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  • Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 7

Provide solution for RD Sharma maths class 12 chapter Straight Line in Space exercise 27.3 question 7

Answers (1)

Answer: The point of intersection is (-1,-6,-12)

 

Hint: Equate the coefficient of vector equation of line.

 

Given: \vec{r}=3 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}+\lambda(\hat{\imath}+2 \hat{\jmath}+2 \hat{k}) \text { and } \vec{r}=5 \hat{\imath}-2 \hat{\jmath}+\mu(3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k})

 

Solution: let the lines are

\begin{aligned} &M: \vec{r}=\overrightarrow{3 t}+\overrightarrow{2 J}-\overrightarrow{4 k}+\lambda(\vec{\imath}+2 \vec{\jmath}+2 \vec{k}) \ \\\\ &N: \vec{r}=\overrightarrow{5 \imath}-\overrightarrow{2 \jmath}+\mu(\overrightarrow{3 l}+\overrightarrow{2 J}+\overrightarrow{6 k}) \end{aligned}

Co-ordinate of any random point on M  are \mathrm{P}(3+\lambda, 2+2 \lambda,-4+2 \lambda) and on N are \mathrm{Q}(5+3 \mu,-2+2 \mu, 6 \mu)

If the lines Mand N intersect then ,they must have a common point on them i.e. P and Q must coincide for some values of λ and μ.

Now,

\begin{aligned} &3+\lambda=5+3 \mu\quad \text {... (i)}\\\\ &2+2 \lambda=-2+2 \mu \quad \text {... (ii)}\\\\ &-4+2 \lambda=6 \mu\quad \text {... (iii)} \end{aligned}

Solving (i) and (ii) ,we get

\begin{aligned} &\lambda=-4 \\\\ &\mu=-2 \end{aligned}

Substitute the value in equation (iii),

\begin{aligned} &-4+2(-4)=6(-2) \\\\ &-4=-12+8 \\\\ &-4=-4 \end{aligned}

So, the given line intersect each other.

Now, point of intersection isP(-1,-6,-12).

 

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