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.