sqrt{5}}$
$\cos θ = \frac{\sqrt{5}}{\sqrt{11}}$
$θ = \cos^{- 1} (\frac{\sqrt{5}}{\sqrt{11}})$

WB JEE 2018

Motion in Plane

143604 Three vectors $A = a \hat{i} + \hat{j} + \hat{k}; B = \hat{i} + b \hat{j} + \hat{k}$ and $C = \hat{i} + \hat{j} + c k$ are mutually perpendicular $(\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $X, Y$ and $Z$ - axis respectively). The respective values of $a, b$ and c are

1

0, 0, 0

2

- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}

3

1, - 1, 1

4

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

Explanation:

B Given that, $A = a \hat{i} + \hat{j} + \hat{k}, B = \hat{i} + b \hat{j} + \hat{k}, C =$ $\hat{i} + \hat{j} + c \hat{k}$ are mutually perpendicular-
So, A.B $= 0$
B. $C = 0$
A. $C = 0$
Now, $a + b + 1 = 0 \dots \dots$ (i)
$1 + b + c = 0$
$a + 1 + c = 0$
On adding equation (i), (ii) and (iii), we get-
$2 (a + b + c) = - 3$
$\Rightarrow a + b + c = - 3 / 2$
By equation (i) and (iv) we get-
$- 1 + c = - 3 / 2$
$c = - 1 / 2$
Similarly, $b = - 1 / 2$
$a = - 1 / 2$

WB JEE 2017

Motion in Plane

143605 If $A = B + C$ have scalar magnitudes of $5, 4, 3$ units respectively, then the angle between $A$ and $C$ is

1

\cos^{- 1} (3 / 5)

2

\cos^{- 1} (4 / 5)

3

π / 2

4

\sin^{- 1} (3 / 4)

Explanation:

A Here, triangle $PQR$ is a given with vector. $A$ , $B, C$ are its adjacent sides.

5, 4 and 3 makes the triangle right angle triangle.
$\cos θ = \frac{QR}{PQ}$
$θ = \cos^{- 1} (\frac{| C |}{| A |})$
$θ = \cos^{- 1} (\frac{3}{5})$

WB JEE 2012

Motion in Plane

143606 The magnitudes of vectors $A, B$ and $C$ are 3,4 and 5 units respectively. If $A + B = C$ , the angle between $A$ and $B$ is

1

\frac{π}{2}

2

\cos^{- 1} (0.6)

3

\tan - 1 (\frac{7}{5})

4

\frac{π}{4}

Explanation:

A Given that, $\vec{A} = 3, \vec{B} = 4, \vec{C} = 5$
Let $θ$ be the angle between them
$\vec{A} + \vec{B} = \vec{C}$
Squaring on both side
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 \vec{A} \vec{B} = | \vec{C} |^{2}$
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 | \vec{A} | | \vec{B} | \cos θ = | \vec{C} |^{2}$
On putting the value, we get-
$(3)^{2} + (4)^{2} + 2 \times 3 \times 4 \cos θ = (5)^{2}$
$25 + 24 \cos θ = 25$
$\cos θ = 0$
$θ = \frac{π}{2}$

AIPMT 1988

Motion in Plane

143602 Consider the vectors $A = \hat{i} + \hat{j} - \hat{k}, B =$ $2 \hat{i} - \hat{j} + \hat{k}$ and $C = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ . What is the value of $C . (A \times B)$ ?

1 1

2 0

3

3 \sqrt{2}

4

18 \sqrt{5}

Explanation:

B Given, $\vec{A} = \hat{i} + \hat{j} - \hat{k}, \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$
$\vec{C} = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$
Now,
$\vec{A} \times \vec{B} = | \begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 11 - 1 \\ 2 - 11 \end{array} |$
$\vec{A} \times \vec{B} = - 3 \hat{j} - 3 \hat{k}$
According to the question,
$\vec{C} \cdot (\vec{A} \times \vec{B}) = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (0 \hat{i} - 3 \hat{j} - 3 \hat{k})$
$= \frac{1}{\sqrt{5}} (6 - 6)$
$= 0$

WB JEE 2020

Motion in Plane

143603 In a triangle $ABC$ , the sides $AB$ and $AC$ are represented by the vectors $3 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Calculate the angle $∠ ABC$ .

