Find a unit vector in the direction of the vector a → = 3 i ^ - 2 j ^ + 6 k ^ .

In order to find a unit vector in the direction of a given vector, that vector must be divided by its magnitude. The magnitude of the vector a → = 3 i ^ - 2 j ^ + 6 k ^ can be given as, a → = 3 2 + - 2 2 + 6 2 ⇒ a → = 9 + 4 + 36 ⇒ a → = 49 ⇒ a → = 7 The unit vector in the direction of vector a → (assume u a → ) is given as, u a → = a → a → ⇒ u a → = 1 7 3 i ^ - 2 j ^ + 6 k ^ ⇒ u a → = 3 7 i ^ - 2 7 j ^ + 6 7 k ^

Find a unit vector in the direction of the vector a → = 3 i ^ - 2 j ^ + 6 k ^ .

In order to find a unit vector in the direction of a given vector, that vector must be divided by its magnitude. The magnitude of the vector a → = 3 i ^ - 2 j ^ + 6 k ^ can be given as, a → = 3 2 + - 2 2 + 6 2 ⇒ a → = 9 + 4 + 36 ⇒ a → = 49 ⇒ a → = 7 The unit vector in the direction of vector a → (assume u a → ) is given as, u a → = a → a → ⇒ u a → = 1 7 3 i ^ - 2 j ^ + 6 k ^ ⇒ u a → = 3 7 i ^ - 2 7 j ^ + 6 7 k ^

Find the direction cosines of the vector i ^ + 2 j ^ + 3 k ^ .

The direction cosines of the vector a → = i ^ + 2 j ^ + 3 k ^ can be obtained as, 1 1 2 + 2 2 + 3 2 , 2 1 2 + 2 2 + 3 2 , 3 1 2 + 2 2 + 3 2 = 1 14 , 2 14 , 3 14

Find the vector in the direction of the vector - i ^ + 2 j ^ + 2 k ^ that has a magnitude of 7 .

Let the given vector be taken as, a → = - i ^ + 2 j ^ + 2 k ^ To find the vector in the direction of the vector a → with magnitude 7 , we first need to find the unit vector in the direction of the vector a → . The unit vector is given as, u a → = a → a → ⇒ u a → = - i ^ + 2 j ^ + 2 k ^ - 1 2 + 2 2 + 2 2 ⇒ u a → = - i ^ + 2 j ^ + 2 k ^ 1 + 4 + 4 ⇒ u a → = - i ^ + 2 j ^ + 2 k ^ 9 ⇒ u a → = - i ^ + 2 j ^ + 2 k ^ 3 ⇒ u a → = - 1 3 i ^ + 2 3 j ^ + 2 3 k ^ Now, we have got a unit vector in the direction of a → . Hence, the vector of magnitude 7 in the direction of a → will be obtained as, 7 u a → = 7 × - 1 3 i ^ + 2 3 j ^ + 2 3 k ^ ⇒ 7 u a → = - 7 3 i ^ + 14 3 j ^ + 14 3 k ^

Find a vector in the direction of the vector a → = i ^ - 2 j ^ that has magnitude of 7 units.

To find a vector in the direction of vector a → with a magnitude of 7 units, we first need to find the unit vector in the direction of vector a → . The unit vector is given as, u a → = a → a → ⇒ u a → = i ^ - 2 j ^ 1 2 + - 2 2 ⇒ u a → = i ^ - 2 j ^ 1 + 4 ⇒ u a → = i ^ - 2 j ^ 5 ⇒ u a → = 1 5 i ^ - 2 5 j ^ Now, we have got a unit vector in the direction of a → . The vector of a magnitude of 7 units in the direction of a → can be obtained as, 7 u a → = 7 1 5 i ^ - 2 5 j ^ ⇒ 7 u a → = 7 5 i ^ - 14 5 j ^

If O P → = 2 i ^ + 3 j ^ - 4 k ^ and O Q → = 5 i ^ + 4 j ^ - 3 k ^ , then find P Q → and the direction cosines of P Q → .

