If r → = 3 i ^ - 4 j ^ + 5 k ^ then find the unit vector in the direction of r → .

Given that, the vector is r → = 3 i ^ - 4 j ^ + 5 k ^ The unit vector in the direction of r → is r → r → . So, the unit vector = 3 i ^ - 4 j ^ + 5 k ^ 3 2 + 4 2 + 5 2 = ± 1 50 3 i ^ - 4 j ^ + 5 k ^

If r → = 3 i ^ - 4 j ^ + 5 k ^ then find the unit vector in the direction of r → .

Given that, the vector is r → = 3 i ^ - 4 j ^ + 5 k ^ The unit vector in the direction of r → is r → r → . So, the unit vector = 3 i ^ - 4 j ^ + 5 k ^ 3 2 + 4 2 + 5 2 = ± 1 50 3 i ^ - 4 j ^ + 5 k ^

Show that a → - 2 b → + 3 c → , - 2 a → + 3 b → - 4 c → and a → - 3 b → + 5 c → are coplanar.

Given that, the vectors are a → - 2 b → + 3 c → , - 2 a → + 3 b → - 4 c → and a → - 3 b → + 5 c → . We know that, For the vectors a ⇀ , b ⇀ , c ⇀ are coplanar vectors then a ⇀ · b ⇀ × c ⇀ = 0 Consider, 1 - 2 3 - 2 3 - 4 1 - 3 5 ⇒ 1 15 - 12 + 2 - 10 + 4 + 3 6 - 3 = 3 - 12 + 9 = 0 .

If a → = 3 i → - j → - 4 k → , b → = - 2 i → + 4 j → - 3 k → then find the magnitude of vector 3 a → - 2 b → .

Given that, the vectors a → = 3 i → - j → - 4 k → , b → = - 2 i → + 4 j → - 3 k → . Consider, 3 a → - 2 b → = 3 3 i → - j → - 4 k → - 2 - 2 i → + 4 j → - 3 k → = 9 i → - 3 j → - 12 k → + 4 i → - 8 j → + 6 k → = 13 i → - 11 j → - 6 k → Magnitude of a vector x i ^ + y j ^ + z k ^ is x 2 + y 2 + z 2 Magnitude of the required vector = 13 2 + 11 2 + 6 2 = 326

Prove that vectors 2 i → - 3 j → + 5 k → and - 2 i → + 2 j → + 2 k → are mutually perpendicular.

Given that, the two vectors are 2 i → - 3 j → + 5 k → and - 2 i → + 2 j → + 2 k → . We know that, cos θ = a ⇀ · b ⇀ a ⇀ b ⇀ , where θ is the angle between the two vectors. ⇒ cos θ = 2 i → - 3 j → + 5 k → · - 2 i → + 2 j → + 2 k → 2 i → - 3 j → + 5 k → - 2 i → + 2 j → + 2 k → ⇒ cos θ = 2 i → - 3 j → + 5 k → · - 2 i → + 2 j → + 2 k → 2 i → - 3 j → + 5 k → - 2 i → + 2 j → + 2 k → = - 4 - 6 + 10 2 2 + 3 2 + 5 2 - 2 2 + 2 2 + 2 2 ⇒ cos θ = 0 ⇒ θ = π 2 Thus the given vectors are perpendicular.

Prove by vector method that cos α + β = cos α · cos β - sin α · sin β .

Draw a graph and plot the vectors O A → , O B → and O C → . Mark the angles α , β as shown in the figure. Let O A → = a → and O B → = b → be two unit vectors. The figure below shows the details. Drop perpendicular from B to the x -axis to meet at point D and let perpendicular from point A meet the x -axis at point C . This forms two triangles O B D and O C A . Now applying the triangle law of vector addition, we get O A → = O C → + C A → ⇒ a → = O C → + C A → . . . 1 O B → = O D → + D B → ⇒ b → = O D → + D B → . . . 2 Since a → , b → are unit vectors and the unit vector along x -axis represents i ^ and the unit vector along y -axis represents j ^ , Using trigonometric relation in triangles O B D and O C A , O C → = cos α i ^ , C A → = - sin α j ^ , O D → = cos β i ^ , D B → = sin β j ^ Substituting these in equation 1 and 2 , a → = cos α i ^ - sin α j ^ b → = cos β i ^ + sin β j ^ Now taking their scalar product, a → · b → = ( cos α i ^ - sin α j ^ ) · ( cos β i ^ + sin β j ^ ) = ( cos α · cos β ) - ( sin α · sin β ) ∵ a → · b → = a → b → cos α + β , a → = b → = 1 , ∴ cos α + β = ( cos α · cos β ) - ( sin α · sin β ) cos α + β = cos ( α ) · cos ( β ) - sin ( α ) · sin ( β ) Hence proved.

