Show that the vectors 2 i ^ + 3 j ^ + 6 k ^ , 3 i ^ - 6 j ^ + 2 k ^ and 6 i ^ + 2 j ^ - 3 k ^ are perpendicular to each other

Let the vectors be a → = 2 i ^ + 3 j ^ + 6 k ^ , b → = 3 i ^ - 6 j ^ + 2 k ^ and c → = 6 i ^ + 2 j ^ - 3 k ^ . We need to check if a → · b → = 0 , b → · c → = 0 and c → · a → = 0 . a → · b → = 2 i ^ + 3 j ^ + 6 k ^ · 3 i ^ - 6 j ^ + 2 k ^ = ( 2 × 3 ) + ( 3 × - 6 ) + ( 6 × 2 ) = 6 - 18 + 12 = 0 b → · c → = 3 i ^ - 6 j ^ + 2 k ^ · 6 i ^ + 2 j ^ - 3 k ^ = ( 3 × 6 ) + ( - 6 × 2 ) + ( 2 × - 3 ) = 18 - 12 - 6 = 0 c → · a → = 6 i ^ + 2 j ^ - 3 k ^ · 2 i ^ + 3 j ^ + 6 k ^ = ( 6 × 2 ) + ( 2 × 3 ) + ( - 3 × 6 ) = 12 + 6 - 18 = 0 Since a → · b → = b → · c → = c → · a → = 0 , the given vectors are perpendicular to each other.

Show that the vectors 2 i ^ + 3 j ^ + 6 k ^ , 3 i ^ - 6 j ^ + 2 k ^ and 6 i ^ + 2 j ^ - 3 k ^ are perpendicular to each other

Let the vectors be a → = 2 i ^ + 3 j ^ + 6 k ^ , b → = 3 i ^ - 6 j ^ + 2 k ^ and c → = 6 i ^ + 2 j ^ - 3 k ^ . We need to check if a → · b → = 0 , b → · c → = 0 and c → · a → = 0 . a → · b → = 2 i ^ + 3 j ^ + 6 k ^ · 3 i ^ - 6 j ^ + 2 k ^ = ( 2 × 3 ) + ( 3 × - 6 ) + ( 6 × 2 ) = 6 - 18 + 12 = 0 b → · c → = 3 i ^ - 6 j ^ + 2 k ^ · 6 i ^ + 2 j ^ - 3 k ^ = ( 3 × 6 ) + ( - 6 × 2 ) + ( 2 × - 3 ) = 18 - 12 - 6 = 0 c → · a → = 6 i ^ + 2 j ^ - 3 k ^ · 2 i ^ + 3 j ^ + 6 k ^ = ( 6 × 2 ) + ( 2 × 3 ) + ( - 3 × 6 ) = 12 + 6 - 18 = 0 Since a → · b → = b → · c → = c → · a → = 0 , the given vectors are perpendicular to each other.

Show that the vectors 3 i ^ + j ^ + 2 k ^ , i ^ - j ^ - k ^ and i ^ + 5 j ^ - 4 k ^ are perpendicular to each other

Let the vectors be a → = 3 i ^ + j ^ + 2 k ^ , b → = i ^ - j ^ - k ^ and c → = i ^ + 5 j ^ - 4 k ^ . We need to check if a → · b → = 0 , b → · c → = 0 and c → · a → = 0 . a → · b → = 3 i ^ + j ^ + 2 k ^ · i ^ - j ^ - k ^ = ( 3 × 1 ) + ( 1 × - 1 ) + ( 2 × - 1 ) = 3 - 1 - 2 = 0 b → · c → = ( i ^ - j ^ - k ^ ) · ( i ^ + 5 j ^ - 4 k ^ ) = ( 1 × 1 ) + ( - 1 × 5 ) + ( - 1 × - 4 ) = 1 - 5 + 4 = 0 c → · a → = i ^ + 5 j ^ - 4 k ^ · 3 i ^ + j ^ + 2 k ^ = ( 1 × 3 ) + ( 5 × 1 ) + ( - 4 × 2 ) = 3 + 5 - 8 = 0 Since a → · b → = b → · c → = c → · a → = 0 , the vectors are perpendicular to each other.

If a → = 3 i ^ + 2 j ^ + 9 k ^ and b → = i ^ + λ j ^ + 3 k ^ , find the value of λ so that a → + b → is perpendicular to a → - b → .

The first step is to find the sum and difference of the two vectors, a → + b → = 3 i ^ + 2 j ^ + 9 k ^ + i ^ + λ j ^ + 3 k ^ = 4 i ^ + 2 + λ j ^ + 12 k ^ a → - b → = 3 i ^ + 2 j ^ + 9 k ^ - i ^ + λ j ^ + 3 k ^ = 2 i ^ + 2 - λ j ^ + 6 k ^ Since the above two vectors are perpendicular, their scalar product is 0 . a → + b → · a → - b → = 0 ⇒ 4 i ^ + 2 + λ j ^ + 12 k ^ · 2 i ^ + 2 - λ j ^ + 6 k ^ = 0 ⇒ 4 × 2 + 2 + λ 2 - λ + 12 × 6 = 0 ⇒ 8 + 2 2 - λ 2 + 72 = 0 ⇒ - λ 2 = - 84 ⇒ λ 2 = 84 ⇒ λ = 84 = ± 2 21 .