1

\cos^{- 1} \sqrt{\frac{5}{11}}

2

\cos^{- 1} \sqrt{\frac{6}{11}}

3

(90^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

4

(180^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

Explanation:

A
Given that, $\vec{A B} = 3 \hat{i} + \hat{j} + \hat{k}, \vec{A C} = \hat{i} + 2 \hat{j} + \hat{k}$
$∴ \vec{CB} = \vec{AB} - \vec{AC}$
$\vec{CB} = 3 \hat{i} + \hat{j} + \hat{k} - (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{CB} = 2 \hat{i} - \hat{j}$
$∠ ABC$ is angle between $\vec{AB}$ and $\vec{CB}$
$∴ \vec{AB} \cdot \vec{CB} = | \vec{AB} | | \vec{CB} | \cos θ$
$(6 - 1) = \sqrt{3^{2} + 1^{2} + 1^{2}} \times \sqrt{(2)^{2} + (- 1)^{2}} \times \cos θ$
$5 = \sqrt{11} \times \sqrt{5} \times \cos θ$
$\cos θ = \frac{5}{\sqrt{11} \times \sqrt{5}}$
$\cos θ = \frac{\sqrt{5}}{\sqrt{11}}$
$θ = \cos^{- 1} (\frac{\sqrt{5}}{\sqrt{11}})$

WB JEE 2018

Motion in Plane

143604 Three vectors $A = a \hat{i} + \hat{j} + \hat{k}; B = \hat{i} + b \hat{j} + \hat{k}$ and $C = \hat{i} + \hat{j} + c k$ are mutually perpendicular $(\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $X, Y$ and $Z$ - axis respectively). The respective values of $a, b$ and c are

1

0, 0, 0

2

- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}

3

1, - 1, 1

4

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

Explanation:

B Given that, $A = a \hat{i} + \hat{j} + \hat{k}, B = \hat{i} + b \hat{j} + \hat{k}, C =$ $\hat{i} + \hat{j} + c \hat{k}$ are mutually perpendicular-
So, A.B $= 0$
B. $C = 0$
A. $C = 0$
Now, $a + b + 1 = 0 \dots \dots$ (i)
$1 + b + c = 0$
$a + 1 + c = 0$
On adding equation (i), (ii) and (iii), we get-
$2 (a + b + c) = - 3$
$\Rightarrow a + b + c = - 3 / 2$
By equation (i) and (iv) we get-
$- 1 + c = - 3 / 2$
$c = - 1 / 2$
Similarly, $b = - 1 / 2$
$a = - 1 / 2$

WB JEE 2017

Motion in Plane

143605 If $A = B + C$ have scalar magnitudes of $5, 4, 3$ units respectively, then the angle between $A$ and $C$ is

1

\cos^{- 1} (3 / 5)

2

\cos^{- 1} (4 / 5)

3

π / 2

4

\sin^{- 1} (3 / 4)

Explanation:

A Here, triangle $PQR$ is a given with vector. $A$ , $B, C$ are its adjacent sides.

5, 4 and 3 makes the triangle right angle triangle.
$\cos θ = \frac{QR}{PQ}$
$θ = \cos^{- 1} (\frac{| C |}{| A |})$
$θ = \cos^{- 1} (\frac{3}{5})$

WB JEE 2012

Motion in Plane

143606 The magnitudes of vectors $A, B$ and $C$ are 3,4 and 5 units respectively. If $A + B = C$ , the angle between $A$ and $B$ is

1

\frac{π}{2}

2

\cos^{- 1} (0.6)

3

\tan - 1 (\frac{7}{5})

4

\frac{π}{4}

Explanation:

A Given that, $\vec{A} = 3, \vec{B} = 4, \vec{C} = 5$
Let $θ$ be the angle between them
$\vec{A} + \vec{B} = \vec{C}$
Squaring on both side
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 \vec{A} \vec{B} = | \vec{C} |^{2}$
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 | \vec{A} | | \vec{B} | \cos θ = | \vec{C} |^{2}$
On putting the value, we get-
$(3)^{2} + (4)^{2} + 2 \times 3 \times 4 \cos θ = (5)^{2}$
$25 + 24 \cos θ = 25$
$\cos θ = 0$
$θ = \frac{π}{2}$

AIPMT 1988

Motion in Plane

143602 Consider the vectors $A = \hat{i} + \hat{j} - \hat{k}, B =$ $2 \hat{i} - \hat{j} + \hat{k}$ and $C = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ . What is the value of $C . (A \times B)$ ?

