Show that vector area of a quadrilateral A B C D is 1 2 A C ¯ × B D ¯ , where A C and B D are its diagonals.

Let a , b , c , d be the position vectors of the points A , B , C , D respectively. A r e a o f q u a d ( A B C D ) = a r ∆ A B D + a r ( ∆ C D B ) Now, a r ( ∆ A B D ) = 1 2 a × b + b × d + d × a a r ( ∆ C D B ) = 1 2 c × d + d × b + b × c ∴ a r e a o f q u a d ( A B C D ) = 1 2 a × b + b × d + d × a + 1 2 c × d + d × b + b × c Since, b × d = - d × b , a × b = - b × a a r e a o f q u a d ( A B C D ) = 1 2 a × b + b × d + d × a + c × d + d × b + b × c = 1 2 a × b + b × d + d × a + c × d - b × d + b × c = 1 2 a × b + d × a + c × d + b × c = 1 2 a × b - a × d + c × d - c × b = 1 2 a × b - d + c × d - b = 1 2 c × d - b - a × d - b = 1 2 c - a × d - b = 1 2 A C × B D Hence, proved.

Show that vector area of a quadrilateral A B C D is 1 2 A C ¯ × B D ¯ , where A C and B D are its diagonals.

Let a , b , c , d be the position vectors of the points A , B , C , D respectively. A r e a o f q u a d ( A B C D ) = a r ∆ A B D + a r ( ∆ C D B ) Now, a r ( ∆ A B D ) = 1 2 a × b + b × d + d × a a r ( ∆ C D B ) = 1 2 c × d + d × b + b × c ∴ a r e a o f q u a d ( A B C D ) = 1 2 a × b + b × d + d × a + 1 2 c × d + d × b + b × c Since, b × d = - d × b , a × b = - b × a a r e a o f q u a d ( A B C D ) = 1 2 a × b + b × d + d × a + c × d + d × b + b × c = 1 2 a × b + b × d + d × a + c × d - b × d + b × c = 1 2 a × b + d × a + c × d + b × c = 1 2 a × b - a × d + c × d - c × b = 1 2 a × b - d + c × d - b = 1 2 c × d - b - a × d - b = 1 2 c - a × d - b = 1 2 A C × B D Hence, proved.

If a ¯ , b ¯ , c ¯ and d ¯ are four distinct vectors such that a ¯ × b ¯ = c ¯ × d ¯ and a ¯ × c ¯ = b ¯ × d ¯ , prove that a ¯ - d ¯ is parallel to b ¯ - c ¯

Given: a ¯ , b ¯ , c ¯ and d ¯ are four distinct vectors. ∴ a ¯ ≠ b ¯ ≠ c ¯ ≠ d ¯ ∴ a ¯ - d ¯ ≠ 0 and b ¯ - c ¯ ≠ 0 ........(i) Now, a ¯ × b ¯ = c ¯ × d ¯ .......(ii) And, a ¯ × c ¯ = b ¯ × d ¯ .........(iii) Subtracting (iii) from (ii), we get a ¯ × b ¯ - a ¯ × c ¯ = c ¯ × d ¯ - b ¯ × d ¯ ⇒ a ¯ × b ¯ - c ¯ = c - b ¯ × d ¯ ⇒ a ¯ × b ¯ - c ¯ = - b ¯ - c × d ¯ ⇒ a ¯ × b ¯ - c ¯ = d ¯ × b ¯ - c ⇒ a ¯ × b ¯ - c ¯ - d ¯ × b ¯ - c = 0 ⇒ a ¯ - d ¯ × b ¯ - c ¯ = 0 Therefore, a ¯ - d ¯ and b ¯ - c ¯ are parallel to each other.

