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 equations of the line in vector and cartesian form.

We know that the vector equation of a line passing through a point with position vector a → and parallel to the vector b → is r → = a → + λ b → Here, a → = 2 i ^ - 3 j ^ + 4 k ^ b → = 3 i ^ + 4 j ^ - 5 k ^ So, the vector equation of the required line is r → = ( 2 i ^ - 3 j ^ + 4 k ^ ) + λ ( 3 i ^ + 4 j ^ - 5 k ^ ) Here, λ is a parameter We know that equation of line in Cartesian form passing through point x 1 , y 1 , z 1 and is in the direction of vector ( a i ^ + b j ^ + c k ^ ) is x - x 1 a = y - y 1 b = z - z 1 c . Hence, the Cartesian form is 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 equations of the line in vector and cartesian form.

We know that the vector equation of a line passing through a point with position vector a → and parallel to the vector b → is r → = a → + λ b → Here, a → = 2 i ^ - 3 j ^ + 4 k ^ b → = 3 i ^ + 4 j ^ - 5 k ^ So, the vector equation of the required line is r → = ( 2 i ^ - 3 j ^ + 4 k ^ ) + λ ( 3 i ^ + 4 j ^ - 5 k ^ ) Here, λ is a parameter We know that equation of line in Cartesian form passing through point x 1 , y 1 , z 1 and is in the direction of vector ( a i ^ + b j ^ + c k ^ ) is x - x 1 a = y - y 1 b = z - z 1 c . Hence, the Cartesian form is x - 2 3 = y + 3 4 = z - 4 - 5

A B C D is a parallelogram. The position vectors of the points A , B and C are respectively, 4 i ^ + 5 j ^ - 10 k ^ , 2 i ^ - 3 j ^ + 4 k ^ and - i ^ + 2 j ^ + k ^ . Find the vector equation of the line B D . Also, reduce it to Cartesian form.

We know that the position vector of the mid-point of a → and b → is a → + b → 2 Let the position vector of point D be x i ^ + y j ^ + z k ^ Position vector of mid-point of A and C = Position vector of mid-point of B and D ∴ ( 4 i ^ + 5 j ^ - 10 k ^ ) + ( - i ^ + 2 j ^ + k ^ ) 2 = ( 2 i ^ - 3 j ^ + 4 k ^ ) + ( x i ^ + y j ^ + z k ^ ) 2 ⇒ 3 2 i ^ + 7 2 j ^ - 9 2 k ^ = x + 2 2 i ^ + - 3 + y 2 j ^ + 4 + z 2 k ^ Comparing the coefficient of i ^ , j ^ and k ^ , we get x + 2 2 = 3 2 ⇒ x = 1 - 3 + y 2 = 7 2 ⇒ y = 10 4 + z 2 = - 9 2 ⇒ z = - 13 Position vector of point D = i ^ + 10 j ^ - 13 k ^ The vector equation of line B D passing through the points with position vectors a → ( B ) and b → ( D ) is r → = a → + λ ( b → - a → ) Here, a → = 2 i ^ - 3 j ^ + 4 k ^ b → = i ^ + 10 j ^ - 13 k ^ Vector equation of the required line is r → = ( 2 i ^ - 3 j ^ + 4 k ^ ) + λ { ( i ^ + 10 j ^ - 13 k ^ ) - ( 2 i ^ - 3 j ^ + 4 k ^ ) } ⇒ r → = ( 2 i ^ - 3 j ^ + 4 k ^ ) + λ ( - i ^ + 13 j ^ - 17 k ^ ) … ( 1 ) Here , λ is a parameter Reducing it to Cartesian form, we get x i ^ + y j ^ + z k ^ = ( 2 i ^ - 3 j ^ + 4 k ^ ) + λ ( - i ^ + 13 j ^ - 17 k ^ ) [Putting r → = x i ^ + y j ^ + z k ^ ] ⇒ x i ^ + y j ^ + z k ^ = ( 2 - λ ) i ^ + ( - 3 + 13 λ ) j ^ + ( 4 - 17 λ ) k ^ Comparing the coefficients of i ^ , j ^ and k ^ , we get x = 2 - λ , y = - 3 + 13 λ , z = 4 - 17 λ ⇒ x - 2 - 1 = λ , y + 3 13 = λ , z - 4 - 17 = λ Hence, the Cartesian form is x - 2 - 1 = y + 3 13 = z - 4 - 17

Find the vector equation for the line which passes through the point ( 1 , 2 , 3 ) and parallel to the vector i ^ - 2 j ^ + 3 k ^ . Reduce the corresponding equation in cartesian from.