1 1

2 0

3

3 \sqrt{2}

4

18 \sqrt{5}

Explanation:

B Given, $\vec{A} = \hat{i} + \hat{j} - \hat{k}, \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$
$\vec{C} = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$
Now,
$\vec{A} \times \vec{B} = | \begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 11 - 1 \\ 2 - 11 \end{array} |$
$\vec{A} \times \vec{B} = - 3 \hat{j} - 3 \hat{k}$
According to the question,
$\vec{C} \cdot (\vec{A} \times \vec{B}) = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (0 \hat{i} - 3 \hat{j} - 3 \hat{k})$
$= \frac{1}{\sqrt{5}} (6 - 6)$
$= 0$

WB JEE 2020

Motion in Plane

143603 In a triangle $ABC$ , the sides $AB$ and $AC$ are represented by the vectors $3 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Calculate the angle $∠ ABC$ .

1

\cos^{- 1} \sqrt{\frac{5}{11}}

2

\cos^{- 1} \sqrt{\frac{6}{11}}

3

(90^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

4

(180^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

Explanation:

A
Given that, $\vec{A B} = 3 \hat{i} + \hat{j} + \hat{k}, \vec{A C} = \hat{i} + 2 \hat{j} + \hat{k}$
$∴ \vec{CB} = \vec{AB} - \vec{AC}$
$\vec{CB} = 3 \hat{i} + \hat{j} + \hat{k} - (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{CB} = 2 \hat{i} - \hat{j}$
$∠ ABC$ is angle between $\vec{AB}$ and $\vec{CB}$
$∴ \vec{AB} \cdot \vec{CB} = | \vec{AB} | | \vec{CB} | \cos θ$
$(6 - 1) = \sqrt{3^{2} + 1^{2} + 1^{2}} \times \sqrt{(2)^{2} + (- 1)^{2}} \times \cos θ$
$5 = \sqrt{11} \times \sqrt{5} \times \cos θ$
$\cos θ = \frac{5}{\sqrt{11} \times \sqrt{5}}$
$\cos θ = \frac{\sqrt{5}}{\sqrt{11}}$
$θ = \cos^{- 1} (\frac{\sqrt{5}}{\sqrt{11}})$

WB JEE 2018

Motion in Plane

143604 Three vectors $A = a \hat{i} + \hat{j} + \hat{k}; B = \hat{i} + b \hat{j} + \hat{k}$ and $C = \hat{i} + \hat{j} + c k$ are mutually perpendicular $(\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $X, Y$ and $Z$ - axis respectively). The respective values of $a, b$ and c are

1

0, 0, 0

2

- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}

3

1, - 1, 1

4

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

Explanation:

B Given that, $A = a \hat{i} + \hat{j} + \hat{k}, B = \hat{i} + b \hat{j} + \hat{k}, C =$ $\hat{i} + \hat{j} + c \hat{k}$ are mutually perpendicular-
So, A.B $= 0$
B. $C = 0$
A. $C = 0$
Now, $a + b + 1 = 0 \dots \dots$ (i)
$1 + b + c = 0$
$a + 1 + c = 0$
On adding equation (i), (ii) and (iii), we get-
$2 (a + b + c) = - 3$
$\Rightarrow a + b + c = - 3 / 2$
By equation (i) and (iv) we get-
$- 1 + c = - 3 / 2$
$c = - 1 / 2$
Similarly, $b = - 1 / 2$
$a = - 1 / 2$

WB JEE 2017

Motion in Plane

143605 If $A = B + C$ have scalar magnitudes of $5, 4, 3$ units respectively, then the angle between $A$ and $C$ is

1

\cos^{- 1} (3 / 5)

2

\cos^{- 1} (4 / 5)

3

π / 2

4

\sin^{- 1} (3 / 4)

Explanation:

A Here, triangle $PQR$ is a given with vector. $A$ , $B, C$ are its adjacent sides.

