If a → = 6 i ^ + 8 j ^ and b → = 3 j ^ + 4 k ^ , find the vector component of a → along b → .

The vector component of a → along b → can be obtained using the formula as a → · b → b → 2 b → . = 6 i ^ + 8 j ^ · 3 j ^ + 4 k ^ 3 2 + 4 2 2 3 j ^ + 4 k ^ = 6 × 0 + 8 × 3 + 0 × 4 9 + 16 2 3 j ^ + 4 k ^ = 24 25 2 3 j ^ + 4 k ^ = 24 25 3 j ^ + 4 k ^

If a → = 6 i ^ + 8 j ^ and b → = 3 j ^ + 4 k ^ , find the vector component of a → along b → .

The vector component of a → along b → can be obtained using the formula as a → · b → b → 2 b → . = 6 i ^ + 8 j ^ · 3 j ^ + 4 k ^ 3 2 + 4 2 2 3 j ^ + 4 k ^ = 6 × 0 + 8 × 3 + 0 × 4 9 + 16 2 3 j ^ + 4 k ^ = 24 25 2 3 j ^ + 4 k ^ = 24 25 3 j ^ + 4 k ^

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

The projections of a line on the axes are 3 , 4 and 2 6 . Find the length of the line.

Given that, projections of line on axes are 3 , 4 and 2 6 Let d l = 3 , d m = 4 and d n = 2 6 , where d is the length of the line and l , m , n are direction cosines of the line. d 2 l 2 + d 2 m 2 + d 2 n 2 = 9 + 16 + 24 ⇒ d 2 l 2 + m 2 + n 2 = 49 ⇒ d 2 = 49 ⇒ d = 7 ∴ Length of line = 7 units

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

Let a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ . Find a vector d → which is perpendicular to both a → and b → and c → · d → = 15 .

Given a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ since d is perpendicular to both a and b , it is parallel to a → × b → . Suppose d = λ ( a → × b → ) for some scalar λ d = λ i ^ j ^ k ^ 1 4 2 3 - 2 7 = λ [ ( 28 + 4 ) i ^ - ( 7 - 6 ) j ^ + ( - 2 - 12 ) k ^ ] = λ [ 32 i ^ - j ^ - 14 k ^ ] c → · d → = 15 (Given) ( 2 i ^ - j ^ + 4 k ^ ) · λ ( 32 i ^ - j ^ - 14 k ^ ) = 15 λ ( 64 + 1 - 56 ) = 15 λ = 5 3 ∴ d → = 5 3 ( 32 i ^ - j ^ - 14 k ^ ) ⇒ d → = 1 3 ( 160 i ^ - 5 j ^ - 70 k ^ )

The two adjacent sides of a parallelogram are 2 i ^ - 4 j ^ + 5 k ^ and i ^ - 2 j ^ - 3 k ^ .Find the unit vector parallel to its diagonal. Also, find its area.

Diagonal of a parallelogram is a → + b → a → + b → = ( 2 + 1 ) i ^ + ( - 4 - 2 ) j ^ + ( 5 - 3 ) k ^ = 3 i ^ - 6 j ^ + 2 k ^ So, the unit vector parallel to diagonal is a → + b → | a → + b → | = 3 i ^ - 6 j ^ + 2 k ^ 3 2 + ( - 6 ) 2 + 2 2 = 3 i ^ - 6 j ^ + 2 k ^ 9 + 36 + 4 = 3 i ^ - 6 j ^ + 2 k ^ 7 = 3 7 i ^ - 6 7 j ^ + 2 7 k ^ Area of a parallelogram = a → × b → a → × b → = i ^ j ^ k ^ 2 - 4 5 1 - 2 - 3 = i ^ ( 12 + 10 ) - j ^ ( - 6 - 5 ) + k ^ ( - 4 + 4 ) = 22 ı ^ + 11 ȷ ^ = 11 ( 2 i ^ + j ^ ) ∴ | a → × b → | = 11 2 2 + 1 2 = 11 5 So, area of parallelogram is 11 5 sq units

For any two vectors a → and b → ,prove that | a → + b → | ≤ | a → | + | b → |

From dot product, | a → + b → | 2 = ( a → + b → ) · ( a → + b → ) = a → · a → + a → · b → + b → · a → + b → . b → ∵ a → · b → = b → · a → = | a → | 2 + 2 a → · b → + | b → | 2 = | a → | 2 + 2 | a → | | b → | | cos θ | + | b → | 2 We know that 0 ≤ cos θ ≤ 1 Considering cos θ ≤ 1 ⇒ 2 | a → | | b → | cos θ ≤ 2 | a → | | b → | ⇒ | a → | 2 + 2 | a → | | b → | | cos θ | + | b → | 2 ≤ | a → | 2 + 2 | a → | | b → | + | b → | 2 ⇒ a → + b → 2 ≤ a → 2 + 2 a → b → + b → 2 ⇒ a → + b → 2 ≤ a → + b → 2 ∴ | a → + b → | ≤ | a → | + | b → | Hence proved.

Find the unit vector in the direction of the vector a → + b → where a → = 2 i ^ - j ^ + 2 k ^ and b → = - i ^ + j ^ - k ^ .