If a → = 4 i ^ + 2 j ^ - k ^ and b → = 5 i ^ + 2 j ^ - 3 k ^ , find the angle between the vectors a → + b → and a → - b →

We need to compute the vectors, a → + b → = 4 i ^ + 2 j ^ - k ^ + 5 i ^ + 2 j ^ - 3 k ^ = 9 i ^ + 4 j ^ - 4 k ^ a → - b → = 4 i ^ + 2 j ^ - k ^ - 5 i ^ + 2 j ^ - 3 k ^ = - i ^ + 2 k ^ The formula to find the angle between two vectors is given by cos θ = a → + b → · a → - b → a → + b → a → - b → . Finding the scalar product, a → + b → · a → - b → = 9 i ^ + 4 j ^ - 4 k ^ · - i ^ + 2 k ^ = 9 × - 1 + 4 × 0 + - 4 × 2 = - 9 + 0 - 8 = - 17 Finding the magnitude of vectors, a → + b → = 9 2 + 4 2 + - 4 2 = 81 + 16 + 16 = 113 a → - b → = - 1 2 + 2 2 = 1 + 4 = 5 Therefore, the angle between the vectors is cos θ = - 17 113 × 5 = - 17 565 ⇒ θ = cos - 1 - 17 565 = π - cos - 1 17 565 .

If a → = 5 i ^ - j ^ + 3 k ^ and b → = i ^ + 3 j ^ - 5 k ^ , then show that the vectors a → + b → and a → - b → are perpendicular

The first step is to compute the sum and difference of the two given vectors, a → + b → = 5 i ^ - j ^ + 3 k ^ + i ^ + 3 j ^ - 5 k ^ = 6 i ^ + 2 j ^ - 2 k ^ a → - b → = 5 i ^ - j ^ + 3 k ^ - i ^ + 3 j ^ - 5 k ^ = 4 i ^ - 4 j ^ + 8 k ^ To show that the above two vectors are perpendicular, their scalar product must be zero. Finding their scalar product, a → + b → · a → - b → = 6 i ^ + 2 j ^ - 2 k ^ · 4 i ^ - 4 j ^ + 8 k ^ = ( 6 × 4 ) + ( 2 × - 4 ) + ( - 2 × 8 ) = 0 Since the scalar product is 0 , they are perpendicular.

For what value of λ , the vectors a → = 2 i ^ + λ j ^ + k ^ and b → = i ^ - 2 j ^ + 3 k ^ perpendicular to each other?

In order to get the values of λ for which the vectors a → = 2 i ^ + λ j ^ + k ^ and b → = i ^ - 2 j ^ + 3 k ^ are perpendicular, the condition is that their dot product must be zero. a → · b → = 0 ⇒ 2 i ^ + λ j ^ + k ^ · i ^ - 2 j ^ + 3 k ^ = 0 ⇒ 2 × 1 + λ × - 2 + 1 × 3 = 0 ⇒ 5 - 2 λ = 0 λ = 5 2 .

If O A → = 2 i ^ - j ^ + k ^ , O B → = i ^ - 3 j ^ - 5 k ^ and O C → = 3 i ^ - 4 j ^ - 4 k ^ , then show that C B → is perpendicular to A C →

The position vectors O A → = 2 i ^ - j ^ + k ^ , O B → = i ^ - 3 j ^ - 5 k ^ and O C → = 2 i ^ - 4 j ^ - 4 k ^ are given in the question, so vectors C B → and A C → can be found out as C B → = O B → - O C → = ( i ^ - 3 j ^ - 5 k ^ ) - ( 3 i ^ - 4 j ^ - 4 k ^ ) = - 2 i ^ + j ^ - k ^ A C → = O C → - O A → = ( 3 i ^ - 4 j ^ - 4 k ^ ) - ( 2 i ^ - j ^ + k ^ ) = i ^ - 3 j ^ - 5 k ^ The scalar product of C B → and A C → must be zero for them to be perpendicular. C B → · A C → = ( - 2 i ^ + j ^ - k ^ ) · ( i ^ - 3 j ^ - 5 k ^ ) = ( - 2 × 1 ) + ( 1 × - 3 ) + ( - 1 × - 5 ) = - 2 - 3 + 5 = 0 Therefore, C B → is perpendicular to A C → .

