If a vector a → is perpendicular to two non-collinear vectors b → and c → , then a → is perpendicular to every vector in the plane of b → and c → .

Given that a → is perpendicular to b → and c → . It means a → · b → = 0 and a → · c → = 0 . . . ( i ) Let r → be some vector in the plane of b → and c → Then, r → , b → , c → are coplanar We know that, three vectors are coplanar if one of them is expressible as linear combination of other two vectors. Let r → = x b → + y c → where x and y are some scalar. r → · a → = ( x b → + y c → ) · a → r → · a → = x b → · a → + y c → · a → = x × 0 + y × 0 r → · a → = 0 Thus a → is perpendicular to every vector in the plane of b → and c →

If a vector a → is perpendicular to two non-collinear vectors b → and c → , then a → is perpendicular to every vector in the plane of b → and c → .

Given that a → is perpendicular to b → and c → . It means a → · b → = 0 and a → · c → = 0 . . . ( i ) Let r → be some vector in the plane of b → and c → Then, r → , b → , c → are coplanar We know that, three vectors are coplanar if one of them is expressible as linear combination of other two vectors. Let r → = x b → + y c → where x and y are some scalar. r → · a → = ( x b → + y c → ) · a → r → · a → = x b → · a → + y c → · a → = x × 0 + y × 0 r → · a → = 0 Thus a → is perpendicular to every vector in the plane of b → and c →

If a → + b → + c → = 0 → , show that the angle θ between the vectors b → and c → is given by cos θ = | a → | 2 - | b → | 2 - | c → | 2 2 | b → | | c → | .

We have a → + b → + c → = 0 → b → + c → = - a → Squaring both the sides. ( b → + c → ) 2 = ( - a → ) 2 b → + c → · b → + c → = a → 2 b → · b → + b → · c → + c → · b → + c → · c → = a → 2 b → · b → + 2 b → · c → + c → · c → = a → 2 b → 2 + c → 2 + 2 b → c → cos θ = a → 2 cos θ = a → 2 - b → 2 - c → 2 2 b → c → Hence, proved.

Let a → = x 2 i ^ + 2 j ^ - 2 k ^ , b → = i ^ - j ^ + k ^ and c → = x 2 i ^ + 5 j ^ - 4 k ^ be three vectors. Find the values of x for which the angle between a → and b → is acute and the angle between b → and c → is obtuse.

Given a → = x 2 i ^ + 2 j ^ - 2 k ^ , b → = i ^ - j ^ + k ^ , c → = x 2 i ^ + 5 j ^ - 4 k ^ Let θ be the angle between a → and b → , then cos θ = a → · b → a → b → = x 2 i ^ + 2 j ^ - 2 k ^ · ( i ^ - j ^ + k ^ ) x 2 2 + ( 2 ) 2 + ( - 2 ) 2 ( 1 ) 2 + ( - 1 ) 2 + ( 1 ) 2 = x 2 ( 1 ) + ( 2 ) ( - 1 ) + ( - 2 ) ( 1 ) x 4 + 4 + 4 1 + 1 + 1 = x 2 - 2 - 2 8 + x 4 3 cos θ = x 2 - 4 3 8 + x 4 Since θ is an acute angle cos θ > 0 x 2 - 4 3 8 + x 4 > 0 x 2 - 4 > 0 ⇒ x < - 2 or x > 2 Again, let ϕ be the angle between b → and c → , cos ϕ = ( i ^ - j ^ + k ^ ) · x 2 i ^ + 5 j ^ - 4 k ^ ( 1 ) 2 + ( - 1 ) 2 + ( 1 ) 2 x 2 2 + ( 5 ) 2 + ( - 4 ) 2 cos ϕ = ( 1 ) x 2 + ( - 1 ) ( 5 ) + ( 1 ) ( - 4 ) 3 x 4 + 25 + 16 = x 2 - 5 - 4 3 x 2 + 41 cos ϕ = x 2 - 9 3 x 2 + 41 Since ϕ is an obtuse angle, so cos ϕ < 0 x 2 - 9 3 x 2 + 41 < 0 x 2 - 9 < 0 ⇒ x > - 3 and x < 3 - 3 < x < - 2 and 2 < x < 3 x ∈ ( - 3 , - 2 ) ∪ ( 2 , 3 )

Find the values of x and y if the vectors a → = 3 i ^ + x j ^ - k ^ and b → = 2 i ^ + j ^ + y k ^ are mutually perpendicular vectors of equal magnitude.