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

Given that, the vectors are 3 i ^ + j ^ + 2 k ^ and 2 i ^ - 2 j ^ + 4 k ^ . We know that, cos θ = a ⇀ · b ⇀ a ⇀ b ⇀ where a ⇀ and b ⇀ are two vectors and the angle between them is θ . ⇒ cos θ = 3 i ^ + j ^ + 2 k ^ · 2 i ^ - 2 j ^ + 4 k ^ 3 i ^ + j ^ + 2 k ^ 2 i ^ - 2 j ^ + 4 k ^ ⇒ cos θ = 6 - 2 + 8 3 2 + 1 2 + 2 2 2 2 + - 2 2 + 4 2 ⇒ cos θ = 12 14 24 = 3 7 3 = 3 7 ⇒ cos θ = 3 7 ⇒ θ = 49 . 106 °

If G is the centroid of a triangle A B C , prove that G A → + G B → + G C → = 0 → .

For any triangle A B C , Given G A → + G B → + G C → = 0 → . Now G is the centroid of triangle A B C , which divides the medians A D , B E and C F in the ratio 2 : 1 ⇒ A G → : G D → = 2 : 1 ⇒ G A → = - 2 3 A D → Similarly, G B → = - 2 3 B E → and G C → = - 2 3 C F → Now, LHS = G A → + G B → + G C → = - 2 3 A D → + B E → + C F → = - 2 3 A B → + 1 2 B C → + B C → + 1 2 C A → + C A → + 1 2 A B → = - 2 3 3 2 ( A B → + B C → + C A → ) = - 1 0 → = 0 → = RHS .

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

The given two vectors are a → = i ^ + j ^ + k ^ and b → = i ^ + 2 j ^ + 3 k ^ . Now we need to find a → + b → and a → - b → . So, a → + b → = i ^ + j ^ + k ^ + i ^ + 2 j ^ + 3 k ^ a → + b → = 2 i ^ + 3 j ^ + 4 k ^ And, a → - b → = i ^ + j ^ + k ^ - i ^ + 2 j ^ + 3 k ^ a → - b → = - j ^ - 2 k ^ Now the cross product of these vectors will be a → + b → × a → - b → = i ^ j ^ k ^ 2 3 4 0 - 1 - 2 ⇒ a → + b → × a → - b → = - 2 i ^ + 4 j ^ - 2 k ^ And the corresponding magnitude will be a → + b → × a → - b → = ( - 2 ) 2 + ( 4 ) 2 + ( - 2 ) 2 ⇒ a → + b → × a → - b → = 4 + 16 + 4 ⇒ a → + b → × a → - b → = 24 ⇒ a → + b → × a → - b → = 6 × 4 ⇒ a → + b → × a → - b → = 2 6 Now we know vector the product a → + b → × a → - b → will be perpendicular to the plane of a → + b → and a → - b → . So the unit vector perpendicular to the plane of a → + b → and a → - b → is given as = a → + b → × a → - b → a → + b → × a → - b → Now substituting the corresponding value, we get = - 2 i ^ + 4 j ^ - 2 k ^ 2 6 = - i ^ + 2 j ^ - k ^ 6 Or, the unit vector can be written as = - 1 6 i ^ + 2 6 j ^ - 1 6 k ^ .

Show that three points A ( - 2 i ^ + 3 j ^ + 5 k ^ ) , B ( i ^ + 2 j ^ + 3 k ^ ) and C ( 7 i ^ - k ^ ) are collinear.

