Find the component form of a ¯ if it lies in X Z plane and makes 45 ° with positive Z -axis and | a ¯ | = 10 .

Let α , β , γ be the direction angles of a . Since a lies in X Z plane, it is perpendicular to Y -axis. ∴ β = 90 ° It is given that, γ = 45 ° ∵ cos 2 α + cos 2 β + cos 2 γ = 1 ⇒ cos 2 α + cos 2 90 ° + cos 2 45 ° = 1 ⇒ cos 2 α + 0 + 1 2 2 = 1 ⇒ cos 2 α = 1 - 1 2 = 1 2 ⇒ cos α = ± 1 2 Unit vector along a is given by a ^ = cos α i ^ + cos β j ^ + cos γ k ^ = ± 1 2 i ^ + 0 j ^ + 1 2 k ^ = ± 1 2 i ^ + 1 2 k ^ ∴ a = a . a ^ = 10 ± 1 2 i ^ + 1 2 k ^ = ± 5 2 i ^ ± 5 2 k ^ Hence, a = ± 5 2 i ^ ± 5 2 k ^

Find the component form of a ¯ if it lies in X Z plane and makes 45 ° with positive Z -axis and | a ¯ | = 10 .

Let α , β , γ be the direction angles of a . Since a lies in X Z plane, it is perpendicular to Y -axis. ∴ β = 90 ° It is given that, γ = 45 ° ∵ cos 2 α + cos 2 β + cos 2 γ = 1 ⇒ cos 2 α + cos 2 90 ° + cos 2 45 ° = 1 ⇒ cos 2 α + 0 + 1 2 2 = 1 ⇒ cos 2 α = 1 - 1 2 = 1 2 ⇒ cos α = ± 1 2 Unit vector along a is given by a ^ = cos α i ^ + cos β j ^ + cos γ k ^ = ± 1 2 i ^ + 0 j ^ + 1 2 k ^ = ± 1 2 i ^ + 1 2 k ^ ∴ a = a . a ^ = 10 ± 1 2 i ^ + 1 2 k ^ = ± 5 2 i ^ ± 5 2 k ^ Hence, a = ± 5 2 i ^ ± 5 2 k ^

Two sides of a parallelogram are 3 i ^ + 4 j ^ - 5 k ^ and - 2 j ^ + 7 k ^ . Find the unit vectors parallel to the diagonals.

The two adjacent sides of a parallelogram are 3 i ^ + 4 j ^ - 5 k ^ and - 2 j ^ + 7 k ^ suppose A B → = 3 i ^ + 4 j ^ - 5 k ^ and B C → = - 2 j ^ + 7 k ^ Then any one diagonal of a parallelogram is A C → = A B → + B C → = 3 i ^ + 4 j ^ - 5 k ^ - 2 j ^ + 7 k ^ = 3 i ^ + 2 j ^ + 2 k ^ Therefore, unit vector along the diagonal is A C → | A C → | = 3 i ^ + 2 j ^ + 2 k ^ 9 + 4 + 4 = 3 i ^ + 2 j ^ + 2 k ^ 17 Another diagonal of a parallelogram is B D → = B A → + A D → B D → = - A B → + B C → = - 3 i ^ - 4 j ^ + 5 k ^ + 2 j ^ - 7 k ^ = - 3 i ^ - 6 j ^ + 12 k ^ = 3 - i ^ - 2 j ^ + 4 k ^ Therefore, unit vector along the diagonal is p → | p → | = 3 - i ^ - 2 j ^ + 4 k ^ 3 1 + 4 + 16 = 3 - i ^ - 2 j ^ + 4 k ^ 21

Express the vector i ^ + 4 j ^ - 4 k ^ as a linear combination of the vectors 2 i ^ - j ^ + 3 k ^ , i ^ - 2 j ^ + 4 k ^ and - i ^ + 3 j ^ - 5 k ^

