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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 8 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 8  Maths Textbook Solution.

Answers (1)

Answer – The answer of the given question \sqrt{13}Units.

(Hint - \mid AB\mid =\sqrt{x^{2}+y^{2}+z^{2}}.

Given – Point with positive vector  \hat{i}+\left ( 6\hat{j}+3\hat{x} \right )and equation line \hat{r}=\hat{j}+2\hat{x}+\gamma \left ( \hat{i}+2\hat{j}+3\hat{x} \right ).

Solution – Let PQ be the perpendicular drawn from P\left ( \hat{i} +6\hat{j}+3\hat{x} \right )   to given line whose endpoint/foot is Q point.

Q is on line.

=\hat{r}=\hat{\jmath}+2 \hat{x}+\gamma(\hat{\imath}+2 \hat{\jmath}+3 \hat{x})

=\gamma \hat{i}+(2 \gamma+1) \hat{j}+(3 \gamma+2) \hat{x} \; is \; the position \; vector \; of \; Q

Hence,

=\overrightarrow{P Q}= Position vector of Q- Position vector of

=\overrightarrow{P Q}=\gamma \hat{\imath}+(2 \gamma+1) \hat{\jmath}+(3 \gamma+2) \hat{x}-(\hat{\imath}+6 \hat{\jmath}+3 \hat{x})

=\overrightarrow{P Q}=(\gamma-1) \hat{\imath}+(2 \gamma+1) \hat{\jmath}+(3 \gamma+2) \hat{x}-\hat{\imath}-6 \hat{\jmath}-3 \hat{x}

=\overrightarrow{P Q}=(\gamma-1) \hat{\imath}+(2 \gamma-5) \hat{\jmath}+(3 \gamma-1) \hat{x}

:: PQ is perpendicular on line

\hat{i}=\hat{j}+2\hat{x}+\gamma \left ( \hat{i}+2\hat{j}+3\hat{x} \right )

:: Their dot product is zero.

Compare given line equation with

\hat{r}=\hat{a}+\gamma \hat{b}

\begin{aligned} &=\overrightarrow{P Q} \cdot \hat{b}=0 \\ \end{aligned}

\begin{aligned} &=\{(\gamma-1) \hat{\imath}+(2 \gamma-5) \hat{\jmath}+(3 \gamma-1) \hat{x}\} \cdot\{(\hat{\imath}+2 \hat{\jmath}+3 \hat{x})\}=0 \\ \end{aligned}

\begin{aligned} &=(\gamma-1)(1)+(2 \gamma-5)(2)+(3 \gamma-1)(3)=0 \\ \end{aligned}

\begin{aligned} &=\gamma-1+4 \gamma-10+9 \gamma-3=0 \\\\ &=14 \gamma-140=0 \\\\ &=\gamma=1 \end{aligned}

Hence, Position vector of Q ,By putting the value of \gamma

                                                                    \begin{aligned} &\qquad \gamma \hat{i}+(2 \gamma+1) \hat{j}+(3 \gamma+2) \hat{x} \\ \end{aligned}

\begin{aligned} &=\hat{\imath}+(2+1) \hat{\jmath}+(3+2) \hat{x} \\ &=\hat{\imath}+3 \hat{\jmath}+5 \hat{x} ; \text { foot of perpendicular } \\ \end{aligned}

Putting the value of γ in PQ.

\begin{aligned} &=\overrightarrow{P Q}=(\gamma-1) \hat{\imath}+(2 \gamma-5) \hat{\jmath}+(3 \gamma-1) \hat{x} \\ &=\overrightarrow{P Q}=(1-1) \hat{\imath}+(2-5) \hat{\jmath}+(3-1) \hat{x} \\ &=\overrightarrow{P Q}=-3 \hat{\jmath}+2 \hat{x} \\ \end{aligned}

Now,  Magnitude of PQ , we know that

\begin{aligned} &|A B|=\sqrt{x^{2}+y^{2}+z^{2}} \\ \end{aligned}

Hence

\begin{aligned} &=|\mathrm{PQ}|=\sqrt{0^{2}+(-3)^{2}+2^{2}} \\\\ &=|\mathrm{PQ}|=\sqrt{13} \text { units. } \end{aligned}

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