A line l passes through point (-1,3,-2) and is perpendicular

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Question:- A line l passes through point (-1,3,-2) and is perpendicular to both the lines \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3} and \dfrac{x+2}{-3}=\dfrac{y-1}{2}=\dfrac{z+1}{5}. Find the vector equation of the line l. Hence, obtain its distance from origin.

Solution:

Point (-1, 3, -2)

Equation of lines

\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3} – – – (i)

\dfrac{x+2}{-3}=\dfrac{y-1}{2}=\dfrac{z+1}{5}  – – – (ii)

Let dr’s of the line l are (a, b, c)

Equation of line passes through the point (-1, 3, -2) is

\dfrac{x+1}{a}=\dfrac{y-3}{b}=\dfrac{z+2}{c} – – -(iii)

Eq (iii) is perpendicular to (i) and (ii)

Hence,

1a+2b+3c=0  – – (iv)

-3a+2b+5c=0 – – (v)

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Solving (iv) and (v) using cross multiplication method

\Rightarrow \dfrac{a}{10-6}=\dfrac{b}{-9-5}=\dfrac{c}{2-(-6)}

\dfrac{a}{4}=\dfrac{b}{-14}=\dfrac{c}{8}

\Rightarrow \dfrac{a}{2}=\dfrac{b}{-7}=\dfrac{c}{4}

Putting the value in (iii)

line l : \dfrac{x+1}{2}=\dfrac{y-3}{-7}=\dfrac{z+2}{4}

Eq of line in vector form

\vec{r}= (-1\hat{i}+3\hat{j}-2\hat{k})+\lamda(2\hat{i}-7\hat{j}+4\hat{k})

Distance of line l from origin

A line l passes through point (-1,3,-2) and is perpendicular

\dfrac{x+1}{2}=\dfrac{y-3}{-7}=\dfrac{z+2}{4}=k(Let)$ $\Rightarrow \dfrac{x+1}{2}=k,\dfrac{y-3}{-7}=k,\dfrac{z+2}{4}=k

\Rightarrow x =2k-1,y=-7k+3,z=4k-2

Point on the line l is D(2k-1,-7k+3, 4k-2)

Dr’s of line OD (2k-1-0,-7k+3-0, 4k-2-0)

=(2k-1,-7k+3, 4k-2)

Since OD is perpendicular line AB

Then 2(2k-1)-7(-7k+3)+ 4(4k-2)=0

\Rightarrow 4k-2+49k-21+16k-8=0

\Rightarrow 69k-31=0

\Rightarrow k = \frac{31}{69}=\frac{5}{33}

Hence point D[2(\frac{31}{69})-1,-7(\frac{31}{69})+3, 4(\frac{31}{69})-2]

\Rightarrow D[\frac{-7}{69},\frac{-10}{69},\frac{-14}{69}]

Distance from origin is OD

OD = \sqrt{(-\frac{7}{69}-0)^2+(-\frac{10}{69}-0)^2+(-\frac{14}{69}-0)^2}

=\sqrt{\frac{49}{(69)^2}+\frac{100}{(69)^2}+\frac{196}{(69)^2}}

=\sqrt{\frac{49+100+196}{(69)^2}}

=\sqrt{\frac{345}{(69)^2}}

=\sqrt{\frac{5}{69}}

Case study three dimension geometry 1
The equation of motion of a missile are x = 3t, y = -4t, z = t where the time

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