5, 4 and 3 makes the triangle right angle triangle.
$\cos θ = \frac{QR}{PQ}$
$θ = \cos^{- 1} (\frac{| C |}{| A |})$
$θ = \cos^{- 1} (\frac{3}{5})$

WB JEE 2012

Motion in Plane

143606 The magnitudes of vectors $A, B$ and $C$ are 3,4 and 5 units respectively. If $A + B = C$ , the angle between $A$ and $B$ is

1

\frac{π}{2}

2

\cos^{- 1} (0.6)

3

\tan - 1 (\frac{7}{5})

4

\frac{π}{4}

Explanation:

A Given that, $\vec{A} = 3, \vec{B} = 4, \vec{C} = 5$
Let $θ$ be the angle between them
$\vec{A} + \vec{B} = \vec{C}$
Squaring on both side
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 \vec{A} \vec{B} = | \vec{C} |^{2}$
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 | \vec{A} | | \vec{B} | \cos θ = | \vec{C} |^{2}$
On putting the value, we get-
$(3)^{2} + (4)^{2} + 2 \times 3 \times 4 \cos θ = (5)^{2}$
$25 + 24 \cos θ = 25$
$\cos θ = 0$
$θ = \frac{π}{2}$

AIPMT 1988

Motion in Plane

143602 Consider the vectors $A = \hat{i} + \hat{j} - \hat{k}, B =$ $2 \hat{i} - \hat{j} + \hat{k}$ and $C = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ . What is the value of $C . (A \times B)$ ?

1 1

2 0

3

3 \sqrt{2}

4

18 \sqrt{5}

Explanation:

B Given, $\vec{A} = \hat{i} + \hat{j} - \hat{k}, \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$
$\vec{C} = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$
Now,
$\vec{A} \times \vec{B} = | \begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 11 - 1 \\ 2 - 11 \end{array} |$
$\vec{A} \times \vec{B} = - 3 \hat{j} - 3 \hat{k}$
According to the question,
$\vec{C} \cdot (\vec{A} \times \vec{B}) = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (0 \hat{i} - 3 \hat{j} - 3 \hat{k})$
$= \frac{1}{\sqrt{5}} (6 - 6)$
$= 0$

WB JEE 2020

Motion in Plane

143603 In a triangle $ABC$ , the sides $AB$ and $AC$ are represented by the vectors $3 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Calculate the angle $∠ ABC$ .

1

\cos^{- 1} \sqrt{\frac{5}{11}}

2

\cos^{- 1} \sqrt{\frac{6}{11}}

3

(90^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

4

(180^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

Explanation:

A
Given that, $\vec{A B} = 3 \hat{i} + \hat{j} + \hat{k}, \vec{A C} = \hat{i} + 2 \hat{j} + \hat{k}$
$∴ \vec{CB} = \vec{AB} - \vec{AC}$
$\vec{CB} = 3 \hat{i} + \hat{j} + \hat{k} - (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{CB} = 2 \hat{i} - \hat{j}$
$∠ ABC$ is angle between $\vec{AB}$ and $\vec{CB}$
$∴ \vec{AB} \cdot \vec{CB} = | \vec{AB} | | \vec{CB} | \cos θ$
$(6 - 1) = \sqrt{3^{2} + 1^{2} + 1^{2}} \times \sqrt{(2)^{2} + (- 1)^{2}} \times \cos θ$
$5 = \sqrt{11} \times \sqrt{5} \times \cos θ$
$\cos θ = \frac{5}{\sqrt{11} \times \sqrt{5}}$
$\cos θ = \frac{\sqrt{5}}{\sqrt{11}}$
$θ = \cos^{- 1} (\frac{\sqrt{5}}{\sqrt{11}})$

WB JEE 2018

Motion in Plane

143604 Three vectors $A = a \hat{i} + \hat{j} + \hat{k}; B = \hat{i} + b \hat{j} + \hat{k}$ and $C = \hat{i} + \hat{j} + c k$ are mutually perpendicular $(\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $X, Y$ and $Z$ - axis respectively). The respective values of $a, b$ and c are

1

0, 0, 0

2

- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}

3

1, - 1, 1

4

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

Explanation:

