What is the angle between the lines (x-2)/(1)=(y+1)/(-2)=(z+2)/(1) and (x-1)/(1)=(2 y+3)/(3)=(z+5)/(2) ?

Correct answer: A. (π)/(2) (x-2)/(1)=(y+1)/(-2)=(z+2)/(1) Direction ratios are proportional to 1,-2,1(x-1)/(1)=(y+(3)/(2))/((3)/(2))=(z+5)/(2) θ= a_1 a_2+b_1 b_2+c_1 c_2 a_1^2+b_1^2+c_1^2 a_2^2+b_2^2+c_2^2 = (1)(1)+(-2)((3)/(2))+(1)(2) 1^2+(-2)^2-1^2 1^2+((3)/(2))^2+(2)^2 θ=(π)/(2) .

NDA 2012

Exam: NDA
Year: 2012
Subject: Mathematics
Chapter: Three Dimensional Geometry

NDA 2012 · Mathematics · Three Dimensional Geometry

What is the angle between the lines (x-2)/(1)=(y+1)/(-2)=(z+2)/(1)...

What is the angle between the lines

\frac{x-2}{1}=\frac{y+1}{-2}=\frac{z+2}{1}

and

\frac{x-1}{1}=\frac{2 y+3}{3}=\frac{z+5}{2}

Options

A $\frac{\pi}{2}$
B $\frac{\pi}{3}$
C $\frac{\pi}{6}$
DNone of the above

Correct answer

A. $\frac{\pi}{2}$

Step-by-step solution

\frac{x-2}{1}=\frac{y+1}{-2}=\frac{z+2}{1}

Direction ratios are proportional to

1,-2,1\frac{x-1}{1}=\frac{y+\frac{3}{2}}{\frac{3}{2}}=\frac{z+5}{2}\cos \theta=\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}=\frac{(1)(1)+(-2)\left(\frac{3}{2}\right)+(1)(2)}{\sqrt{1^2+(-2)^2-1^2} \sqrt{1^2+\left(\frac{3}{2}\right)^2+(2)^2}}\Rightarrow \quad \theta=\frac{\pi}{2}

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