मराठी

The Perpendicular Distance of the Point P (1, 2, 3) from the Line X − 6 3 = Y − 7 2 = Z − 7 − 2 (A) 7 (B) 5 (C) 0 (D) None of These

Advertisements
Advertisements

प्रश्न

The perpendicular distance of the point P (1, 2, 3) from the line \[\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{- 2}\] is 

 

पर्याय

  • 7

  • 5

  • 0

  •  none of these 

MCQ
Advertisements

उत्तर

 7
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). Also, the line is parallel to the vector 

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

Now,

\[\overrightarrow{PQ} = 5 \hat{i} + 5 \hat{j} + 4 \hat{k} \]

\[\therefore \overrightarrow{b} \times \overrightarrow{PQ} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 3 & 2 & - 2 \\ 5 & 5 & 4\end{vmatrix}\]

\[ = 18 \hat{i}  - 22 \hat{j}  + 5 \hat{k}  \]

\[ \Rightarrow \left| \overrightarrow{b} \times \overrightarrow{PQ} \right| = \sqrt{{18}^2 + \left( - 22 \right)^2 + 5^2}\]

\[ = \sqrt{324 + 484 + 25}\]

\[ = \sqrt{833}\]

\[ \therefore d = \frac{\left| \overrightarrow{b} \times \overrightarrow{PQ} \right|}{\left| \overrightarrow{b} \right|}\]

\[ = \frac{\sqrt{833}}{\sqrt{17}}\]

\[ = \sqrt{49}\]

\[ = 7\]

 

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 27: Straight Line in Space - MCQ [पृष्ठ ४३]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 27 Straight Line in Space
MCQ | Q 6 | पृष्ठ ४३

व्हिडिओ ट्यूटोरियलVIEW ALL [4]

संबंधित प्रश्‍न

 

Find the value of p, so that the lines `l_1:(1-x)/3=(7y-14)/p=(z-3)/2 and l_2=(7-7x)/3p=(y-5)/1=(6-z)/5 ` are perpendicular to each other. Also find the equations of a line passing through a point (3, 2, – 4) and parallel to line l1.

 

Find the vector and cartesian equations of the line passing through the point (2, 1, 3) and perpendicular to the lines

`(x-1)/1=(y-2)/2=(z-3)/3 and x/(-3)=y/2=z/5`


Show that the three lines with direction cosines `12/13, (-3)/13, (-4)/13;  4/13, 12/13, 3/13;  3/13, (-4)/13, 12/13 ` are mutually perpendicular.


Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).


ABCD is a parallelogram. The position vectors of the points AB and C are respectively, \[4 \hat{ i} + 5 \hat{j} -10 \hat{k} , 2 \hat{i} - 3 \hat{j} + 4 \hat{k}  \text{ and } - \hat{i} + 2 \hat{j} + \hat{k} .\]  Find the vector equation of the line BD. Also, reduce it to cartesian form.


Show that the points whose position vectors are  \[- 2 \hat{i} + 3 \hat{j} , \hat{i} + 2 \hat{j} + 3 \hat{k}  \text{ and }  7 \text{ i}  - \text{ k} \]  are collinear.


Find the vector equation of the line passing through the point A(1, 2, –1) and parallel to the line 5x – 25 = 14 – 7y = 35z.


Find the angle between the following pair of line: 

\[\overrightarrow{r} = \left( 3 \hat{i} + 2 \hat{j} - 4 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 5 \hat{j} - 2 \hat{k}  \right) + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\]


Find the angle between the following pair of line:

\[\frac{x - 5}{1} = \frac{2y + 6}{- 2} = \frac{z - 3}{1} \text{  and  } \frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 6}{5}\]


Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0) and (1, 2, −1), (2, 1, 1).


If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{2 \lambda} = \frac{z - 3}{2} \text{     and     } \frac{x - 1}{3\lambda} = \frac{y - 1}{1} = \frac{z - 6}{- 5}\]  are perpendicular, find the value of λ.


If the coordinates of the points ABCD be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD


Show that the lines \[\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7} \text{           and                  } \frac{x - 2}{1} = \frac{y - 4}{3} = \frac{z - 6}{5}\]   intersect. Find their point of intersection.


Prove that the line \[\vec{r} = \left( \hat{i }+ \hat{j }- \hat{k} \right) + \lambda\left( 3 \hat{i} - \hat{j} \right) \text{ and } \vec{r} = \left( 4 \hat{i} - \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{k} \right)\] intersect and find their point of intersection.


Determine whether the following pair of lines intersect or not: 

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


Find the foot of the perpendicular drawn from the point  \[\hat{i} + 6 \hat{j} + 3 \hat{k} \]  to the line  \[\overrightarrow{r} = \hat{j} + 2 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) .\]  Also, find the length of the perpendicular


Find the foot of the perpendicular from (1, 2, −3) to the line \[\frac{x + 1}{2} = \frac{y - 3}{- 2} = \frac{z}{- 1} .\]


Find the equation of line passing through the points A (0, 6, −9) and B (−3, −6, 3). If D is the foot of perpendicular drawn from a point C (7, 4, −1) on the line AB, then find the coordinates of the point D and the equation of line CD


Find the shortest distance between the following pairs of lines whose vector are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} + \hat{j} - \hat{k} + \mu\left( 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \right)\]


Find the shortest distance between the following pairs of lines whose cartesian equations are: \[\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} and \frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 5}{5}\] 


Find the shortest distance between the following pairs of lines whose cartesian equations are : \[\frac{x - 1}{- 1} = \frac{y + 2}{1} = \frac{z - 3}{- 2} \text{ and } \frac{x - 1}{1} = \frac{y + 1}{2} = \frac{z + 1}{- 2}\]


By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]


Write the cartesian and vector equations of X-axis.

 

Write the cartesian and vector equations of Y-axis.

 

The equations of a line are given by \[\frac{4 - x}{3} = \frac{y + 3}{3} = \frac{z + 2}{6} .\]  Write the direction cosines of a line parallel to this line.


If a line makes angles α, β and γ with the axes respectively, then cos 2 α + cos 2 β + cos 2 γ =


If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are

 


The shortest distance between the lines  \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\] 

 

 

 

 


If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k. 


The equation 4x2 + 4xy + y2 = 0 represents two ______ 


Find the separate equations of the lines given by x2 + 2xy tan α − y2 = 0 


If slopes of lines represented by kx2 - 4xy + y2 = 0 differ by 2, then k = ______ 


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`


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×