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  • Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 13 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 27 Straight Line in Space Exercise 27.4 Question 13 Maths Textbook Solution.

Answers (1)

Answer – The answer of the given question is 7 units.

(Hint – By using cross product).

Given – Point (2,4,-1) and equation of line \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}

Solution – Let Q be a point through which line passes. Thus for given equation of line co-ordinates of Q is Q(-5,-3,6).

Hence,

Line parallel to \vec{b}=\hat{i}+4\hat{j}-9\hat{k}

Now,

\begin{aligned} &=\overrightarrow{P Q}=(-5 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})-(2 \hat{\imath}+4 \hat{\jmath}-\hat{k}) \\ &=\overrightarrow{P Q}=(-7 \hat{\imath}-7 \hat{\jmath}+7 \hat{k}) \\ \end{aligned}

Now, let’s find cross product of the two vectors.

=\vec{b} \times \overrightarrow{P Q}=\begin{vmatrix} \hat{i} & \hat{\hat{j}} & \hat{k}\\ 1 & 4 &-9 \\ -7 &-7 &7 \end{vmatrix}

                            

                        

\begin{aligned} &=\vec{b} \times \overrightarrow{P Q}=-35 \hat{\imath}+56 \hat{\jmath}+21 \hat{k} \\\\ &=\text { The magnitude of this cross product } \\\\ &=|\vec{b} \times \overrightarrow{P Q}|=\sqrt{1225+3136+441} \\\\ &=|\vec{b} \times \overrightarrow{P Q}|=\sqrt{4802} \end{aligned}

And magnitude of \vec{b}

\begin{aligned} &=|\vec{b}|=\sqrt{1+16+81} \\ &=|\vec{b}|=\sqrt{98} \\ \end{aligned}

Thus, distance of Point from line is

\begin{aligned} &=\mathrm{d}=\frac{|\vec{b} \times \overrightarrow{P Q}|}{|\vec{b}|} \\ &=\mathrm{d}=\frac{\sqrt{4802}}{\sqrt{98}} \\ &=\mathrm{d}=7 \text { units. } \end{aligned}

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