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  • Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (iv) maths

Explain solution RD Sharma class 12 chapter Straight Line in Space exercise 27.5 question 1 sub question (iv) maths

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Answer: d=\frac{8}{\sqrt{29}} \text { units }

Hint: Using formula  =\frac{|(\vec{c}-\vec{a})(\vec{b} \times \vec{d})|}{|(\vec{b} \times \vec{d})|}   and convert the given equation in normal forms.

 

Given: \overrightarrow{\mathrm{\gamma}}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-t) \hat{k} \text { and } \overrightarrow{\mathrm{\gamma}}=(s+) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k}

Solution:

\begin{array}{ll} \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+t(-\hat{\imath}+\hat{\jmath}-2 \hat{k}) & {[\overrightarrow{\mathrm{\gamma}}=\vec{a}+\lambda \vec{b}]} \\\\ \overrightarrow{\mathrm{\gamma}}=(\hat{\imath}-\hat{\jmath}-\hat{k})+s(\hat{\imath}+2 \hat{\jmath}-2 \hat{k}) & {[\overrightarrow{\mathrm{\gamma}}=\vec{c}+\lambda \vec{d}]} \end{array}

\vec{b} \times \vec{d}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{array}\right|

          =\hat{\imath}[1(-2)-(-2)(2)]-\hat{\jmath}[(-1)(-2)-(1)(-2)]+\hat{k}[(-1)(2)-(1)(1)]

\begin{aligned} &\vec{b} \times \vec{d}=2 \hat{\imath}-4 \hat{\jmath}-3 \hat{k} \\\\ &|\vec{b} \times \vec{d}|=\sqrt{(2)^{2}+(-4)^{2}+(-3)^{2}} \Rightarrow \sqrt{9+16+4} \Rightarrow \sqrt{29} \end{aligned}

\begin{aligned} (\vec{c}-\vec{a}) &=(\hat{\jmath}-4 \hat{k}) \\\\ d &=\frac{|(\hat{\jmath}-4 \hat{k})(2 \hat{\imath}-4 \hat{\jmath}-3 \hat{k})|}{|\sqrt{29}|} \Rightarrow \frac{|0-4+12|}{|\sqrt{29}|} \Rightarrow \frac{8}{\sqrt{29}} \end{aligned}

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