Given vectors are a → = 2 i ^ - j ^ + 2 k ^ and b → = - i ^ + j ^ - k ^ a → + b → = 2 i ^ - j ^ + 2 k ^ + - i ^ + j ^ - k ^ = i ^ + k ^ | a → + b → | = ( - 1 ) 2 + 1 2 = 2 Unit vector in the direction of a vector p → is p → p → . Therefore, unit vector in the direction of a → + b → is = i ^ + k ^ 2

For any two vectors a → and b → ,prove that | a → + b → | ≤ | a → | + | b → |

From dot product, | a → + b → | 2 = ( a → + b → ) · ( a → + b → ) = a → · a → + a → · b → + b → · a → + b → . b → ∵ a → · b → = b → · a → = | a → | 2 + 2 a → · b → + | b → | 2 = | a → | 2 + 2 | a → | | b → | | cos θ | + | b → | 2 We know that 0 ≤ cos θ ≤ 1 Considering cos θ ≤ 1 ⇒ 2 | a → | | b → | cos θ ≤ 2 | a → | | b → | ⇒ | a → | 2 + 2 | a → | | b → | | cos θ | + | b → | 2 ≤ | a → | 2 + 2 | a → | | b → | + | b → | 2 ⇒ a → + b → 2 ≤ a → 2 + 2 a → b → + b → 2 ⇒ a → + b → 2 ≤ a → + b → 2 ∴ | a → + b → | ≤ | a → | + | b → | Hence proved.

The two adjacent sides of a parallelogram are 2 i ^ - 4 j ^ + 5 k ^ and i ^ - 2 j ^ - 3 k ^ .Find the unit vector parallel to its diagonal. Also, find its area.

Diagonal of a parallelogram is a → + b → a → + b → = ( 2 + 1 ) i ^ + ( - 4 - 2 ) j ^ + ( 5 - 3 ) k ^ = 3 i ^ - 6 j ^ + 2 k ^ So, the unit vector parallel to diagonal is a → + b → | a → + b → | = 3 i ^ - 6 j ^ + 2 k ^ 3 2 + ( - 6 ) 2 + 2 2 = 3 i ^ - 6 j ^ + 2 k ^ 9 + 36 + 4 = 3 i ^ - 6 j ^ + 2 k ^ 7 = 3 7 i ^ - 6 7 j ^ + 2 7 k ^ Area of a parallelogram = a → × b → a → × b → = i ^ j ^ k ^ 2 - 4 5 1 - 2 - 3 = i ^ ( 12 + 10 ) - j ^ ( - 6 - 5 ) + k ^ ( - 4 + 4 ) = 22 ı ^ + 11 ȷ ^ = 11 ( 2 i ^ + j ^ ) ∴ | a → × b → | = 11 2 2 + 1 2 = 11 5 So, area of parallelogram is 11 5 sq units

Let a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ . Find a vector d → which is perpendicular to both a → and b → and c → · d → = 15 .

Given a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ since d is perpendicular to both a and b , it is parallel to a → × b → . Suppose d = λ ( a → × b → ) for some scalar λ d = λ i ^ j ^ k ^ 1 4 2 3 - 2 7 = λ [ ( 28 + 4 ) i ^ - ( 7 - 6 ) j ^ + ( - 2 - 12 ) k ^ ] = λ [ 32 i ^ - j ^ - 14 k ^ ] c → · d → = 15 (Given) ( 2 i ^ - j ^ + 4 k ^ ) · λ ( 32 i ^ - j ^ - 14 k ^ ) = 15 λ ( 64 + 1 - 56 ) = 15 λ = 5 3 ∴ d → = 5 3 ( 32 i ^ - j ^ - 14 k ^ ) ⇒ d → = 1 3 ( 160 i ^ - 5 j ^ - 70 k ^ )

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

The projections of a line on the axes are 3 , 4 and 2 6 . Find the length of the line.

Given that, projections of line on axes are 3 , 4 and 2 6 Let d l = 3 , d m = 4 and d n = 2 6 , where d is the length of the line and l , m , n are direction cosines of the line. d 2 l 2 + d 2 m 2 + d 2 n 2 = 9 + 16 + 24 ⇒ d 2 l 2 + m 2 + n 2 = 49 ⇒ d 2 = 49 ⇒ d = 7 ∴ Length of line = 7 units

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

Prove that, ( a → + b → ) · ( a → + b → ) = | a → | 2 + | b → | 2 if and only if a → , b → are perpendicular, given a → ≠ 0 → , b → ≠ 0 → .

To prove: ( a → + b → ) · ( a → + b → ) = | a → | 2 + | b → | 2 if and only if a → and b → are perpendicular. Consider ( a → + b → ) · ( a → + b → ) = | a → | 2 + | b → | 2 Applying distributive property of dot product over addition a → + b → · a → + b → = a → · a → + b → + b → · a → + b → = a → · a → + a → . b → + b → · a → + b → · b → = | a → | 2 + | b → | 2 ⇒ | a → | 2 + 2 a → . b → + | b → | 2 = | a → | 2 + | b → | 2 (dot product is commutative i.e., b → · a → = a → · b → ) ⇒ 2 a → . b → = 0 ⇒ a → · b → = 0 Therefore, a → and b → are perpendicular.