Let a ¯ = 2 i ^ - j ^ + 3 k ^ b ¯ = i ^ − 2 j ^ + 4 k ^ , c ¯ = − i ^ + 3 j ^ − 5 k ^ , p ¯ = i ^ + 4 j ^ − 4 k ^ , Suppose p ¯ = x a ¯ + y b ¯ + z c ¯ Then, i + 4 j − 4 k = x ( 2 i - j + 3 k ) + y ( i - 2 j + 4 k ) + z ( − i + 3 j − 5 k ) ∴ i + 4 j − 4 k = ( 2 x + y − z ) i + ( − x − 2 y + 3 z ) j + ( 3 x + 4 y − 5 z ) k By equality vectors, 2 x + y − z = 1 − x − 2 y + 3 z = 4 3 x + 4 y − 5 z = − 4 We have to solve these equations by using Cramer's Rule. D = 2 1 - 1 - 1 - 2 3 3 4 - 5 = 2 ( 10 − 12 ) − 1 ( 5 − 9 ) − 1 ( − 4 + 6 ) = − 4 + 4 − 2 = - 2 ≠ 0 D x = 1 1 - 1 4 - 2 3 - 4 4 - 5 = 1 10 − 12 − 1 − 20 + 12 − 1 16 − 8 = − 2 + 8 − 8 = - 2 D y = 2 1 - 1 - 1 4 3 3 - 4 - 5 = 2 − 20 + 12 − 1 5 − 9 − 1 4 − 12 = − 16 + 4 + 8 = − 4 D z = 2 1 1 - 1 - 2 4 3 4 - 4 = 2 8 − 16 − 1 4 − 12 + 1 − 4 + 6 = − 16 + 8 + 2 = - 6 ∴ x = D x D = - 2 - 2 = 1 ∴ y = D y D = − 4 - 2 = 2 ∴ z = D z D = − 6 - 2 = 3 Hence, p ¯ = a ¯ + 2 b ¯ + 3 c ¯ .

The position vectors of three consecutive vertices of a parallelogram are i ^ + j ^ + k ^ , i ^ + 3 j ^ + 5 k ^ and 7 i ^ + 9 j ^ + 11 k ^ . Find the position vector of the fourth vertex.

Let A B C D be a parallelogram. Let a , b , c , d be the position vectors of the vertices A , B , C , D of the parallelogram. Where, a = i ^ + j ^ + k ^ , b = i ^ + 3 j ^ + 5 k ^ , c = 7 i ^ + 9 j ^ + 11 k ^ We know that the diagonals of a parallelogram bisect each other. Since the mid-point of the two position vectors x and y is given by x + y 2 . So, 1 2 a + c , 1 2 b + d ⇒ d = a + c - b = i ^ + j ^ + k ^ + 7 i ^ + 9 j ^ + 11 k ^ - i ^ - 3 j ^ - 5 k ^ = 7 i ^ + 7 j ^ + 7 k ^ = 7 i ^ + j ^ + k ^ Hence, the position vector of the fourth vertex is 7 i ^ + j ^ + k ^ .

A point P with position vector - 14 i ^ + 39 j ^ + 2 k ^ 5 divides the line joining A ( - 1 , 6 , 5 ) and B in the ratio 3 : 2 then find the point B .

Let A , B and P have position vectors a , b and p respectively. Then a = - i ^ + 6 j ^ + 5 k ^ , p = - 14 i ^ + 39 j ^ + 2 k ^ 5 Now, P divides A B internally in the ratio 3 : 2 . ∴ p = 3 b + 2 a 5 ⇒ 5 p = 3 b + 2 a ⇒ 3 b = 5 p - 2 a = 5 - 14 i ^ + 39 j ^ + 2 k ^ 5 - 2 - i ^ + 6 j ^ + 5 k ^ = - 14 i ^ + 39 j ^ + 2 k ^ + 2 i ^ - 12 j ^ - 10 k ^ = - 12 i ^ + 27 j ^ + 18 k ^ ∴ b = - 12 i ^ + 27 j ^ + 18 k ^ 3 = - 4 i ^ + 9 j ^ + 6 k ^ ∴ Coordinates of B are ( − 4 , 9 , 6 ) .

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 .