If a ¯ = i ^ + j ^ + k ^ and c ¯ = j ^ - k ^ , find a vector b ¯ satisfying a ¯ × b ¯ = c ¯ and a ¯ · b ¯ = 3

Given, a ¯ = i ^ + j ^ + k ^ and c ¯ = j ^ - k ^ , Let b ¯ = x i ^ + y j ^ + z k ^ Then a ¯ · b ¯ = 3 gives i ^ + j ^ + k ^ . x i ^ + y j ^ + z k ^ = 3 ⇒ x + y + z = 3 .......(i) Also, a ¯ × b ¯ = c ¯ ∴ j ^ - k ^ = ı ^ j ^ k ^ 1 1 1 x y z = z - y i ^ - z - x j ^ + y - x k ^ = z - y i ^ + x - z j ^ + y - x k ^ By equality of vectors, z - y = 0 ......(ii) x - z = 1 ......(ii) y - x = - 1 .........(iv) From (ii), y = z From (iii), x = z + 1 Substituting values of x and y in (i), we get z + 1 + z + z = 3 ⇒ 3 z = 2 ⇒ z = 2 3 ∴ y = z = 2 3 ∴ x = 1 + z = 1 + 2 3 = 3 + 2 3 = 5 3 Hence, b = 5 3 i ^ + 2 3 j ^ + 2 3 j ^ = 1 3 5 i ^ + 2 j ^ + 2 k ^

Find a ¯ , if a ¯ × i ^ + 2 a - 5 j ^ = 0 .

Let a = x i ^ + y j ^ + z k ^ Then a × i ^ = x i ^ + y j ^ + z k ^ × i ^ = x ( i × i ^ ) + y j ^ × i ^ + z k ^ × i ^ = z j ^ - y k ^ ∵ i ^ × i ^ = 0 , j ^ × i ^ = - k ^ , k ^ × i ^ = j ^ It is given that, a ¯ × i ^ + 2 a - 5 j ^ = 0 ⇒ z j ^ - y k ^ + 2 x i ^ + y j ^ + z k ^ - 5 j ^ = 0 ⇒ z j ^ - y k ^ + 2 x i ^ + 2 y j ^ + 2 z k ^ - 5 j ^ = 0 ⇒ 2 x i ^ + 2 y + z - 5 j ^ + 2 z - y k ^ = 0 By equality of vectors, 2 x = 0 ⇒ x = 0 2 y + z - 5 = 0 . . . . . . . . . ( i ) 2 z - y = 0 ⇒ y = 2 z . . . . . . . ( ii ) Substituting y = 2 z in ( i ) , we get 2 2 z + z - 5 = 0 ⇒ 4 z + z - 5 = 0 ⇒ 5 z - 5 = 0 ⇒ z = 1 From ( ii ) , y = 2 1 = 2 Hence, x = 0 , y = 2 , z = 1

If | a ¯ · b ¯ | = | a ¯ × b ¯ | and a ¯ · b ¯ < 0 , then find the angle between a ¯ and b ¯

Let θ be the angle between a ¯ and b ¯ . Then, | a ¯ · b ¯ | = | a ¯ × b ¯ | gives | a b cos θ | = | a b sin θ | ⇒ - a b cos θ = a b sin θ ∵ a . b < 0 ⇒ - 1 cos θ = sin θ ⇒ tan θ = - 1 ⇒ tan θ = - tan π 4 = tan π - π 4 ⇒ tan θ = tan 3 π 4 ⇒ θ = 3 π 4 Hence, the angle between a ¯ and b ¯ is 3 π 4 .

If a line makes angles 90 ° , 135 ° , 45 ° with X , Y and Z axes respectively, then find its direction cosines.

Direction cosines of a line making angle α with X - axis, angle β with Y - axis and angle γ with Z - axis are l , m , n , where l = cos α , m = cos β , n = cos γ . Here, α = 90 ° , β = 135 ° , γ = 45 ° So, direction cosines are l = cos 90 ° = 0 m = cos 135 ° = cos 180 ° - 45 ° = - cos 45 ° = - 1 2 n = cos 45 ° = 1 2 ∴ Direction cosines are l , m , n = 0 , - 1 2 , 1 2 .