Prove that the vectors 2 i ^ - j ^ + k ^ and i ^ - 3 j ^ - 5 k ^ are at right angles.

Let the vectors be a → = 2 i ^ - j ^ + k ^ and b → = i ^ - 3 j ^ - 5 k ^ . The condition for perpendicular vectors is a → · b → = 0 . So, applying the condition, a → · b → = 2 i ^ - j ^ + k ^ · i ^ - 3 j ^ - 5 k ^ = ( 2 × 1 ) + ( - 1 × - 3 ) + ( 1 × - 5 ) = 2 + 3 - 5 = 0 Since the condition is true, the vectors are at right angles.

Prove that the vectors 2 i ^ - j ^ + k ^ and i ^ - 3 j ^ - 5 k ^ are at right angles.

Let the vectors be a → = 2 i ^ - j ^ + k ^ and b → = i ^ - 3 j ^ - 5 k ^ . The condition for perpendicular vectors is a → · b → = 0 . So, applying the condition, a → · b → = 2 i ^ - j ^ + k ^ · i ^ - 3 j ^ - 5 k ^ = ( 2 × 1 ) + ( - 1 × - 3 ) + ( 1 × - 5 ) = 2 + 3 - 5 = 0 Since the condition is true, the vectors are at right angles.

Prove that the vectors 2 i ^ + 5 j ^ - k ^ and 3 i ^ - j ^ + k ^ are at right angles

Let the vectors be a → = 2 i ^ + 5 j ^ - k ^ and b → = 3 i ^ - j ^ + k ^ . In order to prove that the vectors are at right angles, we have to prove that they are perpendicular. The condition for perpendicular vectors is a → · b → = 0 . So, applying the condition, a → · b → = 2 i ^ + 5 j ^ - k ^ · 3 i ^ - j ^ + k ^ = ( 2 × 3 ) + ( 5 × - 1 ) + ( - 1 × 1 ) = 6 - 5 - 1 = 0 Since the condition is true, the vectors are at right angles.

Find the angle between the vectors 3 i ^ + 4 j ^ and 2 j ^ - 5 k ^

Let the vectors be a → = 3 i ^ + 4 j ^ and b → = 2 j ^ - 5 k ^ . The formula to find the angle between two vectors is given by cos θ = a → · b → a → b → . The scalar product of the vectors can be found as a → · b → = 3 i ^ + 4 j ^ · 2 j ^ - 5 k ^ = 3 × 0 + 4 × 2 + 0 × - 5 = 0 + 8 + 0 = 8 The magnitude of the vectors is obtained as a → = 3 2 + 4 2 = 9 + 16 = 25 = 5 b → = 2 2 + - 5 2 = 4 + 25 = 29 Substituting these values in the equation, cos θ = 8 5 × 29 θ = cos - 1 8 5 29 .

Find the angle between the vectors 3 i ^ + 4 j ^ + 5 k ^ and i ^ + j ^ + k ^ . Also find the sine of the angle between them

Let the vectors be a → = 3 i ^ + 4 j ^ + 5 k ^ and b → = i ^ + j ^ + k ^ . The formula to find the angle between two vectors is given by cos θ = a → · b → a → b → . The scalar product of the vectors can be found as a → · b → = 3 i ^ + 4 j ^ + 5 k ^ · i ^ + j ^ + k ^ = 3 × 1 + 4 × 1 + 5 × 1 = 3 + 4 + 5 = 12 The magnitude of the vectors is obtained as a → = 3 2 + 4 2 + 5 2 = 9 + 16 + 25 = 50 b → = 1 2 + 1 2 + 1 2 = 1 + 1 + 1 = 3 Substituting these values in the equation, cos θ = 12 50 × 3 = 12 5 2 × 3 × 2 × 3 2 × 3 = 12 6 5 × 6 = 2 6 5 θ = cos - 1 2 6 5 The sine of the angle between the vectors is given by sin θ = sin cos - 1 2 6 5 There is an expression, sin cos - 1 x = 1 - x 2 . Using this, we can find the sine of the angle. sin θ = sin cos - 1 2 6 5 ⇒ 1 - 2 6 5 2 = 1 - 4 × 6 25 = 1 - 24 25 = 1 25 = 1 5 .

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

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

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 5

EASY

JEE Advanced

IMPORTANT

Earn 100

Prove that the vectors $2 \hat{i} - \hat{j} + \hat{k}$ and $\hat{i} - 3 \hat{j} - 5 \hat{k}$ are at right angles.

Important Questions on Scalar or Dot Product of Two Vectors

EASY

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 5

Prove that the vectors $2 \hat{i} + 5 \hat{j} - \hat{k}$ and $3 \hat{i} - \hat{j} + \hat{k}$ are at right angles

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 6

Find the angle between the vectors

3 \hat{i} + 4 \hat{j}

and

2 \hat{j} - 5 \hat{k}

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 7

Find the angle between the vectors

3 \hat{i} + 4 \hat{j} + 5 \hat{k}

and

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

. Also find the sine of the angle between them

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 $6 \hat{i} + 3 \hat{j} + 2 \hat{k}, 2 \hat{i} - 6 \hat{j} + 3 \hat{k}$ and $- 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$ 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 8

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

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