B Given that, $A = a \hat{i} + \hat{j} + \hat{k}, B = \hat{i} + b \hat{j} + \hat{k}, C =$ $\hat{i} + \hat{j} + c \hat{k}$ are mutually perpendicular-
So, A.B $= 0$
B. $C = 0$
A. $C = 0$
Now, $a + b + 1 = 0 \dots \dots$ (i)
$1 + b + c = 0$
$a + 1 + c = 0$
On adding equation (i), (ii) and (iii), we get-
$2 (a + b + c) = - 3$
$\Rightarrow a + b + c = - 3 / 2$
By equation (i) and (iv) we get-
$- 1 + c = - 3 / 2$
$c = - 1 / 2$
Similarly, $b = - 1 / 2$
$a = - 1 / 2$

WB JEE 2017

Motion in Plane

143605 If $A = B + C$ have scalar magnitudes of $5, 4, 3$ units respectively, then the angle between $A$ and $C$ is

1

\cos^{- 1} (3 / 5)

2

\cos^{- 1} (4 / 5)

3

π / 2

4

\sin^{- 1} (3 / 4)

Explanation:

A Here, triangle $PQR$ is a given with vector. $A$ , $B, C$ are its adjacent sides.

5, 4 and 3 makes the triangle right angle triangle.
$\cos θ = \frac{QR}{PQ}$
$θ = \cos^{- 1} (\frac{| C |}{| A |})$
$θ = \cos^{- 1} (\frac{3}{5})$

WB JEE 2012

Motion in Plane

143606 The magnitudes of vectors $A, B$ and $C$ are 3,4 and 5 units respectively. If $A + B = C$ , the angle between $A$ and $B$ is

1

\frac{π}{2}

2

\cos^{- 1} (0.6)

3

\tan - 1 (\frac{7}{5})

4

\frac{π}{4}

Explanation:

A Given that, $\vec{A} = 3, \vec{B} = 4, \vec{C} = 5$
Let $θ$ be the angle between them
$\vec{A} + \vec{B} = \vec{C}$
Squaring on both side
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 \vec{A} \vec{B} = | \vec{C} |^{2}$
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 | \vec{A} | | \vec{B} | \cos θ = | \vec{C} |^{2}$
On putting the value, we get-
$(3)^{2} + (4)^{2} + 2 \times 3 \times 4 \cos θ = (5)^{2}$
$25 + 24 \cos θ = 25$
$\cos θ = 0$
$θ = \frac{π}{2}$

AIPMT 1988

Motion in Plane

143602 Consider the vectors $A = \hat{i} + \hat{j} - \hat{k}, B =$ $2 \hat{i} - \hat{j} + \hat{k}$ and $C = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$ . What is the value of $C . (A \times B)$ ?

1 1

2 0

3

3 \sqrt{2}

4

18 \sqrt{5}

Explanation:

B Given, $\vec{A} = \hat{i} + \hat{j} - \hat{k}, \vec{B} = 2 \hat{i} - \hat{j} + \hat{k}$
$\vec{C} = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k})$
Now,
$\vec{A} \times \vec{B} = | \begin{array}{rrr} \hat{i} \hat{j} \hat{k} \\ 11 - 1 \\ 2 - 11 \end{array} |$
$\vec{A} \times \vec{B} = - 3 \hat{j} - 3 \hat{k}$
According to the question,
$\vec{C} \cdot (\vec{A} \times \vec{B}) = \frac{1}{\sqrt{5}} (\hat{i} - 2 \hat{j} + 2 \hat{k}) \cdot (0 \hat{i} - 3 \hat{j} - 3 \hat{k})$
$= \frac{1}{\sqrt{5}} (6 - 6)$
$= 0$

WB JEE 2020

Motion in Plane

143603 In a triangle $ABC$ , the sides $AB$ and $AC$ are represented by the vectors $3 \hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + \hat{k}$ respectively. Calculate the angle $∠ ABC$ .

