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By-computing-shortest-distance-determine-whether-following-pairs-lines-intersect-or-not-r-i-j-k-3-i-j-r-4-i-k-2-i-3-k

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प्रश्न

By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \overrightarrow{r} = \left( 4 \hat{i} - \hat{k}  \right) + \mu\left( 2 \hat{i}  + 3 \hat{k} \right)\] 

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उत्तर

\[\vec{r} = \left( \hat{i} + \hat{j} - \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and }  \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k}  \right)\] 

Comparing the given equations with the equations 

\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and }  \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] 

we get ,

\[\overrightarrow{a_1} = \hat{i} + \hat{j} - \hat{k} \]

\[ \overrightarrow{a_2} = 4 \hat{i} - \hat{k} \]

\[ \overrightarrow{b_1} = 3 \hat{i} - \hat{j} \]

\[ \overrightarrow{b_2} = 2 \hat{i} + 3 \hat{k} \]

\[ \therefore \overrightarrow{a_2} - \overrightarrow{a_1} = 3 \hat{i} - \hat{j} \]

\[\text{ and }  \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 3 & - 1 & 0 \\ 2 & 0 & 3\end{vmatrix}\]

\[ = - 3 \hat{i}  - 9 \hat{j}  + 2 \hat{k} \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \vec{b_1} \times \overrightarrow{b_2} \right) = \left( 3 \hat{i}  - \hat{j}  \right) . \left( - 3 \hat{i}  - 9 \hat{j}  + 2 \hat{k} \right)\]

\[ = - 9 + 9\]

\[ = 0\]

\[\text{ We observe } \]

\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = 0\]

\[\text{ Thus, the given lines intersect  } .\]

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पाठ 27: Straight Line in Space - Exercise 28.5 [पृष्ठ ३८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 27 Straight Line in Space
Exercise 28.5 | Q 3.2 | पृष्ठ ३८

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

Find the Cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by `(x+3)/3 = (y-4)/5 = (z+8)/6`.


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Find the vector equation of the line passing through the points (−1, 0, 2) and (3, 4, 6).


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Find in vector form as well as in cartesian form, the equation of the line passing through the points A (1, 2, −1) and B (2, 1, 1).


Find the vector equation for the line which passes through the point (1, 2, 3) and parallel to the vector \[\hat{i} - 2 \hat{j} + 3 \hat{k} .\]  Reduce the corresponding equation in cartesian from.


The cartesian equations of a line are x = ay + bz = cy + d. Find its direction ratios and reduce it to vector form. 


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. 


Find the angle between the following pair of line:

\[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{- 3} \text { and } \frac{x + 3}{- 1} = \frac{y - 5}{8} = \frac{z - 1}{4}\]


Find the angle between the following pair of line:

\[\frac{- x + 2}{- 2} = \frac{y - 1}{7} = \frac{z + 3}{- 3} \text{  and  } \frac{x + 2}{- 1} = \frac{2y - 8}{4} = \frac{z - 5}{4}\]


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\[\frac{x + 2}{2} = \frac{2y - 7}{6} = \frac{5 - z}{6}\]  Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.  


Determine whether the following pair of lines intersect or not: 

\[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} - \hat{j} \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]


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By computing the shortest distance determine whether the following pairs of lines intersect or not  : \[\overrightarrow{r} = \left( \hat{i} - \hat{j} \right) + \lambda\left( 2 \hat{i} + \hat{k}  \right) \text{ and }  \overrightarrow{r} = \left( 2 \hat{i} - \hat{j}  \right) + \mu\left( \hat{i} + \hat{j} - \hat{k} \right)\]


Find the shortest distance between the lines \[\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \text{ and }  \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}\]


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Write the angle between the lines \[\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z - 2}{1} \text{ and } \frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{3} .\]


Write the value of λ for which the lines  \[\frac{x - 3}{- 3} = \frac{y + 2}{2\lambda} = \frac{z + 4}{2} \text{ and } \frac{x + 1}{3\lambda} = \frac{y - 2}{1} = \frac{z + 6}{- 5}\]  are perpendicular to each other.


Write the condition for the lines  \[\vec{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and  } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] to be intersecting.


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`(x + 3)/3 = (y - 4)/5 = (z + 8)/6`


Find the vector equation of the lines passing through the point having position vector `(-hati - hatj + 2hatk)` and parallel to the line `vecr = (hati + 2hatj + 3hatk) + λ(3hati + 2hatj + hatk)`.


Find the equations of the diagonals of the parallelogram PQRS whose vertices are P(4, 2, – 6), Q(5, – 3, 1), R(12, 4, 5) and S(11, 9, – 2). Use these equations to find the point of intersection of diagonals.


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