# 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