1

\cos^{- 1} \sqrt{\frac{5}{11}}

2

\cos^{- 1} \sqrt{\frac{6}{11}}

3

(90^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

4

(180^{\circ} - \cos^{- 1} \sqrt{\frac{5}{11}})

Explanation:

A
Given that, $\vec{A B} = 3 \hat{i} + \hat{j} + \hat{k}, \vec{A C} = \hat{i} + 2 \hat{j} + \hat{k}$
$∴ \vec{CB} = \vec{AB} - \vec{AC}$
$\vec{CB} = 3 \hat{i} + \hat{j} + \hat{k} - (\hat{i} + 2 \hat{j} + \hat{k})$
$\vec{CB} = 2 \hat{i} - \hat{j}$
$∠ ABC$ is angle between $\vec{AB}$ and $\vec{CB}$
$∴ \vec{AB} \cdot \vec{CB} = | \vec{AB} | | \vec{CB} | \cos θ$
$(6 - 1) = \sqrt{3^{2} + 1^{2} + 1^{2}} \times \sqrt{(2)^{2} + (- 1)^{2}} \times \cos θ$
$5 = \sqrt{11} \times \sqrt{5} \times \cos θ$
$\cos θ = \frac{5}{\sqrt{11} \times \sqrt{5}}$
$\cos θ = \frac{\sqrt{5}}{\sqrt{11}}$
$θ = \cos^{- 1} (\frac{\sqrt{5}}{\sqrt{11}})$

WB JEE 2018

Motion in Plane

143604 Three vectors $A = a \hat{i} + \hat{j} + \hat{k}; B = \hat{i} + b \hat{j} + \hat{k}$ and $C = \hat{i} + \hat{j} + c k$ are mutually perpendicular $(\hat{i}, \hat{j}$ and $\hat{k}$ are unit vectors along $X, Y$ and $Z$ - axis respectively). The respective values of $a, b$ and c are

1

0, 0, 0

2

- \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}

3

1, - 1, 1

4

\frac{1}{2}, \frac{1}{2}, \frac{1}{2}

Explanation:

B Given that, $A = a \hat{i} + \hat{j} + \hat{k}, B = \hat{i} + b \hat{j} + \hat{k}, C =$ $\hat{i} + \hat{j} + c \hat{k}$ are mutually perpendicular-
So, A.B $= 0$
B. $C = 0$
A. $C = 0$
Now, $a + b + 1 = 0 \dots \dots$ (i)
$1 + b + c = 0$
$a + 1 + c = 0$
On adding equation (i), (ii) and (iii), we get-
$2 (a + b + c) = - 3$
$\Rightarrow a + b + c = - 3 / 2$
By equation (i) and (iv) we get-
$- 1 + c = - 3 / 2$
$c = - 1 / 2$
Similarly, $b = - 1 / 2$
$a = - 1 / 2$

WB JEE 2017

Motion in Plane

143605 If $A = B + C$ have scalar magnitudes of $5, 4, 3$ units respectively, then the angle between $A$ and $C$ is

1

\cos^{- 1} (3 / 5)

2

\cos^{- 1} (4 / 5)

3

π / 2

4

\sin^{- 1} (3 / 4)

Explanation:

A Here, triangle $PQR$ is a given with vector. $A$ , $B, C$ are its adjacent sides.

5, 4 and 3 makes the triangle right angle triangle.
$\cos θ = \frac{QR}{PQ}$
$θ = \cos^{- 1} (\frac{| C |}{| A |})$
$θ = \cos^{- 1} (\frac{3}{5})$

WB JEE 2012

Motion in Plane

143606 The magnitudes of vectors $A, B$ and $C$ are 3,4 and 5 units respectively. If $A + B = C$ , the angle between $A$ and $B$ is

1

\frac{π}{2}

2

\cos^{- 1} (0.6)

3

\tan - 1 (\frac{7}{5})

4

\frac{π}{4}

Explanation:

A Given that, $\vec{A} = 3, \vec{B} = 4, \vec{C} = 5$
Let $θ$ be the angle between them
$\vec{A} + \vec{B} = \vec{C}$
Squaring on both side
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 \vec{A} \vec{B} = | \vec{C} |^{2}$
$| \vec{A} |^{2} + | \vec{B} |^{2} + 2 | \vec{A} | | \vec{B} | \cos θ = | \vec{C} |^{2}$
On putting the value, we get-
$(3)^{2} + (4)^{2} + 2 \times 3 \times 4 \cos θ = (5)^{2}$
$25 + 24 \cos θ = 25$
$\cos θ = 0$
$θ = \frac{π}{2}$

AIPMT 1988