Here a → and b → are mutually perpendicular, then a → · b → = 0 ( 3 i ^ + x j ^ - k ^ ) · ( 2 i ^ + j ^ + y k ^ ) = 0 ( 3 ) ( 2 ) + ( x ) ( 1 ) + ( - 1 ) ( y ) = 0 x - y = - 6 ...(i) Also | a → | = | b → | x 2 + 10 = 5 + y 2 ( x + y ) ( x - y ) = - 5 ...(ii) Solving (i) and (ii), x + x + 6 x - x + 6 = - 5 2 x + 6 - 6 = - 5 2 x = 5 6 - 6 ⇒ x = - 31 12 Put value of x in equation (i), we get y = 41 12

If a → and b → are two non-collinear unit vectors such that | a → + b → | = 3 , find ( 2 a → - 5 b → ) · ( 3 a → + b → ) .

Given a → and b → are unit vectors Thus, | a → | = | b → | = 1 and | a → + b → | = 3 Squaring both the sides, a → + b → 2 = ( 3 ) 2 a → + b → · a → + b → = 3 a → · a → + a → · b → + b → · a → + b → · b → = 3 a → 2 + b → 2 + 2 a → · b → = 3 1 + 1 + 2 a → · b → = 3 a → · b → = 1 2 Now, ( 2 a → - 5 b → ) · ( 3 a → + b → ) = 2 × 3 a → · a → + 2 a → · b → - 5 × 3 b → · a → - 5 b → · b → = 6 a → 2 + 2 a → · b → - 15 a → · b → - 5 b → 2 = 6 ( 1 ) 2 - 13 1 2 - 5 ( 1 ) 2 = - 11 2

If a → , b → are two vectors such that | a → + b → | = | b → | , then prove that a → + 2 b → is perpendicular to a → .

Given | a → + b → | = b → Squaring on both sides | a → + b → | 2 = | b → | 2 ⇒ ( a → + b → ) · ( a → + b → ) = b → · b → ⇒ a → . a → + a → · b → + b → · a → + b → · b → = b → · b → ⇒ a → . a → + a → · b → + a → · b → + b → · b → = b → · b → ⇒ a → . a → + 2 a → · b → = 0 ⇒ a → · ( a → + 2 b → ) = 0 Hence, proved that a → perpendicular to ( a → + 2 b → ) .

In a triangle O A B , ∠ A O B = 90 ° . If P and Q are the points of trisection of A B → , prove that O P 2 + O Q 2 = 5 9 A B 2 .

In a triangle A O B , ∠ A O B = 90 ° and P and Q are the points of trisection of A B . Let the position vector of O A → = a → , O B → = b → Since P , Q are the points of trisection of A B , so A P : P B = 1 : 2 , A Q : Q B = 2 : 1 Position vector of O P → = 2 a → + b → 3 Using section formulae Position vector of O Q → = a → + 2 b → 3 Now, O P 2 + O Q 2 = O P → 2 + O Q → 2 = 2 a → + b → 3 2 a → + b → 3 + a → + 2 b → 3 a → + 2 b → 3 = 4 | a → | 2 + 4 a → ⋅ b → + | b → | 2 + | a → | 2 + 4 a → ⋅ b → + 4 | b → | 2 9 ∵ ∠ A O B = 90 ° ⇒ a → · b → = 0 = 5 | a → | 2 + 5 | b → | 2 9 = 5 9 A B → 2 = 5 9 A B 2 Hence, proved.