A line passes through the point with position vector 2 i ^ - 3 j ^ + 4 k ^ and is in the direction of 3 i ^ + 4 j ^ - 5 k ^ . Find the equation of the line in vector and Cartesian forms.

We know that if a line passes through the point with position vector a → = x 1 i ^ + y 1 j ^ + z 1 k ^ and parallel vector is b → = A i ^ + B j ^ + C k ^ , then the equation of the line in vector form is r → = a → + λ b → and in Cartesian form is x - x 1 A = y - y 1 B = z - z 1 C . Now we have, a → = 2 i ^ - 3 j ^ + 4 k ^ & b → = 3 i ^ + 4 j ^ - 5 k ^ . So the equation of the line in vector form, r → = 2 i ^ - 3 j ^ + 4 k ^ + λ 3 i ^ + 4 j ^ - 5 k ^ In Cartesian form, x - 2 3 = y + 3 4 = z - 4 - 5

A line passes through the point with position vector 2 i ^ - 3 j ^ + 4 k ^ and is in the direction of 3 i ^ + 4 j ^ - 5 k ^ . Find the equation of the line in vector and Cartesian forms.

We know that if a line passes through the point with position vector a → = x 1 i ^ + y 1 j ^ + z 1 k ^ and parallel vector is b → = A i ^ + B j ^ + C k ^ , then the equation of the line in vector form is r → = a → + λ b → and in Cartesian form is x - x 1 A = y - y 1 B = z - z 1 C . Now we have, a → = 2 i ^ - 3 j ^ + 4 k ^ & b → = 3 i ^ + 4 j ^ - 5 k ^ . So the equation of the line in vector form, r → = 2 i ^ - 3 j ^ + 4 k ^ + λ 3 i ^ + 4 j ^ - 5 k ^ In Cartesian form, x - 2 3 = y + 3 4 = z - 4 - 5

Find the angle between the pair of lines r → = 3 i ^ + 2 j ^ - 4 k ^ + λ i ^ + 2 j ^ + 2 k ^ and r → = 5 i ^ - 2 j ^ + μ 3 i ^ + 2 j ^ + 6 k ^ .

Given pairs of lines are r → = 3 i ^ + 2 j ^ - 4 k ^ + λ i ^ + 2 j ^ + 2 k ^ . . . i r → = 5 i ^ - 2 j ^ + μ 3 i ^ + 2 j ^ + 6 k ^ . . . ii Clearly the two equations i and ( ii ) are straight lines in the direction of i ^ + 2 j ^ + 2 k ^ and 3 i ^ + 2 j ^ + 6 k ^ respectively. Now, i ^ + 2 j ^ + 2 k ^ = 1 2 + 2 2 + 2 2 = 3 3 i ^ + 2 j ^ + 6 k ^ = 3 2 + 2 2 + 6 2 = 7 Let ϕ be the angle between the lines i and ii , we get i ^ + 2 j ^ + 2 k ^ · 3 i ^ + 2 j ^ + 6 k ^ = i ^ + 2 j ^ + 2 k ^ · 3 i ^ + 2 j ^ + 6 k ^ · cos ϕ ⇒ 3 + 4 + 12 = 3 × 7 × cos ϕ ⇒ cos ϕ = 19 21 ⇒ ϕ = cos - 1 19 21 .

Show that the lines x - 1 2 = y - 2 3 = z - 3 4 and x - 4 5 = y - 1 2 = z intersect each other. Also, find the point of their intersection.

Given equation of two lines are x - 1 2 = y - 2 3 = z - 3 4 . . . ii x - 4 5 = y - 1 2 = z . . . i Let x - 1 2 = y - 2 3 = z - 3 4 = λ 1 And x - 4 5 = y - 1 2 = z = λ 2 So a point on the first line is 2 λ 1 + 1 , 3 λ 1 + 2 , 4 λ 1 + 3 And, a point on the second line is 5 λ 2 + 4 , 2 λ 2 + 1 , λ 2 If they intersect they should have a common point. 2 λ 1 + 1 = 5 λ 2 + 4 ⇒ 2 λ 1 - 5 λ 2 = 3 . . . iii 3 λ 1 + 2 = 2 λ 2 + 1 ⇒ 3 λ 1 - 2 λ 2 = - 1 . . . iv Solving above equations we get λ 2 = - 1 ⇒ λ 1 = - 1 Substituting for the z - coordinate, we get 4 λ 1 + 3 = - 1 and λ 2 = - 1 So, the lines intersect and their point of intersection is - 1 , - 1 , - 1 .

Find the magnitude of the shortest distance between the lines x 2 = y - 3 = z 1 and x - 2 3 = y - 1 - 5 = z + 2 2 .

When two lines are in the form of x - x 1 a 1 = y - y 1 b 1 = z - z 1 c 1 and x - x 2 a 2 = y - y 2 b 2 = z - z 2 c 2 Then, the shortest distance between these lines is x 2 - x 1 y 2 - y 1 z 2 - z 1 a 1 b 1 c 1 a 2 b 2 c 2 b 1 c 2 - b 2 c 1 2 + c 1 a 2 - c 2 a 1 2 + a 1 b 2 - a 2 b 1 2 Now, the given lines are x - 0 2 = y - 0 - 3 = z - 0 1 and x - 2 3 = y - 1 - 5 = z + 2 2 Let the shortest distance be ' d ' between these two lines Then, d = 2 - 0 1 - 0 - 2 - 0 2 - 3 1 3 - 5 2 - 6 + 5 2 + 3 - 4 2 + - 10 + 9 2 = - 1 1 + 1 + 1 = 1 3 .

Find the image of the point 1 , 2 , 3 in the line x - 6 3 = y - 7 2 = z - 7 - 2 .

Given Cartesian form of line as: x - 6 3 = y - 7 2 = z - 7 - 2 = k [say] So, any general point on the line is 3 k + 6 , 2 k + 7 , - 2 k + 7 Let the coordinates of the foot of the perpendicular drawn P 1 , 2 , 3 on the given line are Q 3 k + 6 , 2 k + 7 , - 2 k + 7 Let R l , m , n be the image of the point with respect to the line x - 6 3 = y - 7 2 = z - 7 - 2 Then direction ratios of P Q are 3 k + 6 - 1 , 2 k + 7 - 2 , - 2 k + 7 - 3 i.e., 3 k + 5 , 2 k + 5 , - 2 k + 4 Since, P Q is perpendicular to the given line, so, 3 3 k + 5 + 2 2 k + 5 - 2 - 2 k + 4 = 0 ⇒ 9 k + 15 + 4 k + 10 + 4 k - 8 = 0 ⇒ 17 k + 17 = 0 ⇒ k = - 1 So, the coordinates of the foot of the perpendicular are: Q - 3 + 6 , - 2 + 7 , 2 + 7 i.e., Q 3 , 5 , 9 So, clearly Q 3 , 5 , 9 is the mid-point of the points P ( 1 , 2 , 3 ) and R ( l , m , n ) i.e., 3 , 5 , 9 = l + 1 2 , 2 + m 2 , 3 + n 2 ⇒ 3 = l + 1 2 , 5 = 2 + m 2 , 9 = 3 + n 2 ⇒ l = 6 - 1 , m = 10 - 2 , n = 18 - 3 ⇒ l = 5 , m = 8 , n = 15 Hence, the required image of the point is R 5 , 8 , 15 .