The x and y components of vector A → are 4 m and 6 m respectively. The x and y components of vector A → + B → are 10 m and 9 m respectively. For the vector B → calculate the following length :

Given x component of A → is 4m and y component is 6m respectively. And, A → + B → has its x component as 10m and y component as 9m respectively. Therefore, the x component of B → is (10-4)m and the y component is (9-6)m respectively. The x and y components are 6m and 3m respectively. So the resultant length of vector b = 6 2 + 3 2 = 45

The x and y components of vector A → are 4 m and 6 m respectively. The x and y components of vector A → + B → are 10 m and 9 m respectively. For the vector B → calculate the following length :

Given x component of A → is 4m and y component is 6m respectively. And, A → + B → has its x component as 10m and y component as 9m respectively. Therefore, the x component of B → is (10-4)m and the y component is (9-6)m respectively. The x and y components are 6m and 3m respectively. So the resultant length of vector b = 6 2 + 3 2 = 45

The x and y components of vector A → are 4 m and 6 m respectively. The x and y components of vector A → + B → are 10 m and 9 m respectively. For the vector B → calculate the following the angle it makes with x -axis :

Given x component of A → is 4m and y component is 6m respectively. And, A → + B → has its x component as 10m and y component as 9m respectively. Therefore, the x component of B → is (10-4)m and the y component is (9-6)m respectively. The x and y components are 6m and 3m respectively. If ∅ is the angle that the resultant makes with x-axis and the vector can be written as x i ^ + y j ^ Then, tan ∅ = y x Therefore, here tan ∅ = 3 6 = 1 2

Find the directional cosines of vector 5 i ^ + 2 j ^ + 6 k ^ . Also write the value of sum of squares of directional cosines of this vector.

If a vector P → = x i ^ + y j ^ + z k ^ makes angle with X , Y and Z axis as α , β and γ then direction cosines of the vector are defined as cos α = x x 2 + y 2 + z 2 , cos β = y x 2 + y 2 + z 2 and cos γ = z x 2 + y 2 + z 2 If the vector is given as 5 i ^ + 2 j ^ + 6 k ^ then put the values of the components in the above equations, cos α = 5 5 2 + 2 2 + 6 2 = 5 65 cos β = 2 5 2 + 2 2 + 6 2 = 2 65 cos γ = 6 5 2 + 2 2 + 6 2 = 6 65 Now, the value of sum of squares of direction cosines are cos 2 α + cos 2 β + cos 2 γ 5 65 2 + 2 65 2 + 6 65 2 = 1

If A → = 6 i ^ – 6 j ^ + 5 k ^ and B → = i ^ + 2 j ^ – k ^ , then find a unit vector parallel to the resultant of A → & B → .

Given, two vectors as A → = 6 i ^ – 6 j ^ + 5 k ^ and B → = i ^ + 2 j ^ – k ^ . Now, resultant of both vectors, i.e., A → + B → = 6 i ^ – 6 j ^ + 5 k ^ + i ^ + 2 j ^ – k ^ A → + B → = 7 i ^ – 4 j ^ + 4 k ^ Unit vector of any vector P → can be find out as, P ^ = P → P → We know that the magnitude of a vector P → = x i ^ + y j ^ + z k ^ is given by, P → = x 2 + y 2 + z 2 . So, unit vector of A → + B → is 7 i ^ – 4 j ^ + 4 k ^ 7 2 + 4 2 + 4 2 = i ^ – j ^ + k ^ 9

If a → and b → are two non collinear unit vectors and a → + b → = 3 , then find the value of a → – b → · 2 a → + b →

Given, a ^ + b ^ = 3 , where a ^ and b ^ are unit vectors. Taking squares both sides, a ^ + b ^ 2 = 3 2 a ^ 2 + b ^ 2 + 2 a ^ b ^ cos θ = 3 ( θ is the angle between the two unit vectors) 1 2 + 1 2 + 2 × 1 × 1 × cos θ = 3 cos θ = 1 2 or θ = 60 ° Now, let's calculate a ^ – b ^ · 2 a ^ + b ^ = a ^ . ( 2 a ^ ) + a ^ . b ^ - b ^ . ( 2 a ^ ) - b ^ . b ^ = a ^ × 2 a ^ × cos 0 ° + a ^ × b ^ × cos 60 ° - b ^ × 2 a ^ × cos 60 ° - a ^ × a ^ × cos 0 ° = 2 + 1 2 - 1 - 1 = 1 2

If A → = 4 i ^ – 2 j ^ + 4 k ^ and B → = – 4 i ^ + 2 j ^ + α k ^ are perpendicular to eachother then find value of α ?

For the two vectors A → and B → the dot product is defined as A → . B → = A → B → cos θ , where θ is the angle between the two vectors. It is given that A → = 4 i ^ – 2 j ^ + 4 k ^ and B → = – 4 i ^ + 2 j ^ + α k ^ and θ = 90 ° . Put these values in the above expression, 4 i ^ – 2 j ^ + 4 k ^ . – 4 i ^ + 2 j ^ + α k ^ = 4 i ^ – 2 j ^ + 4 k ^ – 4 i ^ + 2 j ^ + α k ^ cos 90 ° 4 ( - 4 ) + ( - 2 ) 2 + 4 ( α ) = 0 α = 5

If vector a ^ + 2 b ^ is perpendicular to vector 5 a ^ – 4 b ^ , then find the angle between a ^ and b ^ .

For the two vectors A → and B → the dot product is defined as: A → . B → = A → B → cos θ , where θ is the angle between the two vectors. It is given that a ^ + 2 b ^ and 5 a ^ – 4 b ^ are perpendicular to each other. If two vectors are perpendicular to each other then their dot product is 0 . Put these values in the above expression, a ^ + 2 b ^ . 5 a ^ – 4 b ^ = 0 a ^ . ( 5 a ^ ) - a ^ . ( 4 b ^ ) + ( 2 b ^ ) . ( 5 a ^ ) - ( 2 b ^ ) . ( 4 b ^ ) = 0 a ^ 5 a ^ cos 0 - a ^ 4 b ^ cos θ + 2 b ^ 5 a ^ cos θ - 2 b ^ 4 b ^ cos 0 = 0 Put a ^ . b ^ = a ^ b ^ cos θ , where θ is the angle between the two vectors. Also put a ^ = 1 and , b ^ = 1 and cos 0 = 1 in the above equation. 1 × 5 × 1 - 1 × 4 × cos θ + 2 × 5 × cos θ - 2 × 4 × 1 = 0 cos θ = 1 2 or θ = 60 °

If A → = 2 i ^ – 2 j ^ – k ^ and B → = i ^ + j ^ , then, Find angle between A → and B → .

For the two vectors A → and B → the dot product is defined as: A → . B → = A → B → cos θ , where θ is the angle between the two vectors. It is given that A → = 2 i ^ – 2 j ^ - k ^ and B → = i ^ + j ^ . Put these values in the above expression, 2 i ^ – 2 j ^ - k ^ . i ^ + j ^ = 2 i ^ – 2 j ^ - k ^ i ^ + j ^ cos θ 2 ( 1 ) + ( - 2 ) 1 + ( - 1 ) ( 0 ) = 2 2 + 2 2 + 1 2 1 2 + 1 2 cos θ cos θ = 0 or θ = 90 °