A B → = ( i ^ + 2 j ^ + 3 k ^ ) - ( - 2 i ^ + 3 j ^ + 5 k ^ ) = 3 i ^ - j ^ - 2 k ^ B C → = ( 7 i ^ - k ^ ) - ( i ^ + 2 j ^ + 3 k ^ ) = 6 i ^ - 2 j ^ - 4 k ^ Now , B C → = 6 i ^ - 2 j ^ - 4 k ^ = 2 ( 3 i ^ - j ^ - 2 k ^ ) B C → = λ A B → , where λ = 2 ∴ B C → and A B → are collinear Since , point B is common to both , A , B and C are collinear .

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 ^

Find the angle between two vectors a → and b → with magnitudes 1 and 2 , respectively, and satisfying a → · b → = 1

The angle between two vectors a → and b → can be find out using the formula, cos θ = a → · b → a → b → In the question, a → · b → = 1 , a → = 1 and b → = 2 is given, so substituting these values, cos θ = 1 1 × 2 cos θ = 1 2 θ = cos - 1 1 2 θ = 60 ° or π 3 .

Show that the angle between any two diagonals of a cube is cos - 1 1 3 .

Let the side of a cube be a O B → = a ı ^ + a ȷ ^ + a k ^ ∴ Direction of O B → = a - 0 3 a 2 , a - 0 3 a 2 , a - 0 3 a 2 ⇒ Direction cosines of O B → = 1 3 , 1 3 , 1 3 | O B → | = 1 3 2 + 1 3 2 + 1 3 2 = 1 Similarly, A C → = - a ı ^ + a ȷ ^ + a k ^ A C → = 0 - a 3 a 2 , a - 0 3 a 2 , a - 0 3 a 2 ⇒ Direction cosines of A C → = - 1 3 , 1 3 , 1 3 So, | A C → | = - 1 3 2 + 1 3 2 + 1 3 2 = 1 LET θ be the angle between two diagonals A C → and O B → Then A C → · O B → = | A C → | | O B → | cos θ ⇒ - 1 3 + 1 3 + 1 3 = 1 . 1 . cos θ ⇒ 1 3 = cos θ ⇒ θ = cos - 1 1 3

Find the area of the parallelogram whose adjacent sides are determined by the vectors a → = 3 i ^ + j ^ + 4 k ^ and b → = i ^ - j ^ + k ^ .

The area of the parallelogram whose adjacent sides are a → and b → is | a → × b → | . We are given that, the adjacent sides are: a → = 3 i ^ + j ^ + 4 k ^ and b → = i ^ - j ^ + k ^ ∴ a → × b → = i ^ j ^ k ^ 3 1 4 1 - 1 1 = i ^ 1 + 4 - j ^ 3 - 4 + k ^ - 3 - 1 = 5 i ^ + j ^ - 4 k ^ And | a → × b → | = 5 i ^ + j ^ - 4 k ^ = 5 2 + 1 2 + - 4 2 = 25 + 1 + 16 = 42 Hence, the area of the given parallelogram is 42 sq. units.

If a → + b → + c → = 0 → , then prove that a → × b → = b → × c → = c → × a → .

Given that, a → + b → + c → = 0 → and To prove: a → × b → = b → × c → = c → × a → Here, a → = - ( b → + c → ) ⇒ a → × b → = - ( b → + c → ) × b → = - b → × b → - c → × b → = 0 → + b → × c → = b → × c → [Using a → × a → = b → × b → = c → × c → = 0 ] Again, b → = - ( a → + c → ) ⇒ a → × b → = - a → × ( a → + c → ) = - a → × a → - a → × c → = 0 - a → × c → = c → × a → Thus, a → × b → = b → × c → = c → × a → Hence, proved.

Find the projection of a → = 2 i ^ + 3 j ^ + 2 k ^ on the vector b → = i ^ + 2 j ^ + k ^ .

The projection of a vector a → on b → is given by a → · b → b → . Therefore, a → · b → b → = 2 i ^ + 3 j ^ + 2 k ^ · i ^ + 2 j ^ + k ^ 1 2 + 2 2 + 1 2 = 2 × 1 + 3 × 2 + 2 × 1 1 + 4 + 1 = 2 + 6 + 2 6 = 10 6