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Please solve RD Sharma class 12 chapter Straight Line in Space exercise 27.3 question 1 maths textbook solution

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Answer: The given line intersect at a point (2, 6, 3).

 

Hint: By putting the value of \gamma.

 

Given:  \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}           …(i)

and \frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}       …(ii)

 

Solution:  \frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}=\gamma  (say)

Then \mathrm{x}=\gamma, y=2 \gamma+2, z=3 \gamma-3

Substituting  x,y,z in eqn (ii)

\frac{\gamma-2}{2}=\frac{2 \gamma+2-6}{3}=\frac{3 \gamma-3-3}{4}

Taking first and second ratios, we get

\begin{aligned} &\frac{\gamma-2}{2}=\frac{2 \gamma-4}{3} \\\\ &\Rightarrow 3 \gamma-6=4 \gamma-8 \\\\ &\Rightarrow \gamma=2 \end{aligned}

Now taking second and third ratios, we get

\begin{aligned} &\frac{2 \gamma-4}{3}=\frac{3 \gamma-6}{4} \\\\ &\Rightarrow 8 \gamma-16=9 \gamma-18 \\\\ &\Rightarrow \gamma=2 \end{aligned}

The line intersects at a point (\gamma, 2 \gamma+2,3 \gamma-3)

taking \gamma=2

Points will be (2, 6, 3).

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