We know that the vector equation of a line passing through a point with position vector a → and parallel to the vector b → is r → = a → + λ b → Here, a → = i ^ + 2 j ^ + 3 k ^ b → = i ^ - 2 j ^ + 3 k ^ Vector equation of the required line is r → = ( i ^ + 2 j ^ + 3 k ^ ) + λ ( i ^ - 2 j ^ + 3 k ^ ) … ( 1 ) Here, λ is a parameter Reducing ( 1 ) to cartesian form, we get x i ^ + y j ^ + z k ^ = ( i ^ + 2 j ^ + 3 k ^ ) + λ ( i ^ - 2 j ^ + 3 k ^ ) [Putting r → = x i ^ + y j ^ + z k ^ ] ⇒ x i ^ + y j ^ + z k ^ = ( 1 + λ ) i ^ + ( 2 - 2 λ ) j ^ + ( 3 + 3 λ ) k ^ Comparing the coefficients of i ^ , j ^ and k ^ , we get x = 1 + λ , y = 2 - 2 λ , z = 3 + 3 λ ⇒ x - 1 = λ , y - 2 - 2 = λ , z - 3 3 = λ Hence, the cartesian form is x - 1 1 = y - 2 - 2 = z - 3 3 .

The cartesian equations of a line are x - 5 3 = y + 4 7 = z - 6 2 . Find a vector equation for the line.

The Cartesian equation of the given line is x - 5 3 = y + 4 7 = z - 6 2 It can be re-written as x - 5 3 = y - ( - 4 ) 7 = z - 6 2 Thus, the given line passes through the point having position vector a → = 5 i ^ - 4 j ^ + 6 k ^ and is parallel to the vector b → = 3 i ^ + 7 j ^ + 2 k ^ We know that the vector equation of a line passing through a point with position vector a → and parallel to the vector b → is r → = a → + λ b → Vector equation of the required line is r → = ( 5 i ^ - 4 j ^ + 6 k ^ ) + λ ( 3 i ^ + 7 j ^ + 2 k ^ ) Here, λ is a parameter

Find the cartesian equation of a line passing through ( 1 , − 1 , 2 ) and parallel to the line whose equations are x - 3 1 = y - 1 2 = z + 1 - 2 . Also, reduce the equation obtained in vector form.

If a line is parallel to x - a l = y - b m = z - c n and passing through the point ( p , q , r ) then equation of the line is given by x - p l = y - q m = z - r n So, the equation of line passing through ( 1 , − 1 , 2 ) and parallel to line x - 3 1 = y - 1 2 = z + 1 - 2 is x - 1 1 = y + 1 2 = z - 2 - 2 We know that the vector equation of a line passing through a point with position vector a → and parallel to the vector m → is r → = a → + λ m → Here, the line is passing through the point ( 1 , − 1 , 2 ) and its direction ratios are proportional to 1 , 2 , - 2 Vector equation of the required line is r → = ( i ^ - j ^ + 2 k ^ ) + λ ( i ^ + 2 j ^ - 2 k ^ )

Find the direction cosines of the line 4 - x 2 = y 6 = 1 - z 3 . Also, reduce it to vector form.

The cartesian equation of the given line is 4 - x 2 = y 6 = 1 - z 3 It can be re-written as x - 4 - 2 = y - 0 6 = z - 1 - 3 This shows that the given line passes through the point ( 4 , 0 , 1 ) and its direction ratios are proportional to - 2 , 6 , - 3 So, its direction cosines are - 2 ( - 2 ) 2 + 6 2 + ( - 3 ) 2 , 6 ( - 2 ) 2 + 6 2 + ( - 3 ) 2 , - 3 ( - 2 ) 2 + 6 2 + ( - 3 ) 2 = - 2 7 , 6 7 , - 3 7 Thus, the given line passes through the point having position vector a → = 4 i ^ + k ^ and is parallel to the vector b → = - 2 i ^ + 6 j ^ - 3 k ^ We know that the vector equation of a line passing through a point with position vector a → and parallel to the vector b → is r → = a → + λ b → Vector equation of the required line is r → = ( 4 i ^ + 0 j ^ + k ^ ) + λ ( - 2 i ^ + 6 j ^ - 3 k ^ ) Here, λ is a parameter