Given: O P → = 2 i ^ + 3 j ^ - 4 k ^ O Q → = 5 i ^ + 4 j ^ - 3 k ^ Now, O P → and O Q → are the position vectors of P → and Q → , respectively. So, we get, P Q → = O Q → - O P → ⇒ P Q → = 5 i ^ + 4 j ^ - 3 k ^ - 2 i ^ + 3 j ^ - 4 k ^ ⇒ P Q → = 5 - 2 i ^ + 4 - 3 j ^ + - 3 + 4 k ^ ⇒ P Q → = 3 i ^ + j ^ + k ^ The direction cosines of a vector x i ^ + y j ^ + z k ^ can be found using, x x 2 + y 2 + z 2 , y x 2 + y 2 + z 2 and z x 2 + y 2 + z 2 . For vector P Q → = 3 i ^ + j ^ + k ^ , the directional cosines can be given as, ⇒ 3 3 2 + 1 2 + 1 2 , 1 3 2 + 1 2 + 1 2 , 1 3 2 + 1 2 + 1 2 = 3 9 + 1 + 1 , 1 9 + 1 + 1 , 1 9 + 1 + 1 = 3 11 , 1 11 , 1 11

The position vectors of two points A and B are i ^ + j ^ + k ^ and 5 i ^ - 3 j ^ + k ^ . Find a unit vector in the direction of A B → and also find the directional cosines of A B → . What angles does A B → make with the three axes?

The position vectors of A → and B → are given as i ^ + j ^ + k ^ and 5 i ^ - 3 j ^ + k ^ . A B → = O B → - O A → A B → = 5 i ^ - 3 j ^ + k ^ - i ^ + j ^ + k ^ = 5 - 1 i ^ + - 3 - 1 j ^ + 1 - 1 k ^ = 4 i ^ - 4 j ^ Unit vector in the direction of A B → can be found out as u A B → = A B → A B → . u A B → = 4 i ^ - 4 j ^ 4 2 + 4 2 ⇒ 4 i ^ - j ^ 32 ⇒ i ^ - j ^ 2 ⇒ i ^ 2 - j ^ 2 The directional cosines are the coefficients of the unit vector of A B → = 1 2 , - 1 2 , 0 . Directional cosines l , m and n of a vector are the angles cos α , cos β and cos γ that the vector makes with the x , y and z axes respectively. So, cos α = 1 2 , cos β = - 1 2 , cos γ = 0 . The angle made by A B → with the x - axis is given by α = cos - 1 1 2 = 45 ° . The angle made by A B → with the y - axis is given by β = cos - 1 - 1 2 = 135 ° . The angle made by A B → with the z - axis is given by γ = cos - 1 0 = 90 ° .

Write the direction ratios of the vector a → = i ^ + j ^ - 2 k ^ and hence calculate its direction cosines.

The direction ratios of the vector a → = i ^ + j ^ - 2 k ^ are the coefficients of the unit vectors i ^ , j ^ and k ^ , respectively. Here, the direction ratios are 1 , 1 , - 2 . The direction cosines of vector a → are l , m and n and are obtained by dividing a , b and c , respectively by the magnitude of vector a → . Magnitude of vector a → is given by a → = 1 2 + 1 2 + - 2 2 = 1 + 1 + 4 = 6 Now, l = 1 a → = 1 6 , m = 1 a → = 1 6 , n = - 2 a → = - 2 6

Find the unit vector in the direction of the vector P Q → , where P and Q are the points 1 , 2 , 3 and 4 , 5 , 6 respectively.

The coordinates of the points P and Q are given. So, the position vector of P and Q can be written as O P → = i ^ + 2 j ^ + 3 k ^ and O Q → = 4 i ^ + 5 j ^ + 6 k ^ , where O is the origin. Therefore, P Q → = O Q → - O P → = 4 i ^ + 5 j ^ + 6 k ^ - i ^ + 2 j ^ + 3 k ^ = 3 i ^ + 3 j ^ + 3 k ^ Unit vector in the direction of vector P Q → is given by P Q ^ = P Q → P Q → = 3 i ^ + 3 j ^ + 3 k ^ 3 2 + 3 2 + 3 2 = 3 i ^ + 3 j ^ + 3 k ^ 27 = i ^ + j ^ + k ^ 3 = i ^ 3 + j ^ 3 + k ^ 3

If P ≡ 1 , 5 , 4 and Q ≡ 4 , 1 , - 2 , find the direction ratios of P Q → .

The coordinates of points P and Q are given, so their position vectors can be found out as O P → = i ^ + 5 j ^ + 4 k ^ and O Q → = 4 i ^ + j ^ - 2 k ^ , where O is the origin. P Q → = 4 i ^ + j ^ - 2 k ^ - i ^ + 5 j ^ + 4 k ^ = 3 i ^ - 4 j ^ - 6 k ^ Direction ratios of the vector P Q → are the coefficients of i ^ , j ^ and k ^ , i.e., 3 , - 4 , - 6 respectively.