Question

Find the vector equation of the line passing through the point (1, 2, -4)and perpendicular to the two lines: \(\frac {x-8}{3}=\frac {y+19}{-16}=\frac {z-10}{7}\) and \(\frac {x-15}{3}=\frac {y-29}{8} =\frac {z-5}{-5}\)

Answer 1

Let the required line be parallel to the vector \(\vec b\) given by,

\(\vec b=b_1\hat i+b_2\hat j +b_3 \hat k\)

The position vector of the point (1, 2, -4) is \(\vec a= \hat i+2\hat j-4\hat k\)

The equation of the line passing through (1, 2,-4) and parallel to vector \(\vec b\) is
\(\vec r=\vec a+λ\vec b\)

⇒\(\vec r\) = (\(\hat i+2\hat j-4\hat k\)) + λ(\(b_1\hat i+b_2\hat j +b_3 \hat k\))      ...(1)

The equations of the lines are

\(\frac {x-8}{3}=\frac {y+19}{-16}=\frac {z-10}{7} \)       ...(2)

\(\frac {x-15}{3}=\frac {y-29}{8}=\frac {z-5}{-5 }\)          ...(3)

Line (1) and line (2) are perpendicular to each other.

∴3b₁-16b₂+7b₃ = 0         ...(4)

Also, line (1) and line (3) are perpendicular to each other.

∴3b₁+8b₂-5b₃ = 0       ...(5)

From equations (4) and (5), we obtain

\(\frac {b_1}{(-16)(-5)-8×7}=\frac {b_2}{7×3-3(-5)} =\frac {b_3}{3×8-3(-16)}\)

⇒ \(\frac {b_1}{24}=\frac  {b_2}{36}=\frac {b_3}{72}\)

⇒ \(\frac {b_1}{2}=\frac  {b_2}{3}=\frac {b_3}{6}\)

∴Direction ratios of \(\vec b\) are 2, 3 and 6.

∴ \(\vec b=2\hat i+3\hat j +6\hat k\)

Substituting \(\vec b=2\hat i+3\hat j +6\hat k\) in equation(1), we obtain

\(\vec r\) = \((\hat i+2\hat j-4\hat k)\) + λ\((2\hat i+3\hat j +6\hat k)\)

This is the equation of the required line.