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Show that the Lines X − 5 7 = Y + 2 − 5 = Z 1 a N D X 1 = Y 2 = Z 3 Are Perpendicular to Each Other.

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Question

Show that the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text { and }\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\]  are perpendicular to each other. 

Sum
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Solution

We have 

\[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \text  {and}  \frac{x}{1} = \frac{y}{2} = \frac{z}{3}\]

These equations can be re-written as 

\[\frac{x - 5}{7} = \frac{y - \left( - 2 \right)}{- 5} = \frac{z - 0}{1} . . . (1) \]

\[\frac{x - 0}{1} = \frac{y - 0}{2} = \frac{z - 0}{3} . . . (2)\]

\[\therefore \overrightarrow{m_1} = \text{ Vector parallel to line } (1) = 7 \hat{i} - 5 \hat{j} + \hat{k}  \]

\[ \overrightarrow{m_2} = \text{ Vector parallel to line } (2) = \hat{i} + 2 \hat{j} + 3 \hat{k}  \]

Now,

\[\overrightarrow{m_1} . \overrightarrow{m_2} = \left( 7 \hat{i} - 5 \hat{j} + \hat{k} \right) . \left( \hat{i}+ 2 \hat{j}  + 3 \hat{k} \right)\]

\[ = 7 - 10 + 3\]

\[ = 0\]

Hence, the given two lines are perpendicular to each other.

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Chapter 27: Straight Line in Space - Exercise 28.2 [Page 16]

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R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 27 Straight Line in Space
Exercise 28.2 | Q 5 | Page 16

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