Advertisements
Advertisements
प्रश्न
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
पर्याय
\[\sqrt{30}\]
\[2\sqrt{30}\]
\[5\sqrt{30}\]
\[3\sqrt{30}\]
Advertisements
उत्तर
\[3\sqrt{30}\]
We have ,
\[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} . . . (1) \]
\[\frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4} . . . (2)\]
We know that line (1) passes through the point (3, 8, 3) and has direction ratios proportional to 3,-1 , 1 .
Its vector equation is ,
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} , \text{ where }\overrightarrow{a_1} = 3 \hat{i} + 8 \hat{j} + 3 \hat{k} \text{ and } \overrightarrow{b_1} = 3 \hat{i} - \hat{j} + \hat{k} .\]
Also, line (2) passes through the point ( -3 , -7 , 6 ) and has direction ratios proportional to -3, 2 4 .
Its vector equation is
\[\overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2} , \text{ where } \overrightarrow{a_2} = - 3 \hat{i} - 7 \hat{j} + 6 \hat{k} \text{ and } \overrightarrow{b_2} = - 3 \hat{i} + 2 \hat{j} + 4 \hat{k} .\]
Now,
\[\overrightarrow{a_2} - \overrightarrow{a_1} = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \overrightarrow{b_1} \times \overrightarrow{b_2} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & - 1 & 1 \\ - 3 & 2 & 4\end{vmatrix}\]
\[ = - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \]
\[ \Rightarrow \left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right| = \sqrt{\left( - 6 \right)^2 + \left( - 15 \right)^2 + 3^2}\]
\[ = \sqrt{36 + 225 + 9}\]
\[ = \sqrt{270}\]
\[\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right) = \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right) . \left( - 6 \hat{i} - 15 \hat{j} + 3 \hat{k} \right)\]
\[ = 36 + 225 + 9\]
\[ = 270\]
The shortest distance between the lines
\[\overrightarrow{r} = \overrightarrow{a_1} + \lambda \overrightarrow{b_1} \text{ and } \overrightarrow{r} = \overrightarrow{a_2} + \mu \overrightarrow{b_2}\] is given by
\[d = \left| \frac{\left( \overrightarrow{a_2} - \overrightarrow{a_1} \right) . \left( \overrightarrow{b_1} \times \overrightarrow{b_2} \right)}{\left| \overrightarrow{b_1} \times \overrightarrow{b_2} \right|} \right|\]
\[ = \left| \frac{270}{\sqrt{270}} \right|\]
\[ = \sqrt{270}\]
\[ = 3\sqrt{30}\]
APPEARS IN
संबंधित प्रश्न
If a line drawn from the point A( 1, 2, 1) is perpendicular to the line joining P(1, 4, 6) and Q(5, 4, 4) then find the co-ordinates of the foot of the perpendicular.
The Cartesian equations of line are 3x+1=6y-2=1-z find its equation in vector form.
The Cartestation equation of line is `(x-6)/2=(y+4)/7=(z-5)/3` find its vector equation.
Find the vector and Cartesian equations of the line through the point (1, 2, −4) and perpendicular to the two lines.
`vecr=(8hati-19hatj+10hatk)+lambda(3hati-16hatj+7hatk) " and "vecr=(15hati+29hatj+5hatk)+mu(3hati+8hatj-5hatk)`
A line passes through (2, −1, 3) and is perpendicular to the lines `vecr=(hati+hatj-hatk)+lambda(2hati-2hatj+hatk) and vecr=(2hati-hatj-3hatk)+mu(hati+2hatj+2hatk)` . Obtain its equation in vector and Cartesian from.
The Cartesian equation of a line is `(x-5)/3 = (y+4)/7 = (z-6)/2` Write its vector form.
Show that the lines `(x-5)/7 = (y + 2)/(-5) = z/1` and `x/1 = y/2 = z/3` are perpendicular to each other.
