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  • Find the vector and cartesian equations of the line which is perpendicular to the lines with equations <img alt="\frac{x + 2}{1} =\frac{ y - 3}{2} = \frac{z + 1}{4}" src="https://learn.careers360.com/latex-image/?%5Cfrac%7Bx%20&plus;%202%

Find the vector and cartesian equations of the line which is perpendicular to the lines with equations

\frac{x + 2}{1} =\frac{ y - 3}{2} = \frac{z + 1}{4} and \frac{x - 1}{2} = \frac{y - 2}{} 3 = \frac{z -3}{} 4

and passes through the point (1, 1, 1). Also find the angle between the given lines.

 

 

 
 
 
 
 

Answers (1)

\\$ The equation of line passing throgh (1,1,1) is $ \\ \frac{x-1}{p}=\frac{y-1}{q}=\frac{z-1}{r} ....(i) \\ $ Given: The line is perpendicular to $ \\\\ \frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4} $ and $ \\ \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \\ \Rightarrow p+2 q+4 r=0 ....(ii) \\ \Rightarrow 2p+3 q+4 r=0 ....(iii)\\ $ Equation (iii) - Equation (ii) $ \\ p+q = 0 \Rightarrow p = -q \\ p= 4r = -q \\ \frac{p}{4}=\frac{q}{-4}=\frac{r}{1} \\ $

\\ $ The equation of line in Cartesian form is: $\frac{x-1}{4}=\frac{y-1}{-4}=\frac{z-1}{1} \\ $ Vector form of line is : $\vec{r}=(\hat{i}+j+k)+\lambda(4 \hat{i}-4 j+k) \\ $ Let $\theta$ be the angle between given two lines $ \\ \cos \theta=\frac{1(2)+2(3)+4(4)}{\sqrt{1+4+16} \sqrt{4+9+16}} \\ =\frac{24}{\sqrt{21} \sqrt{29}} \\ \quad \theta=\cos ^{-1}\left(\frac{24}{\sqrt{21} \sqrt{29}}\right)$

Posted by

Safeer PP

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