Question

The angle between two lines x +1 =y + 3 =z - 4 and $\frac {x-4}{1}$ = $\frac {y+2}{2}$ = $\frac {z+1}{2}$ is

• A. cos-1$(\frac {4}{9})$
• B. cos-1$(\frac {2}{9})$
• C. cos-1$(\frac {1}{9})$
• D. cos-1$(\frac {5}{9})$

Answer 1

Answer: (D)

cos-1$(\frac {5}{9})$

Answer 2

To find the angle between two lines, we can use the direction vectors of the lines.For the first line, x + 1 = y + 3 = z - 4,the vector form is
r1 = (x, y, z) = (-1, -3, 4) + t(1, 1, 1)
The direction vector for this line is (1, 1, 1).
For the second line, $\frac {x-4}{1}$ = $\frac {y+2}{2}$ = $\frac {z+1}{2}$
the vector form is r2 = (x, y, z) = (4, -4, -$\frac {1}{2}$) + s(1, 2, $\frac {1}{2}$)
The direction vector for this line is (1, 2, $\frac {1}{2}$).To find the angle between two vectors, use the dot product formula:
cosθ =$\frac {u · v}{|u| |v|}$
calculate the dot product and magnitudes:
u · v = (1, 1, 1) · (1, 2, $\frac {1}{2}$) = 1 + 2 + $\frac {1}{2}$ = $\frac {9}{2}$
|u| = $\sqrt {1² + 1² + 1²}$ = $\sqrt 3$
|v| = $\sqrt {1² + 1² + (\frac {1}{2})^2}$ = $\sqrt {1 + 4 + \frac {1}{4}}$ = $\sqrt {\frac {17}{4}}$ = $\frac {\sqrt {17}}{2}$
cosθ = $\frac {u · v}{|u| |v|}$ =$\frac {\frac {9}{2}}{\sqrt 3 \times \frac {\sqrt {17}}{2}}$ = $\frac {9}{2}$ x $\frac {2}{\sqrt 3 .\sqrt {17}}$= $\frac {9}{\sqrt 3 .\sqrt {17}}$ = $\frac {9}{\sqrt {51}}$
To determine the angle θ, we need to find the inverse cosine (arccos) of cosθ
θ = arccos$(\frac {9}{\sqrt {51}})$
Therefore, the angle between the two lines is equal to cos-1$(\frac {5}{9})$.