Find the vector equation of the lines which passes through the point with position vector `4hati - hatj +2hatk` and is in the direction of `-2hati + hatj + hatk`
ABCD is a parallelogram. The position vectors of the points A, B 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.
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.
Find the cartesian equation of a line passing through (1, −1, 2) and parallel to the line whose equations are \[\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z + 1}{- 2}\] Also, reduce the equation obtained in vector 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 cartesian and vector equations of a line which passes through the point (1, 2, 3) and is parallel to the line \[\frac{- x - 2}{1} = \frac{y + 3}{7} = \frac{2z - 6}{3} .\]
The cartesian equation of a line are 3x + 1 = 6y − 2 = 1 − z. Find the fixed point through which it passes, its direction ratios and also its vector equation.
Find the angle between the following pair of line:
\[\overrightarrow{r} = \lambda\left( \hat{i} + \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 2 \hat{j} + \mu\left\{ \left( \sqrt{3} - 1 \right) \hat{i} - \left( \sqrt{3} + 1 \right) \hat{j} + 4 \hat{k} \right\}\]
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 value of λ so that the following lines are perpendicular to each other. \[\frac{x - 5}{5\lambda + 2} = \frac{2 - y}{5} = \frac{1 - z}{- 1}, \frac{x}{1} = \frac{2y + 1}{4\lambda} = \frac{1 - z}{- 3}\]
Show that the lines \[\frac{x}{1} = \frac{y - 2}{2} = \frac{z + 3}{3} \text{ and } \frac{x - 2}{2} = \frac{y - 6}{3} = \frac{z - 3}{4}\] intersect and find their point of intersection.
Show that the lines \[\vec{r} = 3 \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( \hat{i} + 2 \hat{j} + 2 \hat{k} \right) \text{ and } \vec{r} = 5 \hat{i} - 2 \hat{j} + \mu\left( 3 \hat{i} + 2 \hat{j} + 6 \hat{k} \right)\] are intersecting. Hence, find their point of intersection.
A (1, 0, 4), B (0, −11, 3), C (2, −3, 1) are three points and D is the foot of perpendicular from A on BC. Find the coordinates of D.
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \lambda - 1 \right) \hat{i} + \left( \lambda + 1 \right) \hat{j} - \left( 1 + \lambda \right) \hat{k} \text{ and } \overrightarrow{r} = \left( 1 - \mu \right) \hat{i} + \left( 2\mu - 1 \right) \hat{j} + \left( \mu + 2 \right) \hat{k} \]
Find the distance between the lines l1 and l2 given by \[\overrightarrow{r} = \hat{i} + 2 \hat{j} - 4 \hat{k} + \lambda\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right) \text{ and } , \overrightarrow{r} = 3 \hat{i} + 3 \hat{j} - 5 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + 6 \hat{k} \right)\]
Cartesian equations of a line AB are \[\frac{2x - 1}{2} = \frac{4 - y}{7} = \frac{z + 1}{2} .\] Write the direction ratios of a line parallel to AB.
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
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 angle between the lines 2x = 3y = −z and 6x = −y = −4z.
The cartesian equations of a line AB are \[\frac{2x - 1}{\sqrt{3}} = \frac{y + 2}{2} = \frac{z - 3}{3} .\] Find the direction cosines of a line parallel to AB.
If the equations of a line AB are
\[\frac{3 - x}{1} = \frac{y + 2}{- 2} = \frac{z - 5}{4},\] write the direction ratios of a line parallel to AB.
The angle between the straight lines \[\frac{x + 1}{2} = \frac{y - 2}{5} = \frac{z + 3}{4} and \frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{- 3}\] is
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
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
Show that the lines \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5} \text { and } \frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.
Choose correct alternatives:
If the equation 4x2 + hxy + y2 = 0 represents two coincident lines, then h = _______
If slopes of lines represented by kx2 - 4xy + y2 = 0 differ by 2, then k = ______
P is a point on the line joining the points A(0, 5, −2) and B(3, −1, 2). If the x-coordinate of P is 6, then its z-coordinate is ______.
