Question

Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines:

Answer 1

The line is passing through the point ( 1,2,−4 ).

The given equations of the lines are,

x−8 3 = y+19 −16 = z−10 7 (1)

x−15 3 = y−29 8 = z−5 −5 (2)

Let b → be the vector. The equation of line which is parallel to the vector is given by,

b → = b → 1 i ^ + b → 2 j ^ + b → 3 k ^

The position of the vector of point ( 1,2,−4 ) is given by,

a → = i ^ +2 j ^ −4 k ^

The formula for the equation of line passing through the point ( x 1 , y 1 , z 1 ) and parallel to vector a 1 i ^ + b 1 j ^ + c 1 k ^ is given by,

a → +λ b →

Substitute the values in the above equation,

r → ⋅( i ^ +2 j ^ −4 k ^ )+λ( b → 1 i ^ + b → 2 j ^ + b → 3 k ^ )(3)

According to the given condition, line (1) and line (3) are perpendicular to each other,

3 b 1 −16 b 2 +7 b 3 =0(4)

Similarly, line (2) and line (3) are perpendicular to each other,

3 b 1 +8 b 2 −5 b 3 =0(5)

From the equation (4) and equation (5),

b 1 ( −16 )( −5 )−( 8×7 ) = b 2 ( 7×3 )−( 3×−5 ) = b 3 ( 8×3 )−( 3×−16 ) b 1 24 = b 2 36 = b 3 72 b 1 2 = b 2 3 = b 3 6

The direction of b → are 2, 3 and 6

b → =2 i ^ +3 j ^ +6 k ^

Substitute the values in equation (3).

r → ⋅( i ^ +2 j ^ −4 k ^ )+λ( 2 i ^ +3 j ^ +6 k ^ )

Thus, the equation of line passing through the point ( 1,2,−4 ) is r → ⋅( i ^ +2 j ^ −4 k ^ )+λ( 2 i ^ +3 j ^ +6 k ^ ).