Advertisements
Advertisements
प्रश्न
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.
Advertisements
उत्तर
The equation of the line through A(3, 4, 1) and B(5, 1, 6) is
`(x-3)/(5-3)=(y-4)/(1-4)=(z-1)/(6-1)`
or
`(x-3)/2=(y-4)/-3=(z-1)/5 .......................(1)`
Any point on (1) is given by (2k + 3, −3k + 4, 5k + 1).
We know that the coordinates of any point on the XZ plane are (x1, 0, z1).
If (2k + 3, −3k + 4, 5k + 1) lies on the XZ plane, then
If (2k + 3, −3k + 4, 5k + 1) lies on the XZ plane, then
−3k + 4 = 0
⇒k=43
Thus, the coordinates of the point where the line joining A and B crosses the XZ plane are
`(2xx4/3+3, 0, 5xx4/3+1)=(17/3, 0, 23/3)`
The vector equation of the XZ plane is `vecr.hatj=1`
The vector equation of the line AB is vecr=(3hati+4hatj+hatk)+lambda(2hati-3hatj+5hatk)
Let θ be the angle between the XZ plane and the line AB.
`therefore sin theta=(vecb vecn)/(|vecb||vecn|)=((2hati-3hatj+5hatk).hatj)/(sqrt(4+9+25)sqrt(0+1+0))`
`sin theta=(0-3+0)/sqrt(38)`
`=>theta=sin^-1 (-3/sqrt38)=-sin^-1 (3/sqrt38)`
APPEARS IN
संबंधित प्रश्न
Find the separate equations of the lines represented by the equation 3x2 – 10xy – 8y2 = 0.
Let `A(bara)` and `B(barb)` be any two points in the space and `R(barr)` be a point on the line segment AB dividing it internally in the ratio m : n, then prove that `bar r=(mbarb+nbara)/(m+n)`. Hence find the position vector of R which divides the line segment joining the points A(1, –2, 1) and B(1, 4, –2) internally in the ratio 2 : 1.
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 and Cartesian equations of a line passing through (1, 2, –4) and perpendicular to the two lines `(x - 8)/3 = (y + 19)/(-16) = (z - 10)/7` and `(x - 15)/3 = (y - 29)/8 = (z - 5)/(-5)`
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 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 angle between the following pair of line:
\[\overrightarrow{r} = \left( 4 \hat{i} - \hat{j} \right) + \lambda\left( \hat{i} + 2 \hat{j} - 2 \hat{k} \right) \text{ and }\overrightarrow{r} = \hat{i} - \hat{j} + 2 \hat{k} - \mu\left( 2 \hat{i} + 4 \hat{j} - 4 \hat{k} \right)\]
Find the angle between the following pair of line:
\[\frac{x - 2}{3} = \frac{y + 3}{- 2}, z = 5 \text{ and } \frac{x + 1}{1} = \frac{2y - 3}{3} = \frac{z - 5}{2}\]
Find the angle between the pairs of lines with direction ratios proportional to 2, 2, 1 and 4, 1, 8 .
Find the angle between the pairs of lines with direction ratios proportional to 1, 2, −2 and −2, 2, 1 .
Find the angle between two lines, one of which has direction ratios 2, 2, 1 while the other one is obtained by joining the points (3, 1, 4) and (7, 2, 12).
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
Determine whether the following pair of lines intersect or not:
\[\frac{x - 1}{3} = \frac{y - 1}{- 1} = \frac{z + 1}{0} and \frac{x - 4}{2} = \frac{y - 0}{0} = \frac{z + 1}{3}\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( 1 - t \right) \hat{i} + \left( t - 2 \right) \hat{j} + \left( 3 - t \right) \hat{k} \text{ and } \overrightarrow{r} = \left( s + 1 \right) \hat{i} + \left( 2s - 1 \right) \hat{j} - \left( 2s + 1 \right) \hat{k} \]
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 shortest distance between the following pairs of lines whose vector are: \[\overrightarrow{r} = \left( \hat{i} + \hat{j} \right) + \lambda\left( 2 \hat{i} - \hat{j} + \hat{k} \right) \text{ and } , \overrightarrow{r} = 2 \hat{i} + \hat{j} - \hat{k} + \mu\left( 3 \hat{i} - 5 \hat{j} + 2 \hat{k} \right)\]
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 Z-axis.
Write the direction cosines of the line whose cartesian equations are 6x − 2 = 3y + 1 = 2z − 4.
Write the angle between the lines 2x = 3y = −z and 6x = −y = −4z.
If a line makes angle \[\frac{\pi}{3} \text{ and } \frac{\pi}{4}\] with x-axis and y-axis respectively, then the angle made by the line with z-axis is
The lines \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3} \text { and } \frac{x - 1}{- 2} = \frac{y - 2}{- 4} = \frac{z - 3}{- 6}\]
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\].
The equation of a line is 2x -2 = 3y +1 = 6z -2 find the direction ratios and also find the vector equation of the line.
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
Find the value of λ, so that the lines `(1-"x")/(3) = (7"y" -14)/(λ) = (z -3)/(2) and (7 -7"x")/(3λ) = ("y" - 5)/(1) = (6 -z)/(5)` are at right angles. Also, find whether the lines are intersecting or not.
Find the vector equation of a line passing through a point with position vector `2hati - hatj + hatk` and parallel to the line joining the points `-hati + 4hatj + hatk` and `-hati + 2hatj + 2hatk`.
A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.
