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  • Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 6

Provide solution for RD Sharma maths class 12 chapter 27 Straight Line in Space exercise Multiple choice question 6

Answers (1)

Answer: Correct answer is A.

Hint: Use vector cross product.

Given: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}

Solution: Since, We have\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}

Let point (1,2,3) be P and the point through which the line passes be Q (6,7,7)\Rightarrow              (Given)

Also, the line is parallel to the vector\Rightarrow

\begin{aligned} &\vec{b}=3 \hat{\imath}+2 \hat{j}-2 \hat{k} \Rightarrow(\text { Given }) \\\\ &\overrightarrow{P Q}=5 \hat{\imath}+5 \hat{j}+4 \hat{k} \end{aligned}

Now,

\vec{b} \times \overrightarrow{P Q}=\left|\begin{array}{ccc} \hat{\imath} & \hat{\jmath} & \hat{k} \\ 3 & 2 & -2 \\ 5 & 5 & 4 \end{array}\right|=18 \hat{\imath}-22 \hat{\jmath}+5 \hat{k}

\begin{aligned} &\Rightarrow|\vec{b} \times \overrightarrow{P Q}|=\sqrt{18^{2}+(-22)^{2}+5^{2}} \\\\ &=\sqrt{324+484+25} \\\\ &=\sqrt{833} \end{aligned}

\begin{aligned} &\therefore d=\frac{|\vec{b} \times \overrightarrow{P Q}|}{|\vec{b}|} \\\\ &=\frac{\sqrt{833}}{\sqrt{17}} \\\\ &=\sqrt{49} \\\\ &=7 \end{aligned}

So, the correct option is (a) 7

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