Show that each of the following three vectors is a unit vector. 1 7 2 i ^ + 3 j ^ + 6 k ^ , 1 7 3 i ^ - 6 j ^ + 2 k ^ , 1 7 6 i ^ + 2 j ^ - 3 k ^ . Also, show that they are mutually perpendicular to each other.

Let the vectors be a → = 1 7 ( 2 i ^ + 3 j ^ + 6 k ^ ) , b → = 1 7 ( 3 i ^ - 6 j ^ + 2 k ^ ) , c → = 1 7 ( 6 i ^ + 2 j ^ - 3 k ^ ) . To show that the three vectors are unit vectors, we need to check if their magnitudes are 1 . a → = 1 7 2 ( 2 2 + 3 2 + 6 2 ) = 1 49 ( 4 + 9 + 36 ) = 1 49 ( 49 ) = 1 b → = 1 7 2 ( 3 2 + ( - 6 ) 2 + 2 2 ) = 1 49 ( 9 + 36 + 4 ) = 1 49 ( 49 ) = 1 c → = 1 7 2 ( 6 2 + 2 2 + ( - 3 ) 2 ) = 1 49 ( 36 + 4 + 9 ) = 1 49 ( 49 ) = 1 Since the magnitudes of all the three vectors are 1 , they are unit vectors. To prove that they are mutually perpendicular to each other, we need to show that a → · b → = b → · c → = c → · a → = 0 . a → · b → = 1 7 2 i ^ + 3 j ^ + 6 k ^ · 1 7 3 i ^ - 6 j ^ + 2 k ^ = 1 49 ( 2 × 3 ) + ( 3 × - 6 ) + ( 6 × 2 ) = 1 49 6 - 18 + 12 = 0 b → · c → = 1 7 3 i ^ - 6 j ^ + 2 k ^ · 1 7 6 i ^ + 2 j ^ - 3 k ^ = 1 49 3 × 6 + - 6 × 2 + 2 × - 3 = 1 49 18 - 12 - 6 = 0 c → · a → = 1 7 6 i ^ + 2 j ^ - 3 k ^ · 1 7 2 i ^ + 3 j ^ + 6 k ^ = 1 49 ( 6 × 2 ) + ( 2 × 3 ) + ( - 3 × 6 ) = 1 49 12 + 6 - 18 = 0 Hence, they are mutually perpendicular to each other.

E M B I B E

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 8

MEDIUM

JEE Advanced

IMPORTANT

Earn 100

Show that the vectors $2 \hat{i} + 3 \hat{j} + 6 \hat{k}, 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$ and $6 \hat{i} + 2 \hat{j} - 3 \hat{k}$ are perpendicular to each other

Important Questions on Scalar or Dot Product of Two Vectors

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 8

Show that the vectors $3 \hat{i} + \hat{j} + 2 \hat{k}, \hat{i} - \hat{j} - \hat{k}$ and $\hat{i} + 5 \hat{j} - 4 \hat{k}$ are perpendicular to each other

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 9

If

\vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k}

and

\vec{b} = \hat{i} + λ \hat{j} + 3 \hat{k}

, find the value of

λ

so that

\vec{a} + \vec{b}

is perpendicular to

\vec{a} - \vec{b}

.

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 10

If

\vec{a} = 4 \hat{i} + 2 \hat{j} - \hat{k}

and

\vec{b} = 5 \hat{i} + 2 \hat{j} - 3 \hat{k}

, find the angle between the vectors

\vec{a} + \vec{b}

and

\vec{a} - \vec{b}

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 11

If

\vec{a} = 5 \hat{i} - \hat{j} + 3 \hat{k}

and

\vec{b} = \hat{i} + 3 \hat{j} - 5 \hat{k}

, then show that the vectors

\vec{a} + \vec{b}

and

\vec{a} - \vec{b}

are perpendicular

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 12

For what value of

λ

, the vectors

\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k}

and

\vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}

perpendicular to each other?

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 13

If

\vec{O A} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{O B} = \hat{i} - 3 \hat{j} - 5 \hat{k}

and

\vec{O C} = 3 \hat{i} - 4 \hat{j} - 4 \hat{k}

, then show that

\vec{C B}

is perpendicular to

\vec{A C}

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 14

If

\vec{a} = 2 \hat{i} + 2 \hat{j} + 3 \hat{k}, \vec{b} = - \hat{i} + 2 \hat{j} + \hat{k}

and

\vec{c} = 3 \hat{i} + \hat{j}

such that

\vec{a} + λ \vec{b}

is perpendicular to

\vec{c}

, then find the value of

λ

.

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 15

Show that each of the following three vectors is a unit vector.

$\frac{1}{7} (2 \hat{i} + 3 \hat{j} + 6 \hat{k}), \frac{1}{7} (3 \hat{i} - 6 \hat{j} + 2 \hat{k}), \frac{1}{7} (6 \hat{i} + 2 \hat{j} - 3 \hat{k})$ . Also, show that they are mutually perpendicular to each other.