Find the value of p such that the lines \(\frac{x-1}{3}=\frac{y+4}{p}=\frac{z-9}{1}\) \(\frac{x+2}{1}=\frac{y-3}{1}=\frac{z-7}{-2}\) are at right angles to each other.

Category: QuestionsFind the value of p such that the lines \(\frac{x-1}{3}=\frac{y+4}{p}=\frac{z-9}{1}\) \(\frac{x+2}{1}=\frac{y-3}{1}=\frac{z-7}{-2}\) are at right angles to each other.
Editor">Editor Staff asked 11 months ago

Find the value of p such that the lines
 
\(\frac{x-1}{3}=\frac{y+4}{p}=\frac{z-9}{1}\)
 
\(\frac{x+2}{1}=\frac{y-3}{1}=\frac{z-7}{-2}\)
 
are at right angles to each other.
 
(a) p=2
 
(b) p=1
 
(c) p=-1
 
(d) p=-2
 
I got this question during an online exam.
 
This intriguing question originated from Three Dimensional Geometry in portion Three Dimensional Geometry of Mathematics – Class 12
NCERT Solutions for Subject Clas 12 Math Select the correct answer from above options 
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1 Answers
Editor">Editor Staff answered 11 months ago

Right answer is (c) p=-1
 
Explanation: The angle between two lines is given by the equation
 
\(cos⁡θ=\left |\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |\)
 
cos⁡90°=\(\left |\frac{3(1)+p(1)+1(-2)}{\sqrt{3^2+p^2+1^2}.\sqrt{1^2+1^2+(-2)^2}}\right |\)
 
0=\(|\frac{p+1}{\sqrt{10+p^2}.√6}|\)
 
0=p+1
 
p=-1