Find the angle between the vectors 3 i ^ + 2 j ^ - 6 k ^ and 4 i ^ - 3 j ^ + k ^

The formula to find the angle between two vectors is given by cos θ = a → · b → a → b → . Let a → = 3 i ^ + 2 j ^ - 6 k ^ and b → = 4 i ^ - 3 j ^ + k ^ . The scalar product of the vectors, a → · b → = 3 i ^ + 2 j ^ - 6 k ^ · 4 i ^ - 3 j ^ + k ^ = 3 × 4 + 2 × - 3 + - 6 × 1 = 12 - 6 - 6 = 0 Since, the numerator in the formula is obtained as zero, cos θ = 0 a → b → = 0 ⇒ θ = cos - 1 0 = π 2 .

Find the angle between the vectors 3 i ^ + 2 j ^ - 6 k ^ and 4 i ^ - 3 j ^ + k ^

The formula to find the angle between two vectors is given by cos θ = a → · b → a → b → . Let a → = 3 i ^ + 2 j ^ - 6 k ^ and b → = 4 i ^ - 3 j ^ + k ^ . The scalar product of the vectors, a → · b → = 3 i ^ + 2 j ^ - 6 k ^ · 4 i ^ - 3 j ^ + k ^ = 3 × 4 + 2 × - 3 + - 6 × 1 = 12 - 6 - 6 = 0 Since, the numerator in the formula is obtained as zero, cos θ = 0 a → b → = 0 ⇒ θ = cos - 1 0 = π 2 .

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

The formula to find the angle between two vectors a → , b → is given by cos θ = a → · b → a → b → . Let a → = 2 i ^ - 3 j ^ + k ^ and b → = 3 i ^ - j ^ - 2 k ^ . The scalar product of the vectors is a → · b → = 2 i ^ - 3 j ^ + k ^ · 3 i ^ - j ^ - 2 k ^ = 2 × 3 + - 3 × - 1 + 1 × - 2 = 6 + 3 - 2 = 7 Now, finding the magnitude of the vectors, a → = 2 2 + - 3 2 + 1 2 = 4 + 9 + 1 = 14 b → = 3 2 + - 1 2 + ( - 2 ) 2 = 9 + 1 + 4 = 14 Substituting all this in the equation, cos θ = 7 14 × 14 = 7 14 = 1 2 ∴ θ = cos - 1 1 2 = π 3 .

Find the angle between the vectors i ^ + j ^ - k ^ and i ^ - j ^ + k ^ .

The formula to find the angle between two vectors is given by cos θ = a → · b → a → b → . Let a → = i ^ + j ^ - k ^ and b → = i ^ - j ^ + k ^ . The scalar product of the vectors is obtained as a → · b → = i ^ + j ^ - k ^ · i ^ - j ^ + k ^ = 1 × 1 + 1 × - 1 + - 1 × 1 = 1 - 1 - 1 = - 1 The magnitude of the vectors is found as a → = 1 2 + 1 2 + ( - 1 ) 2 = 1 + 1 + 1 = 3 b → = 1 2 + ( - 1 ) 2 + 1 2 = 1 + 1 + 1 = 3 Substituting the values in the equation, cos θ = - 1 3 × 3 = - 1 3 θ = cos - 1 - 1 3 ⇒ π - cos - 1 1 3 .

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

The formula to find the angle between two vectors is given by cos θ = a → · b → a → b → . Let a → = i ^ - 2 j ^ + 3 k ^ and b → = 3 i ^ - 2 j ^ + k ^ . The scalar product of the vectors can be found as a → · b → = i ^ - 2 j ^ + 3 k ^ · 3 i ^ - 2 j ^ + k ^ = 1 × 3 + - 2 × - 2 + 3 × 1 = 3 + 4 + 3 = 10 The magnitude of the vectors is obtained as a → = 1 2 + ( - 2 ) 2 + 3 2 = 1 + 4 + 9 = 14 b → = 3 2 + ( - 2 ) 2 + 1 2 = 9 + 4 + 1 = 14 Substituting these values in the equation, cos θ = 10 14 × 14 = 10 14 = 5 7 θ = cos - 1 5 7 .

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.

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 4

EASY

JEE Advanced

IMPORTANT

Earn 100

Find the angle between the vectors $3 \hat{i} + 2 \hat{j} - 6 \hat{k}$ and $4 \hat{i} - 3 \hat{j} + \hat{k}$

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 4

Find the angle between the vectors

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

and

3 \hat{i} - \hat{j} - 2 \hat{k}

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 4

Find the angle between the vectors

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

and

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

.

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 4

Find the angle between the vectors

\hat{i} - 2 \hat{j} + 3 \hat{k}

and

3 \hat{i} - 2 \hat{j} + \hat{k}

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

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