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Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.5 question 7 sub question (ii)

Answers (1)

Answer: \mathrm{d}=2 \sqrt{29} \text { units }

Hint: using expression \left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|

\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}+a_{2} b_{1}\right)^{2}}

Given: \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \quad \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}

Solution:

It is known that the shortest distance between the two lines,

\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \quad \text { and } \quad \frac{x-x_{2}}{a_{2}}=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}}

 Comparing the given equation we get,

\begin{array}{lll} x_{1}=-1, & y_{1}=-1, & z_{1}=-1 \\\\ a_{1}=7, & b_{1}=-6, & c_{1}=1 \\\\ x_{2}=3, & y_{2}=5, & z_{2}=7 \\\\ a_{2}=1, & b_{2}=-2, & c_{2}=1 \end{array}

Then

\begin{aligned} &\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|=\left|\begin{array}{ccc} 4 & 6 & 8 \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{array}\right| \\\\ &=4(-6+2)-6(7-1)+8(-14+6)=-116 \end{aligned}

\begin{aligned} &\sqrt{\left.b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{2}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}} \\\\ &=\sqrt{(-6+2)^{2}+(1+7)^{2}+(-14+6)^{2}} \\\\ &=\sqrt{16+36+64} \end{aligned}

\begin{aligned} &=\sqrt{116} \\\\ &=2 \sqrt{29} \end{aligned}

Substituting all the values in equation we get

d=\frac{-116}{2 \sqrt{29}}=\frac{-58}{\sqrt{29}}=\frac{-2 \times 29}{\sqrt{29}}=-2 \sqrt{29}

Since the distance is always non-negative the distance between the given line is

=2 \sqrt{29} \text { units }

 

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