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.
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
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)`
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 of a line passing through the point with position vector \[\hat{i} - 2 \hat{j} - 3 \hat{k}\] and parallel to the line joining the points with position vectors \[\hat{i} - \hat{j} + 4 \hat{k} \text{ and } 2 \hat{i} + \hat{j} + 2 \hat{k} .\] Also, find the cartesian equivalent of this equation.
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 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 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).
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 from (0, 2, 7) on the line \[\frac{x + 2}{- 1} = \frac{y - 1}{3} = \frac{z - 3}{- 2} .\]
Find the shortest distance between the following pairs of lines whose vector equations are: \[\overrightarrow{r} = \left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \lambda\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) \text{ and } \overrightarrow{r} = \left( 2 \hat{i} + 4 \hat{j} + 5 \hat{k} \right) + \mu\left( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \right)\]
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: \[\frac{x - 1}{2} = \frac{y + 1}{3} = z \text{ and } \frac{x + 1}{5} = \frac{y - 2}{1}; z = 2\]
By computing the shortest distance determine whether the following pairs of lines intersect or not: \[\frac{x - 5}{4} = \frac{y - 7}{- 5} = \frac{z + 3}{- 5} \text{ and } \frac{x - 8}{7} = \frac{y - 7}{1} = \frac{z - 5}{3}\]
Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines
(i) (0, 0, 0) and (1, 0, 2)
Find the shortest distance between the lines \[\overrightarrow{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + \lambda\left( \hat{i} - 3 \hat{j} + 2 \hat{k} \right) \text{ and } \overrightarrow{r} = 4 \hat{i} + 5 \hat{j} + 6 \hat{k} + \mu\left( 2 \hat{i} + 3 \hat{j} + \hat{k} \right)\]
Write the cartesian and vector equations of Z-axis.
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 angle between the lines
The equation of the line passing through the points \[a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \text{ and } b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k} \] is
If the direction ratios of a line are proportional to 1, −3, 2, then its direction cosines are
The shortest distance between the lines \[\frac{x - 3}{3} = \frac{y - 8}{- 1} = \frac{z - 3}{1} \text{ and }, \frac{x + 3}{- 3} = \frac{y + 7}{2} = \frac{z - 6}{4}\]
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.
If y – 2x – k = 0 touches the conic 3x2 – 5y2 = 15, find the value of k.
The separate equations of the lines represented by `3x^2 - 2sqrt(3)xy - 3y^2` = 0 are ______
The equation 4x2 + 4xy + y2 = 0 represents two ______
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`
